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Math 017 Materials With Exercises

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<strong>Math</strong> <strong>017</strong><strong>Materials</strong><strong>With</strong><strong>Exercises</strong>


July 2012


TABLE OF CONTENTSLesson 1Variables and algebraic expressions; Evaluation of algebraic expressions……………………….……..3Lesson 2Algebraic expressions and their evaluations; Order of operations…………………......………….…...15Lesson 3Equivalent algebraic expressions……………………………………………………………………....24Lesson 4Operations on power expressions with non-negative integer exponents………………………..……...38Lesson 5Multiplication of algebraic expressions; The Distributive Law; Factorization of a commonfactor; Factorization of 1; Simplification of algebraic fractions with the use of factorization………50Lesson 6Addition and subtraction of algebraic expressions…………………………………………………..…61Lesson 7Evaluation of more complicated algebraic expressions; Substitution of not only numbers but alsoalgebraic expressions……………………………………………………………………………..…….70Lesson 8Generalities on equations; Solving linear equations in one unknown……………………………….…77Lesson 9Solving linear equations involving fractions; Literal expressions: solving equations for a givenvariable………………………………………………………………………………………………....88Lesson 10Linear Inequalities; Graphing sets of the typex a , x a on a number line…………………….….95Lesson 11Recognizing and matching patterns; Writing expressions in a prescribed way; Definition ofa linear equation; Factorization of the difference of squares……….……………...………………....105Appendix ASolutions to all exercises……………………………………………………………..……………….116Appendix BSample tests……………………….……………….……………………………………………….....133Appendix CSample tests solutions……………….………………...……………………………………………....1411


Help available: You can find help in the Learning Lab for math in room B2-36 weekdays and inroom B1-28 Monday- Thursday evenings and Saturdays on Main Campus. During fall and springsemesters, free, peer tutoring is available beginning with the second week of classes for all currentCCP students and free, weekly workshops begin in the third week of the semester. The peer tutors areexperienced CCP students who have taken many of the courses in which they tutor. <strong>Math</strong> specialistsalso tutor as well as lead workshops. Check at Regional Centers for days and times of services. Also,during summer sessions, offerings may vary.2July 2012


Lesson 1__________________________________________________________________________________Topics: Variables and algebraic expressions; Evaluation of algebraic expressions.__________________________________________________________________________________Variables and algebraic expressions as symbolic representations of numbersSuppose that you thought of a number but you did not tell me what it was. I can think about yournumber as a number x . The symbol x is an example of a variable.VariableA variable is a symbol that represents an unknown number.The choice of the name of a variable is arbitrary. One can as well call it n , m or . We treatvariables as if they were numbers. We can, for example, add numbers to variables: m 3, orsubtract other variables from them:m . We can multiply them: 4 m ; divide: ; raise to anymgiven power:2m ; and then, if we wish, add all of the expressions together:resulting expressions are called algebraic expressions. 24 m m . ThemAlgebraic ExpressionAn algebraic expression is a number, variable or combination ofthe two connected by some mathematical operations likeaddition, subtraction, multiplication, division, or exponentiation.Notice that numbers and variables are also examples of algebraic expressions. We can refer to 3,x , ory as algebraic expressions. Just like 4 5,2 5,or 3 2 1 are numbers (written in a „complicated‟ way,but numbers), algebraic expressions 4 m,x y or a n b are symbolic representations of numbers.Both variables and algebraic expressions can be thought of as unknown numbers.Correct language and conventions used when forming algebraic expressionsAlgebraic expressions are read using the same terminology as in arithmetic. For example, A 5 can be2read as “A plus 5” or “the sum of A and 5”; y can be read as “ y raised to the second power” or “ ysquared”; x is read “minus x ” or “the opposite of x ”. The following convention is commonlyadopted to indicate multiplication of a number and a variable, or multiplication of variables.To denote the operation of multiplication, the sign of multiplication between a number and avariable or between two variables or expressions does not have to be explicitly displayed, so forexample,2A means 2 times Axy means x times y,y(a+b) means y times the quantity a+b3


According to the above convention the following is true.x = 1∙ x = 1xAlthough x 1x, it is customary to write x instead of 1x (just like any time we want to write 4, wejust write 4 not 1 4 ). The following is also true.– x = – 1∙ x = – 1xAgain, it is customary to write – x instead of –1x.When forming algebraic expressions, we place parentheses according to the notational conventionadopted in arithmetic.• Any time two operation signs are next to each other, parentheses are needed.For example, we write:we do not write:a ( b)a b,a ( b)a ba ( b)a bLikewise, when multiplying a and b , we write a( b). The parentheses are needed even ifthe multiplication sign is not explicitly displayed. Notice that if the parentheses are omitted,the expression changes its meaning from multiplication of a and b to subtraction a b .• Exponents pertain only to „the closest‟ expression. You might, for example, recall that toindicate that the entire fraction is raised to a given power, we use parentheses. We write 2 . 5Similarly, whenever you exponentiate any algebraic expression that is not represented bya single symbol, you must use parentheses:2x ,2y do not require parentheses.xy2 ,2( x) ,Example 1.1 How are the following expressions read?2a) x b) xz c)md) x e) xxf) y2( y x) . On the other handSolution:a) „ x raised to the second power‟ or „ x squared‟b) „ x times z ‟, or „ the product of x and z ‟, or „ x multiplied by z ‟c) „ x raised to the third power‟ or „ x cubed‟d) „ x raised to the m-th power‟3x225 ,4


e) „minus x ‟ or „the opposite of x ‟f) „ x divided by y‟ or „the quotient of x and y‟Example 1.2 In the following expressions parentheses are needed. Explain why they are needed.na) a ( c)b) ( a)Solution:a) Any time two operation signs are next to each other, parentheses are needed.In this case the “ ” sign is followed by “−” sign.b) <strong>With</strong>out the parentheses, only a would be raised to the n th power. <strong>With</strong>parentheses, we raise a to the n-th power. The two statements have differentmeaning, thus parentheses are needed.Algebraic expressions allow us to express mathematical ideas in a general formAlgebraic expressions allow us to write mathematical ideas in symbols, without using specificnumbers. For example, the area of a square is equal to the square of the length of its side. Not everysquare is going to have the same size, so we use a variable to represent the length of a side. If we2denote s to be a side of a square, then the area of the square can be expressed as s s sExample 1.3 Let x and y denote two different numbers. Express the following statements usingalgebraic symbols.a) The sum of x and yb) The difference between x and yc) The product of x and yd) The quotient of x and ySolution:a) x yb) x yc) xyd) x y or equivalentlyxyExample 1.4 Find the algebraic expressions representing the opposite of the following expressions.Do not simplify.a) x b) xSolution:Recall that to find the opposite of a number, we must multiply the number by 1, thusa) The opposite of x is 1 x xb) The opposite of x is 1(x) (x). Please, notice that since the minussign is directly followed by another minus sign, the parentheses are needed.5


Example 1.5 Use the letter x to represent a number and write the following statements as algebraicexpressions.a) Double a numberb) Two thirds of a numberc) A quantity increased by 3Solution:a) 2 x2b) x3c) x 3Example 1.6 Write the following statements as algebraic expressions.a) x subtracted from Ab) x added to Ac) x multiplied by yd) ba raised to the sixteenth powerSolution:a) A x Notice „the reversed order‟ of variables (if you were asked to subtract3 from 6, you would write 6 3, similarly we write A x ).b) x ( A)Notice the use of parentheses: a plus sign, followed by a minussign, requires parentheses.c) ( x)(y)or x( y)Notice the use of parentheses. A multiplication sign(even if not explicitly displayed), followed by a minus sign, requires parentheses.16 a ad) Parentheses are needed to indicate that the entire expression is b braised to the sixteenth power.Evaluation of algebraic expressionsAs mentioned, we can use variables and algebraic expressions to describe certain quantitativerelationships without information about their specific values. For example, suppose that in a certainstore, a cake costs 5 dollars. Let x be a variable that represents the number of cakes we plan to buy inthat store. To calculate how much we will pay for such a purchase, we multiply the price of one cake, 5dollars, by the number of cakes we buy, x . Thus, we can express the cost as 5x 5x. The algebraicexpression 5 x represents the cost of x cakes bought at 5 dollars each. From now on, any time weknow the number of cakes we wish to buy, i.e. we know the value of x , we can find the price byevaluating the expression 5 x .To Evaluate an ExpressionTo evaluate an algebraic expression means to find its value oncewe know the values of its variables. Each variable has to bereplaced by its value and the resulting numerical expression hasto be calculated.6


For example,Evaluate 5 x when x 10 (in the above example it would mean finding the price of 10 cakes at 5dollars each).5x 510 50(The price of 10 cakes is 50 dollars.)Notice, that what we did was to substitute 10 for x in the expression 5 x . We could do that because x isequal to 10. The following fundamental principle underlies the process of evaluation.If two quantities are equal, you can always substitute one for the other.“Equals can be substituted for equals”1Can any algebraic expression be evaluated? Let us consider evaluation of when a 0 . If weareplace a by its value 0, the operation we would have to perform is division by 0, but division by1 xzero is not defined, so cannot be evaluated if a 0 . Expressions like , or x 0 are undefined.a 0Example 1.7 Rewrite each expression replacing variables with their values and then evaluate, ifpossible. If evaluation is not possible, explain why it is not possible.a) x 6 , if x 5b) C , if C 2c) 2 x , if x 3nd) 2 , if n 31e) y x , if x 1and y 3f)1x 2, if x 2Solution:Each variable should be replaced by its assumed value and the obtained numericalsentence has to be evaluated. Please, pay attention to the way parentheses are used.a) x 6 5 6 1b) C ( 2) 2c) 2x 2( 3) 6Remember that 2 x means multiplication of 2 and x .nd) 2 23 81 1 1e) y x ( 1) 1 13 3 31 1 1f) The expression cannot be evaluated, since division byx 2 2 2 0zero is undefined.7


Common mistakes and misconceptionsMistake 1.1When x 1, and you are asked to evaluate x , you must be careful to write x (1). Donot forget to recopy the minus sign before 1. It is incorrect to evaluate x by simply writing x 1.Mistake 1.2One should NOT think that x always represents a negative quantity. It depends on the valueof x . If, for example, x 1, then x ( 1)1. So we see that if x is a negative value, x actually represents a positive quantity.Mistake 1.3mWhen writing a , do NOT place m at the same level as a but slightly higher. Otherwise,becomes am . These two expressions have different meanings.Mistake 1.42The expression x is NOT read as „ x two‟ or „two x ‟. Instead, read it as „ x raised to thesecond power‟ or „ x squared‟.ma<strong>Exercises</strong> with Answers (For answers see Appendix A)<strong>Exercises</strong> 20-36 will help to review basic arithmetic operations using integers, rational numbers(fractions), decimals.Ex.1 Fill in blanks using the words: „variable(s)‟, „algebraic expression(s)‟, and„number(s)‟ asappropriate.2 a bc33x 2 , y , , ( 2a 1)are examples of ___________________ .2 x , y,a,b,c are examples of ________________ but also examples of __________________ .Variables represent ________________.If we know the value of x , we can evaluate 3x 2 , and as a result we get a ______________ .Ex.2How are the following expressions read?23a) a b) a c)e) yf) cd g) b12amd) 2a 2h) x5Ex.3 Rewrite the following expressions, inserting a multiplication sign whenever multiplication isimplied. Whenever there is no operation of multiplication, clearly say so using the phrase “there is nomultiplication performed in this expression”.a) 7 nb) 5kmc) x y8


3xd) x( y)e) 2f) 2x yz w(t)Ex.4 The operation that is indicated in the algebraic expression a b is, of course, addition. Namethe operation that is to be performed in the following algebraic expressions.qa) ab b) c)sd) 3 xe) 3 xf) 3( x)Ex.5 In the following expressions parentheses are needed. Explain why they are needed.a) x ( b) m b) n c) 3b( c)4d) ( a)e) y( x)f) a ( b)85xEx.6Ex.7Ex.8Determine which expression is raised to the n-th power.nna) ( s)b) sc)d)g)nst e)x )n( sth)x n f)ynxy si)n( st)n x y x ( y s)Fill in the blanks.a) It is customary to write ________ instead of 1 x .b) It is customary to write ________ instead of 1 x .c) When one writes ab , it is understood that the operation that is to be performed is___________ .Fill in the blanks with numbers to make the statement true.a) ______ x xb) ______ x xc) ______ x 0nEx.9 Write an algebraic expression representing the opposite number of (do not removeparentheses).a) xb)c)x 3 d)yEx.10 Use the letter y to represent a number and write the following phrases as algebraicexpressions.a) Half of a numberb) Three fourths of a numberc) A quantity increased by 5x 3y 3x y9


d) A number subtracted from ve) A quantity squaredf) Three more than a numberg) A number decreased by xh) The product of x and a numberi) A number doubledEx.11 Write the following phrases as algebraic expressions. Remember to place parentheses whenneeded (place them only when needed). Do not simplify.a) The sum of a and bb) The difference of a and bc) The product of a and bd) The opposite of Cf) The opposite of C ag) The opposite of bh) The product of v , t , and pi) The quotient of c and Bj) x raised to m -th powerxk) raised to m -th poweryEx.12 Give your answer in the form of an algebraic expression.a) Carlos is x years old at this moment. How old will Carlos be in 10 years?b) An item in a store costs x dollars. What is the price of the item, if after a discount, its pricewas reduced to two thirds of the original one?c) You have x dollars to divide equally among 3 kids. How much will each child get?d) You have $100 to divide equally among x kids. How much will each child get?e) Charles bought 2 more lamps for his apartment today. If there are x lamps in Charles‟apartment now, how many lamps were in his apartment before the purchase?f) There are 30 books on each shelf. How many books are on x shelves?g) There are x students in a classroom. How many students are still in the classroom, if 3students leave?h) John is 5 years older than Tom. If John is x years old, how old is Tom?Ex.13 Let d be a variable representing the distance driven by a car, and let t represent the time it tookto drive that distance. Write the following phrase as an algebraic expression: The distance divided bytime.Ex.14 Let m be a variable representing the mass of a given body, and let a represent its acceleration.Use m and a to write the following phrase as an algebraic expression: The product of the mass of abody and its acceleration.Ex.15 Let h be a variable representing the height of a triangle, and b represent the base of the triangle.Use h and b to write the following statement as an algebraic expression: One half of the product of thebase of a triangle and its height.10


Ex.16 Let x 3. Rewrite the expression replacing the variable with its value and evaluate, if possible.If evaluation is not possible, explain why it is not possible.a) x 5b) x 2xc) d) 4 x3e)26x f) xEx.17 If x 0 the expression 1 1cannot be evaluated. Why not? Can x x 5 be evaluated whenx 0? What if x 5? Find another example of an algebraic expression and a value of a variable(s)for which evaluation is not possible.Ex.18 Let x 0. Rewrite the expression replacing the variable with its value and evaluate, ifpossible. If evaluation is not possible, explain why it is not possible.a) 3 xb) x 24xc) d) x 7e)2x 3f) x0Ex.19 Let x 2 . Rewrite the expression replacing the variable with its value and evaluate, if possible.Otherwise, write “undefined”.xa) 33b) xxc) xEx.20 Evaluate A,ifa) A 2b) A 2Ex.21 Substitute x 6 in the following expressions and then evaluate, if possible. Otherwise, write“undefined”.a) x 8b) 10 xc) 4 xd) x 6e) 2 x 6Ex.22 Substitute x 2in the following expressions and then evaluate, if possible. Otherwise, write“undefined”.a) 2 xb) 2 xc) 2 xd) 5 x 4e) 6 x 10 x11


Ex.23 Substitute x 10 in the following expressions and then evaluate, if possible. Otherwise, write“undefined”.a) 3 xb) 5x 200xc)d) x24e) 5 xf) xEx.24 Substitute x 12in the following expressions and then evaluate, if possible. Otherwise, write“undefined”.xa) 1000xb) 6c) 5xd)e) 24 xf)6x 122Ex.25 Substitute x in the following expressions and then evaluate, if possible. Otherwise, write3“undefined”.51a) xb) x 3525c) x d) x712e) 2 xf) x 3x2Ex.26 Substitute“undefined”.3a) x101b) x 73x in the following expressions and then evaluate, if possible. Otherwise, write5121c) 2 x51d) 1 x 4e)1 x 322Ex.27 Substitute x in the following expressions and then evaluate, if possible. Otherwise, write7“undefined”.a) 2 xb) 7x145c) xd) x328


e) x5f) x2Ex.28 Substitute“undefined”.a)3x in the following expressions and then evaluate, if possible. Otherwise, write42x b) x xc)213xe) 3431d) 1 x 8Ex.29 Substitute x 0. 2 in the following expressions and then evaluate, if possible. Otherwise, write“undefined”.a) x 3. 21b) 35 .01 xxc) d) 40x4e) 0 .3xf)Ex.30 Substitute x 0. 6 in the following expressions and then evaluate, if possible. Otherwise,write “undefined”.a) x 4. 5b) 2 . 7 x1.2c)d) 600x x3e) 0 .001xf) xEx 31 Substitute x 1. 5 in the following expressions and then evaluate, if possible. Otherwise, write“undefined”.a) x 0. 08b) 3x 0. 4c) x 0. 15d) 0.2x 30e)xEx.32 If possible, evaluate x y when3 2a) x , y 5 32 9b) x , y 7 14c) x 0.2,y 1.08f) x0 x0.0413


Ex.33 If possible, evaluate x y when3 2a) x , y 5 32 9b) x , y 7 14c) x 0.2,y 1.08Ex.34 If possible, evaluate xy when2 22a) x , y 11 99b) x 4,y 10c) x 0.2,y 0. 01Ex.35 If possible, evaluate yx when2a) x ,11b)y229910y 0.x 4,y c) x 0.2, 01mEx.36 If possible, evaluate ( x) whena) x 10,m 7b) x 2,m 41c) x ,2m 3d) x 0.1,m 5Ex.37 Use the letter x to represent a number and write the following statements as algebraic1expressions. Then evaluate each expression when x .2a) A number doubledb) Three fourths of a numberc) A number raised to the second powerEx.38 Evaluate t , when t 1,t 1Based on your results, which of the following are true?a) t is always positiveb) t is always negativec) t may be positive or negative depending on the value of t14


Lesson 2__________________________________________________________________________________Topics: Algebraic expressions and their evaluations; Order of operations.__________________________________________________________________________________Recall the order of operations.The order of operationsWhen evaluating arithmetic expressions, the order of operations is1) Perform all operations in parentheses first.2) Do any work with exponents.3) Perform all multiplications and divisions, working from left to right.4) Perform all additions and subtractions, working from left to right.If a numerical expression includes a fraction bar, perform all calculations above and belowthe fraction bar before dividing the top by the bottom number.Let x be an unknown number. If we increase the number by 1, the resulting number can berepresented by x 1. Now, suppose that after adding 1, we multiply some other number, called y , bythe result of the addition. We may now write the expression y ( x 1). Let us analyze why we mustplace parentheses around x 1. If an algebraic expression involves more than one mathematicaloperation, then the order of operations is followed. Thus, if we simply write yx 1(without usingparentheses), we would only be multiplying y and x, rather than y times the entire quantity x 1.According to the order of operations, using parentheses in y ( x 1)indicates that we add 1 to x first,and then multiply the result by y. Expressions yx 1and y ( x 1)have entirely different meanings,and only y ( x 1)correctly represents the result of operations performed in this example.The order of operations is used when algebraic expressions are evaluated.2For example, let us evaluate 2 x x when x 3.2 x 2 x replace each x with 3223 3 raise 3 to the second power2 9 3 multiply it by 218 3 subtract 3 from the result15Example 2.1 Rewrite the following expressions and circle the arithmetic operation together with itsoperands that has to be performed first. Write the name of the operation next to yourexpression.4a) 3 5yb) 2xSolution:15


a) We perform multiplication before subtraction3 − 5y multiplicationb) We exponentiate before multiplication2 x 4 exponentiationExample 2.2 List, according to the order of operations, all the operations together with operands thatare to be performed in the following expressions.a) 4 3xb) 4 8a16Solution:a) There are two operations in 4 3x, multiplication and addition. According tothe order of operations, multiplication should be performed first. Thus, the answer is:Multiply 3 by x , then add 4.b) There are two operations in 4 8a, division and multiplication. According tothe order of operations, they should be performed as they appear reading from the left tothe right. Thus, the answer is: Divide 4 by 8 first, and then multiply by a .Example 2.3 Write the algebraic expressions representing the followinga) a b raised to the seventh powerb) a b subtracted from yc) 8 times the quantity a bd) the opposite of a bSolution:Parentheses must be used in each of these examples.7a) ( a b) We learned that the exponent pertains only to the closest77expression, so if we write a b instead of ( a b) , only b would be raised to thepower 7. This rule can be viewed as a consequence of the order of operations.7Exponentiation should be performed before addition, hence in a b , b is raised to theseventh power, and then the result is added to a . To ensure that b is first added to a ,we need to use parentheses, and only then the sum is raised to the seventh power.b) y ( a b)Notice the order of expressions: “subtract from y” indicates thaty is written first and we subtract from it.c) 8(a b)Thanks to parentheses, we add first, and then multiply.d) ( a b)Taking the opposite means multiplication by 1. To ensureaddition first, we need parentheses.Example 2.4 Use the letter x to represent a number and write the following as algebraic expressions.a) Add three to a number, and then divide it by zb) Seven more than one third of a numberc) A quantity decreased by 9, and then multiplied by Ad) A number cubed, and then decreased by y


Solution:Please, notice the use of parentheses in the examples below.x 3a) ( x 3) z orz1b) x 73c) ( x 9)A3d) x yExample 2.5 Determine if, in the following algebraic expressions, parentheses are necessary, i.e. theychange or do not change the order of operations. To this end determine if the firstoperation that should be performed is the same as if the expression were written withoutany parentheses.a) ( a b) cb) ( a b)cSolution:a) In the expression ( a b) c the first operation that should be performed isthe addition of a and b. If we rewrite the same expression but without parentheses, weget a b c and the operation of addition of a and b is the first operation as well (sinceaddition and subtraction are always performed in the order as they appear from the leftto the right). Thus, we conclude, that parentheses do not change the order of operationsin this case.b) In the expression ( a b)c the addition of a and b should be performed first,but if we “drop” the parentheses the resulting expression is a bc and thus we wouldfirst multiply b and c , and then add a . Parentheses are needed since they do change theorder of operations.Example 2.6 Rewrite the expression replacing variables with their values. Then, evaluate, if possible.If evaluation is not possible, explain why it is not possible.x 1a) if x 2x 21b) 3 y x if x 1and y 3mc) ( n 3)if m 2 and n 5Solution:Each variable has to be replaced by its given value and the resulting numericalexpression has to be evaluated. Please, pay attention to the way parentheses are used.x 1 2 1 3a) , but division by zero is not defined. The expressionx 2 2 2 0cannot be evaluated. The operation is “undefined”.1b) 3y x 3 ( 1) 11 232 2c) ( n 3)m ( 5 3) ( 2) 417


j) A number multiplied by the sum of the same number and 5k) The opposite of a number, then raised to one hundred and twenty first powerl) Square a number, and then take the opposite of itEx.4 Let C be a variable representing the temperature in Celsius. Write the following phrase as analgebraic expression: Nine fifths of the Celsius temperature plus 32.Ex.5 Let L be a variable representing the length of a rectangle, and W its width. Use L and W to writethe following phrase as an algebraic expression: The sum of the length of a rectangle and its width,then multiplied by 2.Ex.6 Let m represent mass and c the speed of light. Use m and c to write the following phrase as analgebraic expression: The product of mass and the square of the speed of light.Ex.7 In the following expressions circle the arithmetic operation, together with its operands, that hasto be performed first. Write the name of the operation next to your expression. For example, in 4 3x,multiplication of 3 and x has to be performed first, thus the answer is4 + 3x multiplication55a) a bb) ( a b)8c) x8d) ( x)a be)cf) a bcg) 4 7yh) 3 a bEx.8 There are two operations in the algebraic expression a 3b, addition and multiplication. Inorder to evaluate a 3b, we would have to perform them according to the order of operations. Firstmultiply 3 and b, and then add a . List, according to the order of operations, the operations that are inthe following algebraic expressions.a 3a) 4 x yb)xsc) ( x 3)yd) 2t22e) 3x f) ( 3x)g)4( a c)h)4a cEx.9 Determine if, in the following algebraic expressions, parentheses are necessary, i.e. theychange or do not change the order of operations. To this end, determine if the first operation thatshould be performed is the same as if the expression were written without any parentheses. If theoperation is different, write “parentheses are needed‟, otherwise rewrite the expression without anychanges.a) ( 2 a) xb) ( c 3) ac) ( 3a) xd) a ( c b)( c d)e) x ( 2ab)f)a19


g)i)4( a 2)h)8y (x)4(ab)j) ( a d) cEx.10 Evaluate, if possible.1a) 2x1, if x 21b) 2a1, if a 21c) 2y1, if y 2d) Did you get the same answer for a, b, and c? Can you explain why it is so?4x3 x14a3 a1e) If 2when x , evaluatewhen a . You should be able tox 12a 12arrive at your answer without performing any evaluation.Ex.11 Let x 3. Rewrite the expression replacing the variable with its value and evaluate, if possible.Otherwise, write “undefined”.a) 2x 52b) 4 xxc)x 32d) ( x)2e) x3 xf)4 xxg) xEx.12 Let x 4 . Rewrite the expression replacing the variable with its value and evaluate, if possible.Otherwise, write “undefined”.xa) 2b)x( 2)c) ( x)d)2 x2Ex.13 Let x 1. Rewrite the expression replacing the variable with its value and evaluate, ifpossible. Otherwise, write “undefined”.a) x xb) x xc) ( x)(x)d)e)2x x220


Ex.14 Substitute“undefined”.a) A1c)1A and then evaluate the following expressions, if possible. Otherwise, write22( A)d)1b) AA Ex.15 Let x 0. 3. Rewrite each expression replacing the variable with its value and evaluate, ifpossible. Otherwise, write “undefined”.a) x 2 xxb) 20.10.3 xc)x 0.3d) 1000x 100x10x3 xEx.16 The expression cannot be evaluated for which of the following values of x and y?y 5Explain why.a) x 3,y 5b) x 3,y 5c) x 3,y 5d) x 3,y 5e) x 3,y 0f) x 0,y 5Ex.17 If possible, evaluate when m 2,n 5. Otherwise, write “undefined”. Before evaluating,rewrite the expressions substituting the numerical values of m and n .a) 2m 3nb) 2m( 3n)c) 2(m 3n)d) ( 2m 3)ne) 2(m 3)n1 4Ex.18 If possible, evaluate when m , n . Otherwise, write “undefined”. Before evaluating,8 5rewrite the expressions substituting the numerical values of m and n .a) 8m 10nb) 10 mnc) 2(n m)2d) 8m n1e) n ( m)83f) n m 102A21


1 2Ex.19 If possible, evaluate when A , B . Otherwise, write “undefined”. Before evaluating,3 3rewrite the expressions substituting the numerical values of A and B .4a) 2A4b) B4c) BA Bd)A BA( B)e)A BEx.20 Evaluate the following expressions:n 3,m 2 . If evaluation is not defined, write “undefined”.a3,4n,ab2,2( ab),n a ,anm1if a 1,b ,3Ex.21 Let x 2 , y 0. 1, and z 1. If possible, evaluate the following expressions. Otherwise,write “undefined”.a) x( z y)b) xz yEx.22 Let a 0.1,b 0.2,c 1. If possible (otherwise, write “undefined”), evaluate thefollowing expressions. Before evaluating, rewrite the expressions substituting the numerical values ofvariables.a) a bcb) a cc)b 10aEx.23 Find the value ofa) a 1b) a 12 a2 2a (2 ) ifEx.24 Find the value of 2 A B , ifa) A 1,B 3b) A 2,B 42c) A 0.3,B 0.7d) A , B 185 43 5e) A 1 , B f) A , B 6 510 7Ex.25 Find the value of ( A 3B), ifa) A 1,B 1b) A 2,B 3c) A 0.1,B 0.222


2d) A 2,B 132 1e) A , B 7 65f) A 4,B 90.1xEx.26 Find the value of , ifya) x 2,y 0. 02b) x 200,y 0. 4c) x 0.1,y 0.2Ex.27 Evaluate the following expressions, if m 1,n 2,and p 3. Before evaluating, rewritethe expressions substituting the numerical values of variables.a) m ( n p)b) m n pEx.28 In the following expressions, identify the first operation that should be performed according tothe order of operations and anytime it is a numerical operation, perform it.a) 3 4 xb) 3 4xc) 3 (4) xd) ( 3 4) x32 3e) x 2f)xEx.29 Write the following phrases as algebraic expressions and then evaluate them when x 3.a) 3 multiplied by x , and then squaredb) 4 subtracted from x , and then divided by 0.2c) 9 divided by x , and then cubed23


Lesson 3__________________________________________________________________________________Topics: Equivalent algebraic expressions.__________________________________________________________________________________Definition of equivalent algebraic expressionsSuppose we wish to write as an algebraic expression “a number x doubled”. Should we write x 2 or2 x ? Because of the commutative property of multiplication, both answers are right. Both have thesame meaning, although they appear to be different. We encounter a similar idea in arithmetic. The2 1fractions and are equivalent, which means that they represent the same number although they „do4 2not look the same‟. Similarly, we would say that x 2 and 2 x are equivalent (we often say equal) andwrite x 2 2x .Equivalent AlgebraicExpressionsTwo algebraic expressions are equivalent if, when evaluated, theyhave the same value for all replacements of the variables.Suppose that two algebraic expressions are equivalent, like the two mentioned above, x 2 and 2 x .What it means, according to the definition, is that if we choose any value of x , let‟s say x 1, andevaluate x 2 12 2, and then evaluate 2 x 21 2, the results must be the same. If we changethe value of x , for example to 4, again two results are equal ( x 2 42 8, and 2 x 24 8 ). Nomatter what the value of x , the two results are always going to be equal. Thus, to determine that twoexpressions are equivalent one would have to evaluate them for all possible sets of values of variables.Since we cannot check all, we cannot prove equivalence by performing evaluation (make sure that youunderstand that even if we determine that two expressions assume the same value for many sets ofvalues of variables, we still cannot claim that the two expressions are equivalent). To prove theequivalence of algebraic expressions, some general rules must be employed.Terms and factorsIn arithmetic, we often refer to numbers that are being added as terms, and to numbers multiplied asfactors. For example, 3 and 4 are terms of addition 3 4 7 , while 3 and 5 are called factors of 15,since35 15. We will now generalize the notions of terms and factors.TermsAlgebraic expressions that are added (or subtracted) arecalled terms. Each sign, + or –, is a part of the term that followsthe sign.In other words, the addition and subtraction signs break the expression into smaller parts, called terms,and so, in 3 x 2xy y there are three terms: 3 x , 2 xy , y . Notice that because y is preceded by a5a5aminus sign, the minus sign is a part of the term: y . The expressions a and are terms in a .d dSome expressions have just one term. For example, both 3 xy and x 3 y have only one term.24


FactorsAlgebraic expressions that are multiplied are called factors.The expression 4 mn has factors 1, 4, m, n and any combination of those, like 4 m , 4 n , and of course,22224 mn . In ab , the expressions a and b are called factors but 1, b, b , ab, ab are also factors of thisexpression. During our study we will be talking about explicit factors. Explicit factors of 4 mn are 4,m and n, i.e. the factors that are separated by the multiplication sign (displayed or not displayed) of an22expression. Explicit factors of ab are a and b . The expressions 2x 3 and x 4 are explicit factorsin ( 2x 3)( x 4).Example 3.1 List all terms of the following expressions.a) 3 y zb)x 3x2 4( z 1)2Solution:a) The terms are 3, y, z . Remember that signs are always a part of terms.We list y as a term (y is preceded by the minus sign).xb) The terms are 3 2 2 x x ,4( z 1), . Notice the minus sign in 3x and ,22and that the expressions 4(x 1)should be viewed as one term.Example 3.2 List all explicit factors of the following multiplication.a) 5 abb) 3(a 1)xSolution:a) Since 5 ab 5 a b, the explicit factors are 5 , a, b.b) Since 3 ( a 1)x 3( a 1)b, the factors are; 3,( a 1)and x . Noticethat when listing factors, the expressions in parentheses are treated as one “unsplitable”expression.Algebra is an abstract generalization of arithmetic, where numbers are ‘replaced’ with variables. Thelaws that are true for numbers also hold for algebraic expressions (recall, algebraic expressions aremerely symbolic representations of numbers).We will discuss some of the laws in the context of equivalent expressions.Commutative property of addition: rearranging terms results in equivalent expressionsWe know that 3 5 and 5 3 are both equal to the same number, 8. It is because the result of additiondoes not depend on the order of numbers that are being added. This property is called the commutativeproperty of addition. Remember, subtraction does not have this property: 5 3 35 . But, if we viewsubtraction as the addition of the opposite number, we get 5 3 5 ( 3) 3 5 . <strong>With</strong> the use ofvariables, we can express the above ideas in a general form (without the use of specific numbers). Forany value of x and y,25


Commutative Propertyof Additionx y y xAlso, since x y x ( y) y x , we haveConsequence of CommutativeProperty of Additionx y y xEquivalently, we can say that, changing the order of terms results in an equivalent expression (anexpression that looks different, but „means‟ the same). For example,The terms of a 3 are a and 3. If we reverse the order of terms, we obtain a 3 3a .2222The terms of a b are a and b . Reversing their order gives us a b b a .The terms of c 2dare c and 2d. Reversing their order gives us c 2 d 2d c.Similarly, if an algebraic expression consists of more than two terms, we can rearrange them in anyorder. For instance, the terms of 3 B C are 3, B , and C . Thus, we can rewrite the expression3 B C as B C 3 or 3 B C :3 B C B C 3 3B C (there are more possible rearrangements).Example 3.3 Using the fact that changing the order of terms results in an equivalent expression,rewrite the following expressions in their equivalent form by rearranging the terms. Usethe equal sign to indicate that the resulting expressions are equivalent.a) 7 a bb) ( b c)2 3aSolution:a) The terms are 7 a and b . We get 7a b b 7ab) The terms of ( b c)2 2 3aare ( b c) and 3 a . If we reverse the order,we get2( b c) 3a 3a ( b c)2Example 3.4 Determine which of the following expressions are equal to 2x 5y 4z.a) 2x 4z 5yb) 5y 4z 2xc) 5y 2x 4zSolution:All of them are. The terms of 2x 5y 4zare 2x, 5yand 4z. As long as the signthat is in front of an expression is not altered, we can rearrange terms. The sign thatgoes before 2 x is plus, and in all expressions (a)-(c) 2 x is also preceded by a plus sign.5 y follows the minus sign and the same is true for all expressions (a)-(c). Finally, 4zispreceded by a plus sign in all these expressions.26


Example 3.5 Replace with expressions such that the resulting statements are true. Use parentheseswhen needed.a) 2x y 5z 2xb) xyz z Solution:a) 2x y 5z y 5z 2x; y 5zb) xyz z(x)y ; (x) yCommutative property of multiplication. Rearranging factors results in equivalent expressionsMultiplication, like addition, is commutative. the result of multiplication does not depend on the order,35 53. In general, we haveCommutative Propertyof Multiplicationxy yx orx y y xThis means that, the rearrangement of the order of factors results in an equivalent expression.For example,3 32a a 2 , x y yx or ( a b)x x(a b).Factors can also be rearranged if we have more than two factors. For example,3ab ba 3,cdef dcef efdc fedc(or any other order of factors of c, d, e, and f)Example 3.6 Rewrite the expressiona different order.a 1y x in two equivalent forms by multiplying its factors inbSolution:We can rewrite the above expression in more than two equivalent forms. For example,a 1 a 1a 1 a 1a 1x y , x y , yx , x y , or y x . Any two would be theb b b bbcorrect answer.Applying rules of operations on fractions results in an equivalent expressionThe rules for addition and subtraction of fractions with common denominators areRule for Addition andSubtraction ofFractions with CommonDenominatorsacbca bca b a b , c 0c c c27


2 x 2 xFor example, and thus also2 x 2 x (notice, that we switched the left side with7 7 77 7 7the right one of the equation, so both equations are true, just like if 2 3 5 then 5 2 3 ). It isalso true thata 3 a 3 2 y 2 y and .x x x x 3 x 3 x 3The rules for multiplication and division of fractions:Rule for Multiplicationand Division ofFractionsa xyaxya xy1x y x yxy, y 0Thus, we should recognize that the following are true:x 1x2 2, 3x3 5 5x1 a b, and ( a b)3 31Finally, you might recall that 2would like to see how to do it) that121 2. Similarly, one may show (ask your instructor, if youRule for Negative Signs inFractionsx x x , y 0y y yExample 3.7 Rewrite each of the following expressions as a sum or a difference of two expressions.a b a b a b a bUse or (we assume that c 0 ) . Indicate with thec c c c c cequal sign that the resulting expressions are equivalent.2 sa)52a 2b)5 xSolution:a)b)2 s 2 s5 5 522a 2 a 2 5 x 5 x 5 x28


ax aExample 3.8 Use the fact that x to rewrite the following expressions as a product of ay ynumerical factor and an algebraic expression.2bx ba) b)5 5Solution:2b2 2a) b ; is a numerical factor.5 5 5x b 1b) ( x b)Notice the use of parentheses. The entire expression in the5 51numerator should be multiplied by . Had we omitted the parentheses, only x would51have been multiplied by . 5Example 3.9 Rewrite each of the following expressions in their equivalent form as a single fraction.6xa) ab) 211yc) 1 ( a 2 1)3a 2bd) x 2 x 2a 1 2a 1 b ce) af) 2 4s4sx xSolution:6 6aa) a 11 11x 2xb) 2 y y221 2 1( a 1)a 1c) ( a 1) Write a 2 1instead of 1(a 2 1)!33 3a 2ba 2bd) x 2 x 2 x 222a 1 2 a a a aa ae) a First replace by , and4s4s4s4s4s4s4s21 a2 aby (equals can be substituted by equals). Then subtract, just as you subtract4s4sfractions.a 1b c 2( a 1)b c 2( a 1) ( b c)f) 2 Notice the use ofx x x x xparentheses.29


Example 3.10 Determine which of the following expressions are equal tob 22 b,cc c2 bc1, (2 b),c. b 2cSolution:The only expression that is not equivalent isb 2c(sinceb 2 2 b ). Todemonstrate that, let us set b 0, c 1 and evaluate2 b 2 0 2,c 1b 2 0 2 2. Since 2 2 , we conclude that 2 b is not equivalent to b 2 .c 1ccThe expression 2 bc is equivalent because of the rules for addition of fractions.c1(2 b)is equivalent because of the rules for multiplication of fractions. b 2isccequivalent because we can replace2 b by its equivalent expression b 2 .Performing numerical operations results in equivalent expressionsThere are many operations one can perform on an algebraic expression to obtain an equivalent one.One of them is performing numerical operations according to the order of operations. For example,2 3x 5 x23x 6xExample 3.11 When possible, perform a numerical operation to create an equivalent expression. If nonumerical operation can be performed, clearly indicate so.4xa) 2 x 1b) c) 12 3x8d)m( 2 3)e)g) ( x)h)m2 3f) 2x ( 3)y i) ( 3m)( 2n)6 x3Solution:a) 2 x 1 21x 1x4x xb) Divide the numerator and denominator by 4 to reduce the fraction.8 2c) 12 3xSince multiplication of 3 and x has to be performed before addition,no numerical operation can be performed.m md) ( 23) 6me) 2 3 Since 3 must be first raised to m -th power, no numerical operation canbe performed.30


f) 2x( 3) y 2( 3)xy 6xyg) ( x) ( 1)(1)x xh) 6 x 2xWe can cancel 3 in the denominator with the factor of 3 in36 32.i) ( 3m)( 2n) 3m(2)n 3( 2)mn 6mnHow to show that two expressions are not equivalentTwo algebraic expressions are equivalent, if for all values of variables they assume the same value.Thus, if we can find just one set of values of variables for which the expressions do not assumethe same value, it is enough to conclude that they are not equivalent.2As an illustration, we will demonstrate that x is not equivalent to 2 x . To this end, we must find somevalue of x that when evaluated, the two expressions assume different values. We will use x 5 (thechoice of the value of x is arbitrary). We evaluate both algebraic expressions.x2 5 2 25and 2x 2510.2Since 25 10 , we conclude that 2 x is not equivalent to x .Notice that there are other values of x for which 2 x is not equal to x 2 , but since we only need one2such value, we already proved that x in not equivalent to 2 x .33Example 3.12 Show that the following two expressions ( 2a ) and 2a are not equivalent byevaluating them when a 1 and demonstrating that the values of the two expressionsare not equal.Solution:33(2a) (21)3 2 8 but32a 3 21 21 2 . Since 8 2 ,not equivalent.3( 2a)and32a areCommon mistakes and misconceptionsMistake 3.1x sWhen writing an expression like, for example, , as a single fraction, make sure that all them msymbols are clearly above the fraction bar. You should not writeThe answer should be written as x s, so the minus sign is included in the numerator.m31


<strong>Exercises</strong> with Answers (For answers see Appendix A)Ex. 1 Write a word to complete each sentence.In the expression 4x 2 2 2y, 4x and 2 y are called ________________ .In the expression4x 2 2y,24x and2 y are called ________________ .Ex.2List all terms of the following expressions.a) 3 xb) ab cdxyc) 2y 2 2 x1d) ( 2 b) z2yEx.3 Is 2+8 equal to 8+2? Is x 8 equal to 8 x ? How about2 cd cd 2a andba2 2 b? Why?Ex.4 a) Evaluate m n and n m when m 2 and n 3. Based on this evaluation, can youdetermine if the two expressions are equivalent?b) Is it true that m n n m?Ex.5 For each of the following expressions- List all its terms- Using the fact that changing the order of terms results in an equivalent expression, rewrite thefollowing expressions in their equivalent form by rearranging the terms. Use the equal sign to indicatethat the resulting expressions are equivalent (for example, the expression A 9 should be rewritten asA 9 9 A).a) 2 m zb) x 2c) 3c 2d)e)c( d y2 f ) f)32 y 2x 22 s 2 ( x y)3Ex.6Fill in the blanks to make a true statement.a) x mn 2 mn 2__________b) 3(2a 3b) 4x 4x__________Ex.7 List all terms, and then, by changing the order of these terms, create two new equivalentexpressions for each of the following .232 3xa) x x xb) a 2bc2Ex.8 For each of the following expressions (1)-(5) find an expression equivalent to it amongexpression (A)-(E). Rewrite each matched pair with the equal sign between them to indicate theirequivalence. (1) s t u(A) t u s(2) t s u(B) s t u(3) u s t(C) t s u(4) u t s(D) s u t(5) s t u(E) s u t32


Ex. 9 Rewrite the following expressions placing the multiplication sign „ ' whenever (according tothe convention) it was omitted. Then, identify all explicit factors.a) 2 ab) 3(a b)c) 3x 2d) 4(x y)(b c)yEx.10 Rewrite each of the following expressions in its equivalent form using xy yx . Use the equalsign to indicate that the resulting expression is equivalent to the original one (for example, theexpression 9Ashould be rewritten as 9A A9). Remember about parentheses.a) mn b) 5 7c) cdd) c( a d)Ex.11 a) Rewrite the expression vst in its equivalent form by changing the order of its factors tocreate three new equivalent expressions. Indicate their equivalence by using the equal sign (forexample, one of the answers might be vst tsv ).b) Repeat the above exercise for v ( x y)t .Ex.12 a) Is AB equivalent to BA ? How about 3ABand 3BA, 3ABand BA (3)? Why?b) Is 3ABequivalent to BA 3? Why? Support your answer by evaluating both expressionswhen A 1 and B 2 .Ex.13 Is ab 2 equivalent to 2 ab ? How about ab 2 and 2 ba?Why? How about( mn 4)( a b)and ( b a)(4 nm)? Why?Ex.14 According to the rules for adding and subtracting fractions, we havea bca b a b (assume c 0 ) Rewrite each of the expressions below as a sum or a difference of twoc c cexpressions. Use equal signs to indicate that the resulting expressions are equivalent to the originalones (for example, the expressiona)c)2 57a 2a ba b a bEx.15 Using the fact that andc c cexpressions as a single fraction. Do not simplify.na)4 425m2nc) 4c 2 4c 22 t2 t 2should be rewritten as t ).33 3 3a 6b)32b cd)2ab ca bc2m 7m n b) 4 4d)A B C2x xacbcacbcand(assume c 0 ), rewrite the following33


Ex.16 Write the following expressions as a single fraction using the fact that4 7 2a) 5 5 527mn 3b) 4x4x4xm 3 tc) s 1 s 1s 1adbdcda b cdEx.17 According to the rule for multiplication of fractions the following is true:3x3 x . We can7 73x 3say that the quotient was written as a product of a numerical factor and an algebraic expression77x . Write the following expressions as a product of a numerical factor and an algebraic expression.2x2x 2a)b)3y 2x2c)3xe) 3 2(a 2b)d)ya 2bf)3x axEx.18 Using the rule for multiplication of fractions a , rewrite each of the followingy yx 2xexpressions in their equivalent form as a single fraction (for example, 2 ). Remember to usey yparentheses when needed.m ma) 3 b) 3 nn b b c) a d) a 4 4 t ae) ( s 4)f) 4 nn 1m n x 1g) 3 h) a 2txEx.19 Students were to write an answer to the following problem: Using algebraic symbols, write anopposite number to ts .Student A gave the answer:Student B gave the answer:st st34


Student C gave the answer:Who was right? Why?s tx x xEx.20 You know that . By placing the minus sign differently, write each expressiony y yin two additional equivalent ways. Use parentheses when needed.2a2a ca) b)b 2dEx.21 Write each of the following expressions in their equivalent form as a single fraction. Do notsimplify.2 42 1a) x yb) x y3 33 31 yc) 2x 7 d) 3 1x y3 3t tx y1 ne) 2 3f) 2t tk t k ta b cd 12 a m ng) 3 2h) 3t txy xyEx.22 Fill in the blanks to make a true statement.3xa) x ____________y3x1c) ____________y y3xb) 3__________ya b 1d) _________y yEx.23 Perform all numerical operations that are possible. If none are possible, write “not possible”.a) 4 a 8b) 4a( 8)8ac) 42d) 8 x2e) 8x f) 4 2xg) ( 4 2) xh) ( 2x)m 1 i) ( 5 2)j) 2x2 2 2mk) 0.1x(10y)l) 5 2m) 10 xn) 2(6 5) x5212yo) 4 xp)5 2xz1 13bdq) x r)2 29ac35


s) ( 0.2xy)( 0.3z)t)4x1Ex.24 Is 1 2 3 x equivalent to x ? Is2 4 1 23x equivalent to x ?2 4Ex.25 Is ( x y)(1 a)equivalent to x y( 1a)? Explain your answer.Ex.26 Replace with expressions such that the resulting statement is true. Use parentheses whenneeded.a) a 2 b c c 2b b) x x 4xy 1c) xd) a 2 4 42x y xe) f) z c c 3 3abg) xyz yzh) a 2Ex.27 Determine which of the following expressions are equivalent to mm m1, , m (1), m 1,11mEx.28 Determine which of the following expressions are equivalent to m n :n m , m( n), m (n), nm , n m, ( 1)n mEx.29 Determine which of the following are equivalent to a c b d . a b c d, a b c d,d b c a,a c b dEx.30 Determine which of the following are equivalent to a,b1 a ,b1 b ,a2a ,2ba .ba, b a bEx.31 Determine which of the following expressions are equivalent to5 x , 5 5 5 10, , ,x 1x x , 5x6 66 6x12 65x .6Ex.32 Determine which of the following expressions are equivalent to x 2 .x , 2 x , 2 x ,2 4x xx , ,28 236


Ex.33 Determine which of the following expressions are equivalent to11 a , 8a 3, 3 a 8, 11 a ,3 8a.3 8a2 2Ex.34 Determine which of the following expressions are equivalent tob 3a6,1(3a b),63ab,63ab ,6 6b 3a,63a b6(3ab)m nEx.35 Determine which of the following are equivalent to .3 3n m m n 1 11 n m ,, m n,( m n) ,3 3 3 3 33 3.16Ex.36. The correct answer to a problem isMary whose answer is2t2 v ?2vt2. John‟s answer is1 vt2 . Is John right? How about244Ex.37 Show that ( x) is not equivalent to x by evaluating both expressions when x 1anddemonstrating that the values are not the same.2 22Ex.38 Show that x y is not equivalent to ( x y) by evaluating both expressions when x 1,y 2 and demonstrating that the values are not the same.Ex.39 Show that m n p is not equivalent to m ( n p)by evaluating both expressions whenm 2, n 5, and p 1 and demonstrating that the values are not the same.mmEx.40 Evaluate ( 1) and 1whena) m 1b) m 3c) m 5d) m 7e) Based on the above, can you determine iff) Evaluateequivalent?m( 1) andm 1 when m 2m( 1) andm1are equivalent?. Can you now determine ifm( 1) andm 1 are37


Lesson 4__________________________________________________________________________________Topics: Operations on power expressions with non-negative integer exponents.__________________________________________________________________________________Exponential notationExponential notation is used to write repeated factors in a compact way. Exponents are another way ofwriting multiplication. For example,2 52 2 2 2 2, 3 2 3 3.25times timesWe will extend this idea to algebraic expressions.Exponential ExpressionThe expression of the form a n is called an exponentialexpression. It is defined as followsa0 1,1a a ,2a a a ,…na a a ... a , for n = 1, 2, 3, … .ntimes(a is repeated as a factor n times).a is called the base, n is called an exponent or power.Notice that, according to the definition any expression raised to the zero-th power is equal to one. Also,a variable that appears to have no exponent is raised to the first power.a 1 = an2Recall that a is read as “ a to the n-th power”. In the case of a , often, instead of “ a to the second3power” we read it as “ a squared”; a is often read as “ a cubed”.You should also remember that:The exponent pertains only to “the closest” number or variable. To apply the exponent to theentire expression we must place parentheses around the expression. The exponent is then placedoutside the parentheses.For example,3in 2b , only b is raised to the third power.3in ( 2b ) , 2bis raised to the third power.orininn x , only x is raised to the n-th power (n pertains only to x not to x)n( x) , x is raised to the n-th power38


The above convention can be interpreted as the consequence of the order of operations. For example,2in ( 2x ) parentheses indicate that multiplication should be performed first, and only then the result israised to the second power. This means that the entire 2 x is raised to the second power. <strong>With</strong>outparentheses, we first exponentiate, and then multiply by 2, so only x is squared. Notice also, that22without this convention, we would have no means to distinguish between, let‟s say, 2x and ( 2x ) , or3( a b) and3a b .Example 4.1 Expand, that is write without exponential notation.3a) ( 5A )b) 5ASolution:3a) The exponent pertains to the entire expression 5 A: (5A) 5A5A5Ab) The exponent pertains only to A : 5A3 5AAA .3Example 4.2 Rewrite using exponential notation whenever it is possible.a) 7 mmmb) aaaa aac) b ( c 2d)bb(2d c)bbyd) y yxSolution:a)b7mmm 7m3tomesb) aaaa aa 4times32times a4 ac) Notice that c 2 d 2d c , thus5( c 2d)bb(2d c)bb bbbbb(c 2d)(c 2d) b ( c 2d5times2 2timesyd) Notice that y y is equivalent tox3y yyy y y y x x xyyy , thusx)2Example 4.3 Evaluate a)0( 2x ) b)02x c)4(mn7 ) 0Solution:a) Remember that any expression raised to the zero-th power is equal to 1, andthat because of parentheses the zero power pertains to the entire expression 2 x ,thus (2x)0 1.b) This time only x is raised to the zero-th power, hence 2x 0 21 2.77 0c) The expression m n is raised to the zero-th power, so 4( mn ) 41 439


Numerical coefficientsNumerical CoefficientIn a product of a number, variables or algebraic expressions, thenumerical factor is called a numerical coefficient.For example, in the expression 2 x , the numerical coefficient (often called the coefficient) is equal to 2.m5The coefficient of 5(a b)is 5 and the coefficient of 4a is 4 . If there is no number in frontof a product, it is implied that the coefficient is 1. If there is a negative sign, it is implied that thecoefficient is 1. For instance, the coefficient of x 2 22y is equal to 1 (because x y 1 x y ). Thecoefficient of a is 1 (because a 1 a).Example 4.4 In the following expressions, identify bases, exponents and numerical coefficients.a)273a b)Solution:a) base: a; exponent: 27; numerical coefficient: 3yzk9b) base: kyz ; exponent: 9; numerical coefficient: 1c) base: x ; exponent: 4; numerical coefficient: 1c)4 xLaws of exponentsConsidera3 a23 25( a ) (a ) ( a a a)(a a) a a a a a aWe have 3 a‟s and another 2 a‟s for a total of 3 2 5 a‟s (to multiplyexponents). The idea can be generalized to obtain3a and2a we addedProduct Rule forExponentsaman amnWhen multiplying exponential expressions with like bases, we add the exponents and keep the10 5 10515common base. For example, x x x x .If we haveOut of 5 repeated a‟s in the numerator, we canceled 3 a‟s for a total 5 3 2 a‟s (to dividewe subtracted exponents). In general one can prove that5a by3a40


Quotient Rule forExponentsaamnmn a , a 0When dividing exponential expressions with like bases, subtract the exponents and keep the15x 15510common base. For example, x x5x32 2 2 26If we have a a a a ( a a)(a a)(a a) a a a a a a aWe can think of this as 3 groups of 2 for a total of 2 3 6 a‟s. In general,Power Rule forExponents(m n mna ) aWhen raising an exponential expression to another power, keep the same base, and multiply the10 10550exponents. For example, x5 x x.33 3Now, consider ab This can be extended toabab ab ( a a a)(b bb) a bProduct to Powers Rule forExponents(n nab) abnAn exponent outside the parentheses applies to all parts of a product inside the parentheses andthus to raise a product to a power, one can equivalently raise each factor to that power.5 5 5For example, 2x 2 x .3Finally,3 a a a a a a a a ,3 b b b b b bbbb 0In general,Quotient to Powers Rule forExponents a b nabnn, b 0An exponent outside the parentheses applies to all parts of a quotient inside the parentheses.Thus, to raise a quotient to a power, one can equivalently raise both numerator and denominator55 x xto that power. For example, .5 3 341


For your convenience, we will display all the rules together.Laws of Exponents1. a 0 12. a 1 a5.m n mn( a ) a6.3.aman mn a4.n nn n n a a( ab) a b7. n b aabmnmn a , a 0, b 0Example 4.5 Perform the indicated operations and simplify.58xa) 3x( 2)( 4x)b) c) ( 2x6 ) 326x57 y 3d) y8 2 3 2( a b)e) (2a b)( a b ) f)3y2( a b)Solution:a)b)c)3x( 2)( 4x) 3( 2)(4)xx 24x58x4 524 3 x x26x3 36 3 3 6 3 63( 2x) 2 ( x ) 8x 8x1855 3 538 y 3 y y y yd) 8 y y8 y0 1y0 18888y y y ye)f)2 3 22 3 2 23125 3( 2ab)(ab ) 2( 1)a a bb 2ab 2ab7( a b)2( a b)31 ( a b)27321 ( a b)2Example 4.6 Rewrite in its equivalent form as a single exponential expression without parentheses.Identify the numerical coefficient of the final expression.32a) ( y)b) ( 2y) 3 2ySolution:a)( y33 3 3 3 3 y) ( 1y) ( 1)y 1y ; the coefficient: 13 2 3 3 2b) ( 2y)2y ( 2)y 2y3 2 82yy32 16y5 16y;the coefficient: 16.Example 4.7 Write as an algebraic expression using parentheses where appropriate, then remove theparentheses and simplify.2a) The product of x and y , then raised to the sixth powerb) The quotient of4 b2 3a and42ab , then raised to the second power42


Solution:626 2 6 6 6 266 12 6 12a) b)( x)(y ) ( 1x)( y ) ( 1)x y 1xy x y2 4ab2 ab32 (4a21b32)21 (4ab1)2 (4ab)22 4 a2b22 16ab2(3xExample 4.8 Simplify the expression3 4x x5 ) 21and then evaluate, when x .2Solution:5 2 2 5 2 5210(3x) 3 ( x ) 9x9x 9x3 4347 7x x x x x1To evaluate, we substitute x .23107 9x33 1 1( 1) 19x 9 9 9 9 . 2 2 2 8 89333Example 4.9 Which of the following expressions are equivalent to4 x7 2y ,( 2 y2 7x ) ,4 y2 5 2x ( ) , xy 7 x4 ,4 y2 7x ?2 3 44xy ySolution:2 5 2 2 5224x( y ) 4xy 4xy10, and hence2 54x( y7 2equivalent. 4yx is equivalent because only the order of factorschanged.2 7 2 2 7 2 7( 2x)y 2 x y 4xy ,77 2 74xyx 4xxy 4xy ,2 3 4 2 342 74xy y 4xy 4xy.)2is not equivalent. All others are7y andExample 4.10 Find the numerical value of such that the following statements are true.a) 4 6 3 7 2b) 36 6 62x has beenSolution:6a) We must express 4 as an exponential expression with the base 2 (to match it26 2 6 12to 2 ). Since 4 2 , we get 4 (2 ) 2 . As a result 2 12 2 , hence 12.b) We must express2Notice that 36 6 , and thus3 736 6 as an exponential expression with the base 6.3 7 2 3 7 6 7 1336 6 (6 ) 6 6 6 6 ; 13.819Example 4.11 Evaluate. a)8097 64 4b)114Solution:43


9 818081a) 9 98097 6 76134 4 4 4 13112b) 4 4 16 .11 11 114 4 4Common mistakes and misconceptionsMistake 4.1There is a difference betweensquared,Mistake 4.2In the expression2 x and2( x) . In22 x) ( x)(x) . Just like 3 2 9( xMistake 4.3Although it is true thatWe recognize that2 x only x is squared, in , while ( 3)2 9 .x7 y 3 , since the bases are not the same, DO NOT add exponents.( b2 2 2ab) a ,( 7 x2 2x) 49 , however,( ab2 2 2 b) a .( 7x22 x) 49 .2( x) , x isMistake 4.4Please, remember2 5 2 5 1x x x x x x251 x8 x25(not x7 x ). In other words, if one of thefactors does not have an explicit exponent, it means it is raised to the first power, and thus onehas to be added.<strong>Exercises</strong> with Answers (For answers see Appendix A)mEx.1 In the expression 3 x , 3 is called the____________________ , m is called the____________or ____________ and x is called the______________ .Ex.2Ex.3Fill in the blanks.a) An expression x raised to the________ power is equal to itself.b) An expression x raised to the________ power is equal to 1.ma) In the expression ab the exponent pertains to ________________.nb) In the expression ( ab) the exponent pertains to ________________.c) In the expressiond) In the expressione) In the expressionc )n( de the exponent pertains to ________________.n(a)the exponent pertains to ________________.n a 2x f) In the expression y the exponent pertains to ________________.mthe exponent pertains to ________________.44


Ex.4Ex.5Ex.6Write the following statements as algebraic expressions using parentheses where appropriate.a) The quotient of a and 2, then raised to the fourth powerb) Two thirds of a , then raised to the third powerc) c cubed, and then divided by 7d) The product of a and 5, then raised to the second powere) Raise a to the second power, and then multiply the result by 8.f) The opposite of x , then raised to the fifth power.g) Raise x to the tenth power, and then take the opposite of the result.In each of the following expressions, identify the base, exponent and numerical coefficient.4ma) 3x b) x2x32c)d) ( a bc)3e)g)m x y 73 3x z 4 w f)h)( x y)47( ab)2Write using exponential notation whenever it is possible.a) 6 6666b) zzzzc) 3 a 3a3a ad) xyxyxxxe) a aaaaf) xxy yyx3 3 3 3g) ( a b)(a b)(a b)h) (2t)(2t)(2t)(2t)5kkki) m nn mj) kn z z z( z)(z)(z)k)l)zzzzz z zxm) x x xn) ( 3x)(x 3)(3 x)2 2a a 2a 3 33 o) 2 p) b b b x x x 1 x y x y x y q)r) ( w 2v)(2v w)(2v w) m m m m s) ( m n)(m n)mnt) ( m p n)(p m n)(n m p)Ex.7Write the following expressions without using exponential notation.55a) ( 4)b) 4c)e)g)3( m)d)3( 2a )f)2( a b)h)3 m32a2a b45


Ex.8Simplify by raising to the indicated power.0a) ( 3x )b)0c) 3 xe)g)0abc f)0ab ch)d)03xa( b c)0(abc)a0b0c00Ex. 9 While copying expressions from a blackboard, John kept forgetting to copy parentheses.33a) (x)John copied as xb)c)d)e)f)g)4( x) John copied as7 (x) John copied as3a ( 2b)John copied as( b3a 2 ) John copied as3a 2(b)John copied asa )m( bc John copied as44 x7 x3a 2b3a 2b3a 2bmabc4 2x2xh) John copied as y yDetermine if what was on the board and what John copied has the same meaning, i.e. are theparentheses necessary. Write “same” or “different” for each.3Ex.10 Evaluate a) 2x b)when x 100. Explain why you did not get the same result.( 2x)3Ex.11 Fill in the blanks, each time using one of the following words: add, subtract, multiply, divide.a) To multiply exponential expressions with the same bases one needs to __________ theexponents.b) To _____________ exponential expressions with the same bases one needs to subtract theirexponents.c) To raise an exponential expression to another power one needs to _________ the exponents.Ex.12 Write as a single exponential expression.3 20a) n nb)(s7 ) 28xc) d) b 7 b2x24e) ( 4x)f) 2mm1g) b2b h) xx209a3 4i) j) x x x4a4a3k) l) x4 22ax5546


m)o)6aa11 3 18 a n) 5(t3 ) 4p)0.5s0.1s( 2x127 ) 3Ex.13 Write as an algebraic expression using parentheses where appropriate, then remove theparentheses and simplify.2a) The product of m and 2m53b) The quotient of 3x and 18x5c) The expression 2a raised to the third powerEx.14 Simplify the following expressions first, and then evaluate when a 2 .7a2 2a) b) 2aa6a1 21 8c) (2a)( a )d) a2( a2 ) 3Ex.15 Simplify the following expressions first, and then evaluate when m 1.3ma) m 3 m 5b)2 2m2 14( m )7 50c)d) 2mm310 xEx.16 Simplify the following expression , and then evaluate when8xa) x 7b) x 72c) x 3d) x 0. 07.Ex.17 Rewrite in its equivalent form as a single exponential expression without parentheses. Identifythe numerical coefficient of the final expression.58a) ( B)b) ( B)c)( B5 3 B)d)2 B ( B)Ex.18 Perform the indicated operations and simplify.a) 4 2 2 x ( 2x)b) ( 4x)( 2x)32 5c) (3x)( 2)( x )d) (3x)(4xx)4ae)2( 2a)g)(4x) x353xf)2(3x)( a)h)22a3247


73a 2xxi) 2aj)2( 42a)x(3xk)4x2 ) 32 1m) vv 7 2 4 o) 3 2x xp) 3 5 2al) a4 an) ( 3aa252 ) 23 x 0.4 Ex.19 Write as an algebraic expression using parentheses where appropriate, then remove theparentheses and simplify.2a) The product of 3xand x , then raised to the third power122b) The quotient of 4a and a , then raised to the second power3c) a raised to seventh power, then multiplied by aEx.20 Perform the indicated operations and simplify.25 3 2 x y a) ( x y )b) x c) 3a ( b5 ) 2d) 3(ab2 ) 232 a 2 3e) 24f) x y( x ) y b 72 ab a2 s xg) h) x 2 62ba b( 4s)22( x) 3 yi) yxj)8( a b)k)42( a b)l)324(m n)( m n13 2a b c5 2a c b424)3Ex.21 Write as an algebraic expression using parentheses where appropriate, then remove theparentheses and simplify.7a) xy raised to the fifth power , then divided by y3b)33ab squared, and then multiplied by bEx.22 Simplify the following expressions, and then evaluate when m 2and n 1.m 2 n52 3a)mnb) ( mn )2m(n )34( m n)c)2( m n)d)( 2m n)048


Ex.23 Circle all expressions that are equivalent to4 y2 2 2,525y 2, y y , 52 y2 . aEx.24 Circle all expressions that are equivalent to3 202012 8aa(a ), ,2020333Ex.25 Circle all expressions that are equivalent to4 x2 3y ,( 2 x2 3y ) ,520.2 2 2 y , yy 55a a3(a )312 84 5, ,20204 y3 2x .2 3 2) y ( 2), 4xy)2 x( x( , 4 ( xy)2 xEx.26 Circle all expressions that are equivalent to2 b6 2 3a c ,2 a2 3 2 3a b c ,112(abc ) ,2 c7cb a146 3 2a b .2 3 6,3 32(a c) 2 bEx.27 Replace with a number so the following are equal.a) 81 4 9b) 7 49c) 16 100 2d) 27 4 3e) 1 7 1 f) 4 2 0.255(0.04)Ex.28 Evaluate the following expressions.24815a)247153 6(12 )c)17126445e)1321 103 3b)293100.5 0.5d)7 2(0.5 )4215f)27649


Lesson 5__________________________________________________________________________________Topics: Multiplication of algebraic expressions; The Distributive Law; Factorization of a commonfactor; Factorization of 1; Simplification of algebraic fractions with the use of factorization.__________________________________________________________________________________This section is about the Distributive Law, the law that relates addition to multiplication. In manycases, it allows us to rewrite the product of expressions as a sum or the sum of expression as a product.The Distributive LawIf 5 girls and 7 boys get 2 cookies each, then the girls get 2 5 10cookies, the boys get 27 14cookies, so the total number of cookies they all receive is25 271014 24Alternatively, one can find the number of cookies by adding the number of girls and boys first,5 7 12and then since each of the children gets 2 cookies, together they get2(5 7) 212 24This property is called The Distributive Law. In general,The Distributive Lawa(b c) ab aca(b c) ab acTo remove parentheses from the expression of the formfactor outside parentheses by each term inside parentheses.a(b + c) or a(b − c), multiply theExample 5.1 Use the Distributive Law to remove parentheses in the following expressions.2a) x(2 x )b) ( x 2y)c) a( b c d)Solution:2212a) x( 2 x ) x 2 x x 2x x3 2x xb) ( x 2y) 1( x 2y) 1x 12y x 2yc) The expression inside the parentheses should be treated not as a sum of threeterms but as a sum of two ( b c)and d , that is b c d ( b c) d . We can thenapply the Distributive Law a(b c d) a((b c) d) a(b c) adIf we apply the Distributive Law once again, this time to a( b c), we geta( b c d) a((b c) d) a(b c) ad ab ac ad .Thus a( b c d) ab ac ad .50


The Distributive Law applied to the product of two sumsWe can apply the Distributive Law to remove parentheses in the product of two sums ( x y)(b c).We need to look at this expression as a sum of two terms in parentheses ( b c)multiplied by theexpression ( x y). In other words, ( x y)„plays the role of a ‟ in the Distributive Law. We mayreplace a in the Distributive Law with ( x y).( x y)(b c) ( x y)b ( x y)cIf we now apply The Distributive Law to ( x y)band ( x y)c, we obtain( x y)(b c) ( x y)b ( x y)c xb yb xc ycThus the following is true.The Distributive LawFor The Product of Two Sums( x y)(b c) xb xc yb ycTo multiply two sums, you need to multiply each term of the first sum by each term of the secondone and then add all the products.Please recall, that terms can be rearranged so, equivalently, we can write( x y)(b c) xb yb xc yc .Any other order of terms would be also correct.Example 5.2 Write an equivalent expression without parentheses.a) ( a b)(2a 3)b) 3x 12xc) ( x 3y z)d) ( b c d e)ae) ( a b)(c d)(e f )Solution:a) Use the Distributive Law for product of two sums.2( a b)(2a 3) a(2a) a(3) b(2a) b(3) 2a 3a 2ab 3bb) Apply the Distributive Law. The order of factors can be changed, and thus3x 12x 2x3x1 2x(3x) (2x)1 6x2 2xc) A minus sign in front of parentheses is equivalent to multiplication by 1 ( x 3y z) ( 1)(x 3y z) 1x1(3y)1(z) x 3y zd) Each term inside parentheses must be multiplied by a .( b c d e)a ba ca da eae) Multiply ( a b)(c d), then the result must be multiplied by ( e f ) .( a b)(c d)(e f ) ( ac ad bc bd)(e f ) ace ade bce bde acf adf bcf bdf51


FactorizationThe Distributive Law lets us change an expression from a product to a sum. When we do that „inreverse‟, the process is called factorization.FactorizationChanging the sum of two or more terms to a product iscalled factorization.When factoring algebraic expressions, we merely rewrite them in a different form. The expressionobtained in the process is equivalent to the original one.Factorization of a common factorConsider 3 A Ax . The expression has two terms: 3 A and Ax . Each has A as its factor. A is acommon factor of both terms. We will factor A from 3 A Ax . 3AAx 3A Ax A A(3 x) A A The first step follows from the Distributive Law. To convince yourself, multiply each term inside 3AAx 3AAxparentheses by A : A A A 3A Ax . The second step is obtained by A A A Acanceling A ‟s . 3 A Ax was originally written as a sum of two terms. By factoring A from 3A + Ax,we are now able to express it in a factored form (that is, as a product of two expressions, rather than asum) 3A Ax A(3 x). Notice that, after factorization, you can always check your answer bymultiplying factors.Example 5.3 Factor2x from the expression3xx5 3 2 2x .Solution:3x5 2x3 x2 x2 3x2 x52x2x3xx22 x2(3x3 2x1)Example 5.4 Factor 2afrom the expression2aa2 3 4a 6 .Solution:2a 4a2 6a32 2a4a 2a 2a2a36a2a 2a(1 2a 3a2)52


Example 5.5 Factor22 2 4 3 3xy from the expression 4xy xy 3xy .Solution:4x2y2 xy43 3xy3224xy xy2 xy2xyxy4233xy2xy3 xy2(4x y2 3x2y)Example 5.6 Factor ( a 2b)from the expression 4x(a 2b) ( a 2b).Solution: 4x(a 2b)a 2b4x( a 2b) ( a 2b) ( a 2b) ( a 2b)(4x1) ( a 2b)a 2ba 2bPlease, notice that 1 and thus 1 appears as one of the terms inside parentheses.a 2bExample 5.7 Factor 41 from the expression1 3 x y .4 4Solution:1 3 1x y 4 4 4 1x4143y4141 1 4 441 x 3441 1 y ( x 3y) 4Example 5.8 Factor 5 from the expression x 5.Solution: x 5 x x 5 5 51 5 5 5 Factorization of −1Consider a b . We can factor 1 a b a b ( 1) ( 1)(a b) (a b) (b a) 11Factorization of 1 causes the signs in front of each term in the original expression to change (in ourexample a changes to a , b changes to b ). Thus, a b ( b a)as we see above.Example 5.9 Factor 1 from the expression2 x y z .Solution:2x y z 1(2x y z) (2x y z)53


Simplification of algebraic fractionsRecall the operation of simplifying fractions.To simplify this fraction we divide the numerator and denominator by their common factor 5. We„cancel 5‟. We can always divide the numerator and denominator by the same non-zeroexpression. This rule, true for numerical fractions, is also true for algebraic fractions, that is fractionswhose numerator and denominator are not necessarily numbers but any algebraic expressions, like for243xa xy xexample , , or . To simplify an algebraic fraction divide the numerator andx 1a(1 a)2xdenominator by all of their common factors. For example, consider3a ( b c)ad, ad 0Since a is the common factor of the numerator and the denominator, we divide both, the numeratorand the denominator, by a . We “cancel a ”.The following rule must always be followed.In algebraic fractions, only factors can be cancelled, but not terms.a xFor example, in the expression , a 0 , a cannot be cancelled. Although a in the denominatoracan be viewed as a factor ( 1 a a ), in the numerator a is used as a term and thus we cannot „cancel3 1 3 1it‟. (Just like we cannot cancel 3 in , 1 ).3 3Application of factorization of a common factor to simplification of algebraic fractionsFactorization can often be used to simplify algebraic fractions. In the expression3a ay, ax 0axa is a factor of the denominator but not of the numerator. Notice, however, that a is a factor of eachterm in the numerator. Thus, we can factor a from the numerator, and then cancel it54


Example 5.10 Simplify the following expressions, if possible. If not possible, explain why it is notpossible.m 2mnxyv 4za)b) c)22mx y 3xy4zSolution:a) One needs to find all common factors of both the denominator and thenumerator. In this case the common factor is m . We will factor m in the numerator andthen cancel it with m in the denominator and arrive at our result..m 2mnm(1 2n) 12nm mb) The common factor is xy . Factor xy in the denominator and cancel it.xy xy 1 x2 y 3xy2 xy(x 3y)x 3yc) It cannot be simplified. In the numerator 4 z is used only as a term, not as afactor.Application of the factorization of −1 to the simplification of algebraic fractions x yConsider , x y 0 .x yThere are no common factors that could be canceled, but we should notice that signs in the numeratorare exactly the opposite of signs in the denominator (x follows a minus sign in the numerator but a plussign in the denominator; y is preceded by a plus sign in the numerator but by a minus sign in thedenominator). Factoring 1 either in the numerator or in the denominator (the choice is arbitrary) willreverse the signs and allow the simplification. x y 1(x y)1(x y)= 1x y x y x yExample 5.11 Simplify the following expressiona 4d. 4d aSolution:One needs to notice „the reversed signs‟ (all terms in the numerators follow a plus sign,while all terms in the denominator follow a minus sign). This requires factorization of 1 (either in the numerator or denominator). We will factor 1 in the denominator.a 4da 4da 4d 1 4d a 1(4d a)1(a 4d)55


Mistake 5.1When factoring 3 from 3x 3y 3,3x 3y 3 3( x y)Instead, 3x 3y 3 3( x y 1).Common mistakes and misconceptionsMistake 5.2x 1 1 . Remember that only factors can be cancelled, not terms.x<strong>Exercises</strong> with Answers (For answers see Appendix A)In all exercises of this lesson, we assume that denominators are different from zero.Ex.1 The Distributive Law states that c( a b) ca cb . Explain how we can use it to write anequivalent expression without parentheses. from the expression ( a b)c .Ex.2Write an equivalent expression without parentheses.L Wa) 2 b) R1 xc) P rtR2 2 r1 d) se) ( 3 xy)f) ( x 2 7z)x5g) c(2a c c )h) ( a a2 2)Ex.3 The Distributive Law applied to the product of two sums states:( x y)(b c) xb yb xc ycCan we instead state it as ( x y)(b c) xb xc yb yc ? Why? Reformulate the Distributive Lawin four different ways.Ex.4Write an equivalent expression without parentheses.54 3a) F 18b) 3y(6y 8y)9242c) ( x 2x)xd) (3c 33d)3223 3 4e) a(a ab ab )f) (2x y xy ) xy723g) ( x y)(z w)h) ( a b)(10 a )53 2i) ( a b g)(c d)j) ( x x x 1)(y 1)1k) (4x 8)(2y 3)l) ( 2a1)(1 b)(c 2)456


Ex.5 First write each of the following statements as an algebraic expression using parentheses whereappropriate, and then rewrite it again in its equivalent form without parentheses.a) The product of x 2 y and 5b) The opposite of 4x1c) The product a and c 1d) The product of 2 y and a 2 b de) The product of x 1and y3 2f) The opposite ofx x2 2x4Ex.6Write the following expressions in five different equivalent ways.a) 3(a b)b) ( 2 y) zEx.7 Circle all expressions that are equivalent to m( n p).mn mp ( n p)m pm nm mn p mp nmEx.8 Circle all expressions that are equivalent to ( x y z).( 1)( x y z)( x y z)(1) 1(x y z)( x y z)1 x y z x y z x y z ( x z y)Ex.9 a) After factoring a common factor from a two term expression, how many terms should youhave inside parentheses?b) After factoring a common factor from a three term expression, how many terms should youhave inside parentheses?c) After factoring a common factor from an m-term expression, how many terms should youhave inside parentheses?Ex.10 Factora) 5 from the expression 5x 5yb) 7 from the expression 7 49a2c) c from the expression 3c 2c6d) y from the expression 8 xy y3e) 11 t from the expression 11t 44tf) 2 from the 2lw 2lh 2wh3 2g) x from the expression 4x 5x 5xEx.11 Factor xy from the following expressions2a) 2xy a xy2 2b) x y xyc) axy xby yxEx.12 Factor 1 from a) 3+ x b) a b 1c)x y za 257


Ex.13 Factor2a) 5 a from the following expression 10a 15a11b) 11 t from the following expression t 2 44t235 3c) 5x from the following expression 15x 5x56d) 4y from the following expression 8xy 4ye) c 2 d 22 2 2 3from the following expression 8c d ac df) x 22 3 23 y from the following expression 3xy 9xyx 2x xg) from the following expression ay y y11 1h) from the following expression sa ba b a bEx.14 Factor2 2a) from the following expression x 2 4 y z3 3 31 1 1b) from the following expression x 5 5 25c) 0 . 6 from the following expression 3.6x 6y 0. 61 5d) from the following expression a b6 62 4 2e) from the following expression x 27 7 7f) 0 . 1 from the following expression 2a 0.3b 0. 13 6g) from the following expression 3xy x 111 11h) 0 . 02 from the following expression 0 .4m nEx.15 Factor a b from the following expressionsa) 6(a b) x(a b)b)4(a b) 3( a b)Ex.16 Factora) 2 xy from the following expression22 2 22xy 2xy 4xyb) a 3 b 33 4 3 7 3 3from the following expression a b 5ba a b5 3 3 2c) 17 xy from the following expression 17x y 34xy 51xyd) m 4 4 3 4 8n from the following expression m n 5nm m n33 7 8e) 4ac from the following expression 16ac 8ac12acd23 2 42f) 7ab from the following expression 35ab 14ab 21abg) x 2yfrom the following expression 2(x 2y) ( x 2y)zh)2( c d) from the following expression52( c d) 4a(c d)3258


Ex.17 Factor a from the following expression a a 2 3Ex.18 From the expression 4 x , factor the following.a) 2 b) xc) 2 xd) 4 xEx.19 List all terms of the denominator and numerator of the following algebraic fraction. For eachsuch term list all its explicit factors. Find all factors that are common to all terms. If you were asked tosimplify the fraction, what would be the expression by which you would divide the numerator anddenominator to simplify it?ta)2t tyx xyb)2ax3abc)2ab a7x Ex.20 In the expression , can x be viewed as a factor of the denominator? Can x be viewed asx 5a factor of the numerator? Can we “cancel x ”. If not, why? If yes, what is the resulting expression?7xEx.21 In the expression , can x be viewed as a factor of the denominator? Can x be viewed as a2x yfactor of the numerator? Can we “cancel x ”. If not, why? If yes, what is the resulting expression?a ( a x)Ex.22 In the expression , can a be viewed as a factor of the denominator? Can a be viewedaas a factor of the numerator? Can we “cancel a ”. If not, why? If yes, what is the resulting expression? abEx.23 In the expression , can ab be viewed as a factor of the denominator? Can ab be2ab cviewed as a factor of the numerator? Can we “cancel ab ”. If not, why? If yes, what is the resultingexpression?Ex.24 Simplify, if possible. Otherwise write “not possible”. Also, name the expression by which youdivide the numerator and denominator.3xy2abca)b)9yx8ab2 ac)2 2b ae)15x(a b)25x45xyd)320yf)2a ( b c)2a59


24x ( 5x)g)xa bi)7(a b)bc ( b e)h)b2x 3j)2xEx.25 Simplify, if possible. Otherwise write “not possible”.2a)b)2x 2y24x 5xc)x3xe)23x 9xa 3bg) 3b aa b ci)a b12x 4k)3xxy xym)3yxo)4uv22 6vu2uv2x2yxy xz3x23x 3xd)6x3 4 4a b 4baf)a2xh)2 52x xxyj)2xy xy 4x12y 8zl)4u 2v sn)s 2v u 2vuxyp)43xy 6x2y2z60


Lesson 6__________________________________________________________________________________Topics: Addition and subtraction of algebraic expressions.__________________________________________________________________________________In this lesson we will learn how and when to perform the operations of addition and subtraction ofalgebraic expressions.Like terms22Consider 3 x 4y z . The expression consists of three terms: 3x , 4y, z . Each term can beviewed as a product of a numerical and non-numerical factor (recall that numerical factors are alsocalled numerical coefficients). And so,3 x has a numerical factor 3, non-numerical factor x22 4y has a numerical factor 4 , non-numerical factor yz has a numerical factor 1, non-numerical factor zLike TermsLike terms are terms that have equal non-numerical factors.What is meant by non-numerical factors being equal is that they are equal (equivalent) as algebraicexpressions. That does not mean that they „look identical‟. You will notice that in the second examplebelow, xy and yx are equivalent even though they do not „look identical‟.Examples:3 a and 7 a are like terms because their non-numerical factors, both a ‟s, are equivalent.4 xy and 3yxare like terms because xy and yx are equivalent.2x and3x are not like terms since2x is not equivalent toAnother way of looking at like terms is that two terms are alike if they can be written as expressions3 2that have the same variables with the same exponents. For example, 5x y and 4 yx 3 y are like terms,because yx 33 24 y can be written as 4xy , and thus both expressions consist of the variable x raised tothe third power and variable y raised to the second power. It should be stressed once again, that beinglike terms does not mean ‘looking identical’.3x .Example 6.1 Circle all expressions that are like2 34yxyx 3 y23 y3 2x .0 .2( xy)2y2 32xySolution:We should circle3yx y2x y2123 23 2 x y ).2 3 4yx (because2 3 3 2 4yx 4xy ) andyx 3 y2(because61


Adding and subtracting like termsIf we are to add or subtract objects they must have the same units. For example, we can add dollars todollars and pounds to pounds, but we cannot add dollars to pounds. If we have 3 oranges and 5oranges, together we have 8 oranges, but if we have 3 oranges and 5 plums, we cannot add themtogether. The same idea applies to addition and subtraction of algebraic expressions. Like, and onlylike, terms can be added or subtracted.We apply the Distributive Law to add like terms.3a 5a (3 5) a 8a(or, 3 oranges and 5 oranges equal 8 oranges)To add (or subtract) like terms, add (or subtract) their numerical coefficients and keep nonnumericalfactors the same.For example,Similarly,3x 8x (3 8)x 5x1 3 1 3 4xy yx xy xy7 7 7 7 72 220.3y 0.1y (0.3 0.1) y 0.2y2The process of adding and subtracting like terms is called combining like terms or collecting liketerms.Example 6.2 If possible, collect like terms. Otherwise, write “not possible”.a) 3x 4x1 2 3 2b) a a2 7c) 0.5mn 0. 7mab 1d) ba3 6Solution:a) 3 x 4x ( 3 4)x x1 2 3 2 1 3 2 7 6 2 1 2b) a a a a a2 7 2 7 1414 14c) Since terms mn and m are not like terms, terms cannot be combined. It is“not possible”.ab 1 1 1 1 1 2 1 1d) ba ab ab ab ab ab3 6 3 6 3 6 6 6 662


Simplification of expressions with two or more terms by collecting like termsLet us examine the following expression 3x 4y 2x 6y. There are two “groups” of like terms inthis expression, x' s and y‟s. By changing the order of terms, we can group all like terms together,collect them, and as a result obtain a simplified expression.3x 4y 2x 6y 3x 2x 4y 6y (3 2) x (4 6) y 5x 2yExample 6.3 Simplify by collecting like terms.a) 3 x a 4x ab) 7y 5 12y y 6Solution:a) 3 x a 4x a 3x 4x a a (3 4)x a a xb) 7y 5 12y y 6 7y12y y 5 6 (7 121)y 1 4y1Simplification of expressions requiring the removal of parenthesesCollecting like terms often has to be preceded by removing parentheses. This is always done accordingto the Distributive Law.4m ( 2 m) 4m 2 m 4m m 2 5m 2 Parentheses following a plus sign areredundant, and thus can be dropped. Eachterm inside parentheses stays the same.3x (5x1) 3x 5x1 2x1Parentheses following a minus sign areremoved by reversing the sign of each terminside parentheses (operation equivalent tomultiplication by 1).a 3(a 2) a 3a 3( 2) a 3a 6 4a 6 We multiply 3 by each term in parentheses,and then collect all like terms.Example 6.4 Write each of the following expressions using algebraic symbols, then rewrite it in itsequivalent form without parentheses and, if possible, collect like terms.a) add 3x 2 and 4x 2b) subtract 3x 2 from 4x 2c) multiply 3x 2 and 4x 2Solution:a) This should be written as 3x 2 ( 4x 2). We remove parentheses (usingthe Distributive Law) and collect like terms ( x's separately, numbers separately) to get3x 2 ( 4x 2) 3x 2 4x 2 x63


) This should be written as 4x 2 (3x 2)Notice that “subtract from”causes us to reverse the order of the expressions. Removing parentheses following aminus sign reverses signs of all terms inside parentheses ( 3 x „becomes‟ 3x, 2„becomes‟ 2 ). After removing parentheses we collect like terms ( x's separately,numbers separately) to get 4x 2 (3x 2) 4x 2 3x 2 7x 4c) This should be written as ( 3x 2)( 4x 2). To remove parentheses we applythe Distributive Law ( 3x 2)( 4x 2) 3x(4x) 3x(2) 2( 4x) 2(2)12x2 6x 8x 4 12x214x 455Example 6.5 Rewrite the expression 3( x 2) (4 x ) in its equivalent form without parenthesesand simplify by collecting like terms.Solution:3( x5 2) (4 x5) 3x5 (3)( 2) 4 x5 3x5 x5 6 4 4x510The square of the sum or the difference of two expressionsThe Distributive Law, together with the ability to collect like terms, allows us to derive two importantformulas.2Consider ( a b) . Our goal is to write the expression in its equivalent form without parentheses.Thus the following is true.( a b)2 use the definition of the square of the expression( a b)(a b)apply the Distributive Law for the product of two sums22a ab ba b collect like terms22a 2ab bThe Square of the Sumof Two Expressions.2 22( a b) a 2ab bNotice that this means that the square of the sum of two terms is not equal to the sum of their2 2 2squares. In other words ( a b) a b .Using exactly the same technique, one can prove thatThe Square of the Differenceof Two Expressions.2 22( a b) a 2ab bAgain, this means that the square of the difference of two terms is not equal to the difference of2 2 2their squares. In other words ( a b) a b .64


Example 6.6 Rewriteterms.( 2 x2 ) in its equivalent form without parentheses. Simplify by collecting likeSolution:22(2 x) (2 x)(2 x) 22 2( x) x(2) x(x) 4 2x 2x x 4 4x x2Common mistakes and misconceptionsMistake 6.122 apples 3apples 5(apples) . Similarly, 2xxInstead,2x 3x 5x(although 2 x2x(3x) 6 ).2 3x 5 .Mistake 6.23 x ( a b) 3x a bThe minus sign pertains to the whole expression ( a b), and all signs must be „changed to theopposite‟, not only the first one, 3 x ( a b) 3x a b .Mistake 6.3Although it is true that( b2 2 2ab) a ,2 2 2( a b) a bMistake 6.4x 2xWhen collecting like terms in you CANNOT multiply the expression by 3 to get rid of3 3x 2x1 2fractions. x 2x(it would be like saying that 12).3 33 3<strong>Exercises</strong> with Answers (For answers see Appendix A)Ex.1Which of the following rows consists of all like terms?a) 1 22 1 x,4x, 0.5x, x2 , 3x263b) 3xy,8yx, 0.6xy,2xy, xy72 3 4c) 5x, 5x, 5x, 5x, 5d) 2a, 5a,7a,29a,a,11aEx.2 Are x and x like terms?Ex.3 Are any of the following like terms: 7 x 2 y ,27xy , and27(xy ) ?65


Ex.4 Circle all terms that are like 6 x 2 y .6 y2 2x 2x 2 y 5 xy0.3yx23Ex.5 Circle all terms that are like a 2 b .7235(ab ) 2 a 22 bab ab772Ex.6Circle all expressions that are likey 3 x 2xxy 3 2 3 y2x 2 y 3 .x y 2 xxyyxxyEx.7Circle all expressions that are like2 5 3a b a b 2 a5 ba 5 b 3 .3 3( ab ) b3 22(ab ) aa3 b 5Ex.8If possible, add (or subtract) the following expressions. Otherwise, write “not possible”.a) 4x 2xb) a ac) y yd) a abe) st tsf) ac cac2 2g) ac 6cah) 7 mhv hmv3 5 3 5i) 2xyz y z x3 2j) 3mn 6mnmk)7 z3 5 4 5xy z xyl) 9a 2 b2 7 ab 2Ex.9 Collect like terms by first factoring x .a) 3x 4xb) 2x x3 72 3xc) x d) 0.3x 0. 5x11 227 2xxe) x f) 9 55233xx 10Ex.10 If possible, collect like terms. Otherwise, write “not possible”.a) 7x 8xb) 2a 7a1 11 3c) x xd) x x2 22 2e) ab 8baf) 4 7x3 42g) 5c 10ch) 0.4xy 0.8yxEx.11 Rewrite by grouping all like terms together. Then collect like terms. For example,3x 7y 2x 2y 3x 2x 7y 2y 5x 9y.a) 3x 8y8x 2xb) 5a 3b 2b 7ac) 2ab 4ba 2 3ab1266


Ex.12 Simplify by collecting all like terms.2a) 3 j 4 j 2 jb) 2 a a3c) 0 .2z 0. 5z zd) 2 7m 43 4 3e) x x xf) 2x y x 2y3x11g) x 1h) 3y x x 4y2 22cd 12 3 2 3 2 3i) dc d cj) a b 4ba 6ab3 3443 7 3 1k) x y 2x 3yx 5xl) ab ba 2 338 19 4 8Ex.13 Students were asked to simplify the expression 2(a b) 3a 3b. The following answerswere given.Student A: a bStudent B: b aStudent C: b aStudent D: a bList all students who gave the right answer.Ex.14 Collect like terms, and only then evaluatea) x 1,y 3b) x 2,1y 5c) x 0.5,y 3.22 y 3x 2y y 4x when3 3Ex.15 Evaluate2 1x x x when a) x 45 2 10b) x 102c) x .3If you have not done this yet, simplify the expression by collecting like terms. Compare the obtainedexpression with the results of the evaluations.Ex.16 Write an equivalent expression without parentheses and then collect like termsa) ( 6a 2) 12ab) 4a 7a 2(8a1)c) y 3(y 2)d) 100(t 0.1) 102t2xe) (2x 1)5f) a 0.1(2 3a)g) ( x 1) xh) ( q 6)( 8) 21q4i) 3a 9b (4a b)5j) 2x 5y 4(3x y)67


682 122k) 6 ( d a) a dl) 3a1(4a 2)3 2m) 3xy 7yx ( xy 3)n) 3cb abc 2( a bc)1 14 33 2o) a b ( a 2b)p) (6x 3x1) (3x 3x 3)3 32 32 3q) (1 4x 6x 7x) (5 4x 6x 7x) r) ( 0.2m 0.03) (2.3m 4)Ex.17 a) Students were asked to write2( A B) in its equivalent form without parentheses and collect2 2like terms of the resulting expression. One student claimed that the answer was A B , the other one22that it was A 2AB B . Was either of them right?2b) If you were asked to write ( A B) in its equivalent form without parentheses and thencollect like terms, what would be your answer?Ex.18 Write an equivalent expression without parentheses, and then collect like terms.4 22a) ( x 2)b) ( 3x1) 1 c) ( a 2)( a 4)d) ( 3b c)b c 2 12e) (3c x) 2(5x c)f) (1 x 2x)(2 x)33 2 2 g) ( a b)(a 2b)h) 2x 3 i) ( a 3b c) ( c a)j) 2(x 2y) 3(5x 3y)2k) 3x (4 x)(x 2)l) 5 a (3 2a)m) 6x 2 ( x 1)(3x 2) 2n)o) 3k 2 3k ( k 2)(2k 3)p) x4 m 2( m 1)422 2 29 ( x 3) 6xEx.19 Write each of the following expressions using algebraic symbols, then rewrite it in itsequivalent form without parentheses and, if necessary, collect like terms.a) Subtract 2xyfrom 3 yxb) The sum of 3x 1and 4x 2c) The difference of 4a 3 2aand 4a3 2d) The difference of a 2 3band 2 b a1 2e) The product of a and 2a3 5f) The product of x 222 y and 3y xg) The sum of mnk , 4 nmk , and 3mnh) The sum of 3 x and 2, then raised to the second poweri) The difference of 2 a and b , then raised to the second powerj) The product of 2 and 4x 1, then added to 5x 2Ex.20 Rewrite the expression a 2(a b) 5bin its equivalent form without parentheses, collectlike terms, and only then evaluate if22


a) a 1,b 23 2b) a , b 14 7c) a 2.4,b 0. 622Ex.21 Rewrite the expression 4x xy 2( yx 12x) 2 in its equivalent form withoutparentheses, collect like terms, and only then evaluate whena) x 1,y 31b) x 2,y 6c) x 0.5,y 0.269


Lesson 7__________________________________________________________________________________Topics:Evaluation of more complicated algebraic expressions; Substitution of not only numbersbut also algebraic expressions.__________________________________________________________________________________Recall,If two quantities are equal, you can always substitute one for the other.“equals can be substituted for equals”Substitution of numbers for entire parts of expressionsAccording to the principle “equals can be substituted for equals” not only can we substitute numbersfor variables, but also for entire „parts of expressions‟.2If, for example, we know that a b 2, we can evaluate ( a b) 3( a b)by substituting the value2 for a b.22( a b) 3( a b) 2 32 4 6 2Example 7.1 Evaluate the expression x y , if 2 x y 0. 2 and z2 0. 4zSolution:x y 0.2 2 z 0.4 4212Using equivalent forms of an algebraic expression for its evaluationIf we are asked to evaluate a „complicated‟ algebraic expression, we may be able to simplify theexpression first, and then use its simplified form to perform the evaluation. Notice, that what we do isreplace („substitute for‟) the original algebraic expression with its equivalent („equal‟) form. Theprinciple “equals can be substituted for equals” is used.Example 7.2 Simplify the expression 3x 3( y x), and then evaluate it when x 3 and y 2.Solution:3x 3( y x) 3x 3y 3x 3x 3x 3y 3yIf y 2, then 3y 3( 2) 6.Sometimes, the only way to perform the evaluation is to first replace the expression with a certainequivalent form of it. Suppose, for example, that we know that x 5, and y z 4 . Is it possible to70


evaluate xy xz ? We will be able to do that if we can find an equivalent form of xy xz , a form thatwould „match‟ the information we have. If we factor x , we obtainxy xz x( y z)Now, we can replace x with 5, and y z with 4,xy xz x( y z) 5(4) 20Example 7.3 Evaluate the following expressions, if a b 1and c 2a) a c bb) 2c ( b a)a b3a 3bc) d)c ccSolution:a) We change the order of terms a c b a b c 12 3b) We remove parentheses, and then change the order of terms.2c ( b a) 2c b a a b 2c 12(2) 14 3a b a b 1c) We write the expression as one fraction c c c 23a 3b3( a b)3( 1)3d) We factor 3 in the numerator c c 2 2mExample 7.4 Evaluate the following expressions, if 4 .n2mna) b)2nmSolution:22m m a) 4 2 162 n n n 1 1b) Notice thatm m 4nn 1 becausem mn1mnm 1 1nnmnmReplacing parts of algebraic expressions with other equivalent expressionsIf we know that x 3mand y 5m, then we can express x + y in terms of m i.e. write it as anexpression that depends only on the variable m . “Equals can be substituted for equals”, and so wereplace x with 3a, y with 5 a .x y 3a 5a 2aThis exhibits the direct relationship between x y and a , and, in turn, allows us to find the value ofx y any time we know the value of a .71


For instance, the area of a rectangle is equal to the product LW where L represents its length and Wits width. Let us consider a rectangle with the length and width equal to each other. Recall that thistype of a rectangle is called a square. Let a denote the side of the square. We have W a,L a . Wecan express the area of a square in terms of a . To this end, we replace both W and L with a :2LW aa aAnd thus the area of a square is equal to the square of its side.Now, if for example, we would like to fine the area of a square, whose side is 2 inches long, wedetermine that a 2 (side is 2 inches long), and the area can be found by evaluatinga2 2 2 4The area of the square is 4 (actually 4 in 2 ).Example 7.5 Ifx z, express the following expressions in terms of z.2a) x b) xSolution:2 2 2a) x ( z) zb) x (z) z22Example 7.6 Express ( m n)(m mn n ) in terms ofa) x , if m 2, and n 3x.b) m , if n m.Solution:a) We replace m with 2, and n with 3 x , and simplify.22222( m n)(m mn n ) (2 3x)(2 2(3x) (3x)) (2 3x)(4 6x 3 x2 (2 3x)(4 6x 9x)b) We replace n with m , and simplify.22222 2 22( m n)(m mn n ) ( m m)(m mm m ) 2m(m m m ) 2mm 2m2)3Example 7.7 If2A vandB 2 2v, express ( A B)(A B ) in terms of v .Solution:We must find an equivalent form of the above expression expressed only in terms of2and B . We will remove parentheses.22 2 2( A B)(A B) A AB BA B A B .22Now, we can replace A with v , B with 2 v to get2 2( A B)(A B) A B v 2v v.2AEx.1 If x2 10, evaluate the following.<strong>Exercises</strong> with Answers (For answers see Appendix A)72


2 1a) 2x b) 30 5Ex.2 If A 3, evaluate the following.5a) 2Ab)c)5 2 2Ad)a bEx.3 Evaluate the following expressions, if 2.ca ba) 2cb)c)2 a b c a b c Ex.4 If x y 2, evaluate.2a) 7x 7yb)c) x yd)32 x c) 3x2 2 ( 2A3 23AA5 ) 2x y 7 7( x y)2Ex.5If1 A27, evaluate.a) 1 Ab)c)1 A1A1d)2AEx.6 Evaluate the following expressions, if 2a b c 7.a) c b 2ab) 2 a b cc) ( 2a b c)d) 2a ( c b)e) a b c aEx.7 Evaluate the following expressions, if GHJ 1.1a) JHGb) GHJc) G ( 2) H(1)Jd)G3(HJ)3aEx.8 Evaluate the following expressions, if 3.b73


a) a b 2bc) a bd) ae)b)aba 22a a f)3bbEx.9Evaluate the following expressions, ifyxa)z2x yc)2zze) xy2xyz1 3.b)xy z1z d) xyf) xy zEx.10 Evaluate the following expressions, if a b 2c d 3.a) a b c db) 2(a d c b)c) 9 ( a b)2 d c2Ex.11 Evaluate the following expressions, if x y 0.2,x z 0. 6x z2a)b) ( x z) x yc)2x yy( x z)x2Ex.12 Evaluate the following expressions, if xy 1,zt 3.a) 4xyztb) zxtyc) xy ztyx tzd)xy3 4 6 2Ex.13 Evaluate the following expressions, if x y z .36 4 3a) z y xb)c)x y33 46 zd)xz3 2 3( y z)26 4 3( 2) y ( 3)x2 2Ex.14 Evaluate the following expressions, if x y 0. 12 222x ya) y 0.3 xb) 0.2 0.21 2 1 21c) x yd)2 22 2x y74


Ex.15 Simplify the following expressions, and then evaluate when x 2,z 3.3x 3za)b) 4x ( z 4x)33aEx.16 Rewrite the expression in terms of x , if it is given that2a 2x. Simplify.Ex.17 Let C 3x,xD . Express the following expression in terms of x . Simplify.2a) CD b) C Dc) 3 C 2 DC 2Dd)42Ex.18 Rewrite the expression a in terms of x for each of the following. Write your answer withoutparentheses.a) a xb) a 5xc) a xd) a 5x3e) a x 1f) a x22Ex.19 Rewrite the expression a 2ab b in terms of x , ifa) a 1 , b xb) a 3x,b 2xxc) a , b x2d) a b xEx.20 Let P 3x2 2x1, Q x 2 , and R 3x1. Express the following in terms of x . Removeparentheses and simplify.a) Q Rb) Pc) 2 Pd) P R2e) Rf) QRg) 3 R Qh)Ex.21 Express the following expression in terms of s , if2P Q3x s , y 2sa) 2 x yb) x 2 yc)32x yd)2xy4. Simplify, if possible.Ex.22 Express ab in terms of x , if5xa ,126b 1265. Simplify.2 2Ex.23 Express the expression ( mn) mn in terms ofa) m, if n m. Simplify.b) s , if m 2 s,n s . Simplify75


Ex.24 Express the expressiona) m s, n 2sb) m 2s2 , n sc) n m sEx.25 Express the expressiona) s , if m 2s,n 3sb) m, if n mc) m, if n m43 2 2m n 2mn in terms of s , if( mn22 n) ( m ) in terms ofEx.26 The following is true:4 4 3 2 2 3 4( a b) a 4ab 6ab 4ab b4a) Express ( a b) in terms of b , if a b .4 3 2 2 3 4b) Express a 4ab 6ab 4ab b in terms of b , if a b . Simplify.c) Compare the results of (a) and (b).4d) Express ( a b) in terms of b , if a b.4 3 2 2 3 4e) Express a 4ab 6ab 4ab b in terms of b , if a b. Simplify.f) Compare the results of (d) and (e).Ex.27 Rewrite the expression4 d 2b x. Simplify.a 2b 3c 4din terms of x , if it is given that a 3c 5xandEx.28 Express the following expression in terms of m , if ab maa) 3bab) b26 d) a3 b 311e) ( 2b)( a)f) a 2bac) bx tEx.29 Express the following expression in terms of y, if y, 2y. Simplify.z za)x 2t x t 2 b) z z z z x t1 2c)d) x tzz zEx.30 Rewrite the expression ( a b)(a b)in terms of x , if it is given that a 5xand b 2 x .Simplify.76


Lesson 8__________________________________________________________________________________Topics: Generalities on equations; Solving linear equations in one unknown.__________________________________________________________________________________This lesson introduces an important algebraic concept: equations.EquationsEquationA mathematical statement consisting of two expressionsseparated by an equal sign is called an equation.The following are examples of equations.2 3 5, 3x 5, x 3 y .We will refer to expressions on the left of the equal sign as the left-hand side of the equation and tothe expressions on the right of the equal sign as the right-hand side of the equation. In x 3 y ,x 3 is the left-hand side of the equation, and y is the right-hand side of the equation. We canalways reverse sides of the equation. For instance, instead of x 3 y , we can write y x 3. Bothstatements have exactly the same meaning.The difference between equations and algebraic expressionsNotice the difference between an equation and algebraic expression. Equations are two algebraicexpressions separated by the equal sign. There is always the left-hand side and the right-hand side ofeach equation. Algebraic expressions are different from equations. For example, x y 3x 2 is anequation, but x y , 3x 2 are simply algebraic expressions.Example 8.1 Determine whether the following mathematical sentences represent an equation or analgebraic expression. Any time you find an equation, circle its left-hand side.a) 4x 2b) 4x 2 7c) x2 3y2 4 xSolution:Only b) and c) are equations. The left hand sides: 4x 2 7, x2 3y2 4 x77


A solution, the solution set, and what it means to solve an equationEquations are often used to state the equality of two expressions containing one or more variables(often called unknowns).The solution setA value of the variable for which the equation is true (that is, avalue for which the left hand side of the equation is equal to theright hand side) is called a solution. The solution set of theequation is the set of all solutions of the equation.For example, x 7 is a solution of the equation x 2 9 (when we replace x with 7, the right-handside is equal to the left-hand side of the equation 7 2 9 ). Later on, we will learn that x 7 is theonly solution of this equation, and thus, since there are no other solutions, x 7 is also the solutionset of x 2 9 . Consider the equation x 2 36. One can check that x 6 is a solution of x2 36(substitute 6 for x to get 6 2 36 ). It is not the solution set, because it is not the only solution of thisequation. For example, x 6is another solution, since ( 6)2 36 . The solution set of an equationconsists of all solutions. Therefore, in the previous example, the solution set would need to includeboth 6 and 6 .To Solve an EquationTo solve an equation is to find the set of all solutions of theequation, or prove that it does not have a solution.3Example 8.2 Determine which of the following numbers are solutions of the equation x 6 x.a) x 2b) x 2Solution:a) Evaluate the left-hand side of the equation when x 2 .x 3 2 3 8 .Evaluate the right-hand side of the equation when x 2 .6 x 6 2 8.Since both the left-hand and right-hand side of the equations are equal, 2 is a solution.b) Evaluate the left-hand side of the equation when x 2.3 3x ( 2) 8Evaluate the right-hand side of the equation when x 2.6 x 6 ( 2) 6 2 4 .Since the left-hand side is not equal to the right-hand side ( 8 4 ), 2 is not asolution.Why do we solve equations? Here is a very basic example. Suppose that you bought some candies andyou did not know how many you bought but you know that you spent 80 cents and that each candycost 2 cents. To find out how many candies you actually bought, you set up an equation 2x 80,where x represents number of candies you bought. If you solve this equation, you would find thatx 40 and this means that you bought 40 candies. Now, let us learn how to solve certain types ofequations.78


Operations that can be performed on any equation without changing the set of solutionsThe following operations can be done to any equation without changing its solution.Any quantity can be added to, or subtracted from both sides of an equation.For example, If x y then x 2 y 2If x y then x 2 y 2Both sides of an equation can be multiplied, or divided by any nonzero quantity.For example, If x y then 2x 2yx yIf x y then 2 2It is important to remember that, any time we perform any operation on one side of an equation,exactly the same operation must be performed on the other side. We always perform the sameoperation on both sides.Solving linear equations in one variableThere are many types of equations, and depending on the type of equation, different solving techniquesare involved. We will learn how to solve linear equations in one variable (the formal definition of alinear equation will be provided later).To solve a linear equation we perform the operations of adding, subtracting, multiplyingor dividing both sides of an equation by suitable quantities with the goal of isolating thevariable (often x) on one side, and „bringing all numbers to the other side‟ of the equation.The numerical value obtained on the other side is the solution.We will learn how to implement the above idea by solving several equations.■ Solve the following equation x 2 7 .We need to isolate x , and to this end, we must somehow „get rid of 2‟ on the left-hand side of theequation. 2 is added to x , so if we subtract 2 (opposite operation to addition), we will get zero, and 2will no longer be on the left-hand side. But if we subtract 2 from the left-hand side of the equation, thesame operation must be performed on the right-side. Thus we need to subtract 2 from both sides. Wecould say that we are „bringing 2 to the other side‟x 2 7Subtract 2 from both sides.x 2 2 7 2Simplify.x 5The solution of x 2 7 is x 5.■79


■ Solve the following equation 3x 6 .3x 6To isolate x , „bring 3 to the other side‟. 3 is multiplied byx , thus we need to divide (opposite operation tomultiplication) both sides by 3.3x6 Simplify.3 3x 2The solution of 3x 6 is x 2 ■x■ Solve the following equation 94 .The solution of 94 x4 94 x 9 44x 36xis x 36 ■.To isolate x multiply both sides by 4 (since x is divided by4 and the opposite operation to multiplication is division).Simplify.■ Solve the following equation 3x 2 5.The goal is to isolate x on one side by „undoing‟ the operations that were performed on x .The unknown x was multiplied by 3, and then 2 was added. To solve for x we reverse the order.Since the last operation was adding 2, we “undo” that first by subtracting 2. Next we have to “undo”multiplication of x by 3, so we divide by 3.3x 2 5Subtract 2 from both sides of the equation.3x 2 2 5 2Simplify.3x 3Divide both sides by 3.3x3 3 3Simplify.x 1The solution of 3x 2 5 is x 1■80


■ Solve the following equation2x 9 34x.2x 9 34xIf x' s appear on both sides, „bring all x' s on one side‟. 4 xis subtracted, thus we need to add (opposite operation) 4 xto both sides.2x 9 4x 34x 4xCollect like terms (on both sides separately!)6x 9 3Now, the equation is in the form we already know how tosolve. We will continue solving it using methods introducedearlier. To isolate x , we first „bring 9 to the other side‟ byadding it to both sides.6x 9 9 39Simplify.6x 12Divide both sides by 6.6x12 Simplify.6 6x 2The solution of 2x 9 34xis x 2 ■■ Solve the following equation 3(x 2) 9.3(x 2) 9If parentheses are involved on one or both sides of theequation, we first remove parentheses by applying theDistributive Law.3x 6 9The equation is in the form we already know how tosolve: add 6 to both sides. The operation of adding 6 to 6 can easily be performed mentally, without recordingit and this is how we will be doing it from now on. Youare encouraged to do the same.3x 15Divide each side by 3.15x Simplify.3x 5The solution of 3(x 2) 9 is x 5. ■Example 8.3 Solve the following equations.a) 98 xb) 4 x 5 15 xSolution:a) x 98 To eliminate the minus sign, multiply (or equivalentlydivide) both sides by 1 (recall that x 1 x ).1(x ) 1(98)Perform the indicated operations.x 98The solution of x 98 is x 98.b) 4 x 5 15 x „Bring all x‟s on one side‟ by subtracting x from both81


sides (subtracting 4 x from both sides would also becorrect).4x x 5 15Collect like terms.3x 5 15Add 5 to both sides3x 15 5 Perform the operation of addition.3x 20 Divide both sides by 3.20x 320The solution of 4 x 5 15 x is x .3Checking solution of a linear equationsIt is always possible to check your solution. To this end, replace the variable in the original equationwith the value of the solution and verify that the left-hand side of the equation is equal to the righthandside. For example, to check that x 5 is indeed a solution of the equation3(x 2) 9, substitute5 for x in the equation.3(5 2) 933 99 9Since the left hand side is equal to the right hand side of the equation, 5 is indeed a solution.Example 8.4 Solve the equation 3 2x 5 and check your solution.Solution:3 2x 5 Add 5 to both sides ( x was multiplied by 2 , and then 5 wassubtracted, so to „undo‟ that, we first add 5)8 2x Divide each side by 2 (divide, since x is multiplied by 2 :there is no subtraction!). Simplify. 4 xThe solution of 3 2x 5 is x 4.To check the answer we replace x with 4 in the original equation 3 2(4) 5,evaluate both sides, 3 3, and conclude that, indeed, 4 is a solution.Linear equations with no solution or a solution set consisting of all real numbersAll of the above equations had exactly one solution. However, there are some linear equations thathave no solutions. The solution set of other equations may consist of all real numbers. We will seebelow an example of an equation that has no solution, followed by an example whose solution consistsof all real numbers.82


■ Solve the following equation 2x 1 2x 5.2x 1 2x 5Subtract 2 x from both sides. 1 5As we know, subtracting 2 x from both sides produces an equation with exactly the same solution setas the original one. But 1 5 is a false statement. No matter what the value of x is, it is never true.In other words, there is no value of x that would satisfy it. We conclude:The equation 2x 1 2x 5 does not have a solution. We write „no solution‟ as the answer. ■■ Solve the following equation 2(x 3) 2x 6 .2(x 3) 2x 6Remove parentheses.2x 6 2x 6Subtract 2 x from both sides. 6 6The statement 6 6is true. It is true regardless of the value of x . It is true if x 1,x 2,orx 1000. It is true for any value of x . All real numbers make it true. Therefore, the solution of2(x 3) 2x 6 is all real numbers. We write „all real numbers‟ as the answer. ■In general, if in the process of solving an equation the variable is cancelled (disappears from theequation), then the equation either has no solution or its solution is all real numbers.If an equation reduces to a false numerical statement, the original equation does not havea solution.If an equation reduces to a true numerical statement, the original equation has a solutionconsisting of all real numbers.Example 8.5 Solve the following equation 4x 4( x 1)Solution:4x 4( x 1)Remove parentheses using the Distributive Law.4x 4x 4 Subtract 4 x from both sides.0 4Since the statement 0 4is false, the equation 4x 4( x 1)has no solutions.Common mistakes and misconceptionsMistake 8.1Algebraic expressions cannot be ‘solved’. To solve means to find all values of variables forwhich the left-hand side of the equation is equal to the right-hand side. So, for example, itwould be meaningless to use the phrase ‘Solve 3 x + 2’.83


Mistake 8.2While solving an equation, DO NOT display the work as a sequence of equations.3x 2 6 3x 4…Each equation must always have a left-hand side and a right-hand side. Any time you performan operation rewrite the whole equation.Mistake 8.3When solving 6x 9 5, first subtract 9. DO NOT divide by 6 as a first step.Mistake 8.4If in the equation 5x 5 4xyou decide to subtract 4 x from both sides, remember that onthe right hand-side you get 0, and you must record that zero: 5x 5 4x 0 (the right-handside DOES NOT „disappear‟).Mistake 8.5DO NOT write ‘no solution’ when you get x 0. 0 IS the solution.Mistake 8.6When asked to solve an equation and after all operations you get 3 5, you need to write thefinal answer: there is no solution (similarly, write all real numbers are solutions, when you get3 3).<strong>Exercises</strong> with Answers (For answers see Appendix A)Ex.1 Fill in the blanks using the following words: „equation‟, „algebraic expression‟, and „solution‟as appropriate.One can solve a(n)____________ but not a(n) __________________ .If the left hand side of an equation is equal to the right hand side of the equation for x 7 , then 7 iscalled a ____________ .The __________(s) of an equation are all values of variables that make the equation true.The statement that contains two quantities separated by an equal sign is called a(n) ___________ .A(n) ______________ always makes the ______________ true.Ex.2 Determine whether the following mathematical sentences represent an equation or an algebraicexpression. In the case of an equation, circle the right-hand side of the equation.a) 5 xb) 5x 2c) x2 36x 1d) x2e) x 4 2xEx.3 Tom found the solution of an equation to be x 3, but the teacher gave as a correct solution3 x . Is Tom‟s answer right? Mary‟s answer to the same question is x 3. Did Mary correctlysolve the equation? Tell why or why not for both Tom and Mary.84


Ex.4 Does x 7 make the statement 2(x 1) x 7 true or false? Does that mean that 7 is, or isnot a solution of 2(x 1) x 7 ?1Ex.5 Determine if any of the following numbers 2, 16, , 2 is a solution of the equation.2 x 4 16. How about the equation x 4 16?Ex.6 Is x 1a solution ofa) ( x)( x) 2b) x x 2Ex.7 Is x 2 a solution ofa) 6 x 36b) 6 x 36c) ( 6)x 362Ex.8 Does a make the following statements true or false?54a)2 3 a b) a 125 5 c) a 12 5d) a 5a25Ex.9 Determine if y 0. 3 is a solution of any of the following equations.a) 0.027 yy 3b) 1 y 726c) ( y 0.7) 13Ex.10 Determine whether the following mathematical sentences represent an equation or an algebraicexpression. In the case of an equation, determine whether or not x 3is a solution of it.xxa)b) 0x 3x 3 c) x x2 12d) x 2 2xe) x 2 2xf) x3 x2 xEx.11 Determine whether the following mathematical sentences represent an equation or an algebraicexpression. In the case of an equation, determine whether or not x 0. 6 is a solution of it.22a) 10xb) 3.6 10xc) (0.5 )2 2 x 0. 01d) ( 0.5 x)12 x 0.6e) 20f) 0. 6 xx85


Ex.12 Determine whether the following mathematical sentences represent an equation or an algebraicexpression. In the case of an equation, guess one of its solutions.a) 2 3 3x 2b) 2x 3x24c) x 5 x 5d) 4x7e) 8xf) 3(x 7) 0Ex.13 Find a number that makes the equation1x1 2a) trueb) falseEx.14 Find a number that is a) a solution of x 5b) not a solution of x 5Ex.15 Check each of the following to determine whether or not it is a solution of the equation( 4 x )(2x 5)( x 3) x 0a) x 3b) x 055c) x d) x 22e) x 4f) x 4Ex.16 Guess four solutions of the equation ( 2 x) 2 x .Ex.17 Mr. X tried to solve the following equation x 2xby dividing each side by x . As a result, heobtained the equation 1 2 and concluded that the equation x 2xhas no solution. The correctsolution of this equation is x 0.What did Mr. X do wrong?Ex.18 Solve the following linear equations and check your solution.a) x 4 1b) 3x18c) 7 x 2xd) 5 4e) 10 0.2xf) 5 2 xg) 4 2xh) x 8Ex.19 Solve the following equations and check your solution.a) 2x 15 7b) 4x 7 5c) 9 32xd) 5x 4 0e) x 13 5f) 10 3x 5Ex.20 Solve the following equations.a) 4x x 12b) 3x 5 4xc) 6x 15xd) 4x 2 3xe) 3 7x xf) 3 14x86


Ex.21 Solve the following equations.a) 4x 3 12b) 0 2(1 x)c) 6 2( x 5)d) ( x 3) 2xe) 2 ( x 3) xf) x 4 (x 2)Ex.22 Determine which of the following equations has no solution, exactly one solution, or a solutionset consisting of all real numbers.a) x 8 x 9b) x 0c) x x 0d) 4(x 2) 2 4x 1 e) 3x1 3x 3 Ex.23 Solve the following equations.a) 6x 12ab) 154c) x 7d) x 01e) a 4 46f) 3x 7 8g) 0.8y 0.1 1.5h) 22 5y yi) x 4x 2xj) x 5 3x7k) 4 x 3 1xl) 2.3a 5 1.8am) 3(2B 5) 16n) 3x 12 7xo) 3(x 2) 6 3xp) x 7 3(x 5)q) 3 7x 2 4xr) 32m 4 6m 6s) ( x 3) 2x 3(1 x)t) 2 (4x1) 7x 4 xu) 4(y 2) (1 y)v) ( x 3) x (2x1)87


Lesson 9__________________________________________________________________________________Topics: Solving linear equations involving fractions; Solving equations for a given variable.__________________________________________________________________________________We will continue solving equations.Solving linear equations involving fractionsIf you are to solve an equation involving fractions, you can eliminate the fractions by applying thefollowing „trick‟: multiply both sides by a common multiple of all denominators of the equation.This step or ‘trick’ is not a requirement; however, if you chose to omit this step, you would have tocontinue to work with fractions in order to solve the equation.■ Solve the following equation4x1523.4x1523In this equation, “15” is a common multiple of bothdenominators (which are 5 and 3). We multiply both sidesby 15. 4x 2 151 15 5 3 Remove parentheses.4x215151 1553Cancel the denominators.34x 15 52Simplify.12x 1510„Bring 15 to the other side‟ by subtracting 15 from bothsides12x 5 Divide each side by 12.5x 12The solution of the equation4x2 51 is x . ■5 3 12Example 9.1 Solve the following equationSolution:3x1x 12 33x1x 1.2 3To get rid of fractions, multiply both sides by 6 (commonmultiple of all denominators on both sides, i.e. 2 and 3).88


■ Solvexa c ya for a .xa c yaBy subtracting ya from both sides, group all terms with a onone side.xa ya cIf a variable for which we solve an equation appears in severalterms, factor it. In this case, factor a .a( x y) cDivide each side by x y (assume x y 0 ).ca x ycThe equation xa c ya is solved for a : a . ■x yExample 9.3 Solve the equationA B2 B for A , and then find the value of A , if B 3 .BSolution:A B BBMultiply each side by B.2A BB BBBSimplify.2A B2 B2Add B to both sides.2A 2B2To find the value of A , if B 3, replace B in A 2B with 3 and evaluate.2A 23 2918.Example 9.4 Solve each equation for the specified variable. Always assume that the denominator(divisor) is different from zero. Simplify your final answer.a 2 2a) b 3b for a b) xy 1 y x for xbSolution:a)ab2 2 b 3bAdd2b to both sides.a 2 2 3b bbCollect like terms.a 24bMultiply each side by b .ba 2b 4bbSimplify.b3a 4bb) xy 1 y xBring all terms with x‟s on one side.91


xy x 1 yFactor x .x( y 1)1yDivide each side by y 11yFactor 1 from the numerator (or you couldx y 1choose to factor 1 from the denominator).1(y 1)x Cancel y 1.y 1x 1<strong>Exercises</strong> with Answers (For answers see Appendix A)When needed, assume that all denominators (divisors) are not equal to zero.Ex.1 Solve the following equations.x 2xa) 5b) 1 8x 42xx c) 3x 1 0d)1 1x53 21 1 3x 2 3e) y yf) 5 5 102 3 4Ex.2 Solve the following equations.a) 8x 2 3x 5( x 2)b) x ( 2 x) 5x12x2 53x3c) 1 a d) 13 62 4e) 4(3x 1)12x 2f) 4x 2 3x (1 5x)g) 2( x 3) 4 x 2h)x 3 1x2 5i) 3 (4 x) 2(x 6) x5 7 3j) x x 8 12 4k) 3(4 7x)15(3 x)l) ( x 2) 2( x 1) 3x 5Ex.3 Let A 2x,B x.Find x , so the following are true (Hint: Substitute algebraicexpressions for A and B . Then solve for x ).a) A B 0b) A 3B1A Bc) 3 8Ex.4 Let P 3(x 1)and Q 4x 5. Find x such that the following is true (Hint: Substitutealgebraic expressions for P and Q . Then solve for x ).a) P Qb) P Q92


Qc) P2Ex.5 Let x 3 a 2, y 2a1,z 2 a . Find a , so the following are true (Hint: Substitutealgebraic expressions for x, y,and z . Then solve for a ).a) x y zb) 2 x y zx y zc) 8 2 4x 1Ex.6 Does the following statement make sense: “Solve 5y2xEx.7 Solve for x . a) 72 xb) bafor x“? Why or why not?Ex.8 Solve for x . a) x 3 8b) x a bEx.9 Solve for x . a) 3x 7 11b) ax b cEx.10 Solve for x . a) 5x 2x 6 0 b) ax bx c 0Ex.11 Solve for x . a) 4x 2( x 1)b) ax b( x c)Ex.12. Solve for the indicated variable. Simplify your answer whenever possible.ba) x a for x b) ac for baa bc) for a d) ax b 4bfor a3b c2ae) 2 3a for u f) abc (ac) for bu2mg) 4 n for m h) 2x y t 3xfor y2n33i) x xy 2x for y j) AX A 1 X for A2k) s( x 1) s for x l) ax by 3ax 4byfor am) 3 ( v t) s for v n) ax x( a 2) 1for xm 2no) 2n m for xxp) mb 2mb 3bfor mEx.13 Solve the following equationx d e forya) d b) x c) yEx.14 Solve the following equation3 at b 2at t for93


a) b b) a c) tEx.15. SolveEx.16. Solvex 6 for x , and then evaluate, if 1a3aa .2 mn 2 for n , and then evaluate, if 43mnmm .Ex.17. If b represents the base of a triangle, h is the height, and A is the area of the triangle, thenbhthe following is true: A .2a) Solve the above formula for hb) Find the height of a triangle with base b 2 inches and the area A 4 square inchesEx.18. If L represents the length of a rectangle and W its width, then the perimeter P of therectangle is given by the formula P 2(L W)a) Solve the above formula for W .b) Find the value of W when P 4,and L 1.c) Find the width of a rectangle with perimeter 10 inches and length equal to 3 inches.94


Lesson 10__________________________________________________________________________________Topics: Linear Inequalities; Graphing sets of the type x a , x a on a number line.__________________________________________________________________________________We will now be discussing inequalities.Meaning and symbolic notation for ‘less (greater) than’, ‘less (greater) than or equal to’Recall thatx < 2 means that a number x is less than 2x > 2 means that a number x is greater than 2.Notice, that x 2 has exactly the same meaning as 2 x , and x 2 the same as 2 x (just as 1 2has exactly the same meaning as 2 > 1).The following symbols are also in usex ≤ 2 means that a number x is less than or equal to 2x ≥ 2 means that a number x is greater than or equal to 2.Again, x 2 is equivalent to 2 x, x 2 is equivalent to 2 x.The only difference between „ x 2‟ and „ x 2‟ is that 2 satisfies the first inequality but does notsatisfy the second one (2 is not less than 2). Similarly, 2 satisfies ‘ x 2’, but does not satisfy ‘ x 2 ’.Example 10.1 Describe the following set of numbers using inequality signs.a) All numbers x that are positiveb) All numbers x that are non-negativec) All numbers x that are at most equal to 9d) All numbers x that are at least equal to 9e) All numbers x that are not less than 9Solution:a) x 0 (zero should not be included)b) x 0 (zero should be included)c) x 9 (at most means that amount or less)d) x 9 (at least means that amount or higher)e) x 9 (not less than includes the given number and higher)Example 10.2 Determine which of the following numbers satisfies the condition x 2. 4, 2.1, 2, 0.9, 0, 1Solution:95


For each number you should check if it satisfies the inequality by replacing x with thisnumber. For example, the number 4 satisfies the inequality 4 2(it might help toplot the 6 points on a number line and note which ones are either equal to 2 or to theleft of 2 ). The final answer is 4, 2.1, 2Graphing of inequalitiesTo Graph an InequalityTo graph an inequality means to plot all numbers satisfyingthe inequality on a number line.Graphing gives a visual representation of a given set of numbers. For example, the graph of x 2consists of all numbers less than 2. These numbers are represented by points that are to the left of 2 ona number line. The number 2 does not belong to the set. We shade all points to the left of 2 and place‘an open circle’ at 2, to indicate that 2 is excluded.The graph of x 2 consists of all numbers greater than or equal to 2, represented by points to the rightof 2 on a number line, including 2. The number 2 does belong to the set. We will shade all points tothe right of 2 and place ‘a closed circle’ at 2, to indicate that 2 is included.A solution, the solution set, and what it means to solve an inequalityThe concept of solution(s) of an inequality is identical to the concept used in equations.Solution Set of an InequalityThe values of the variables for which the inequality is true arecalled solutions. The solution set of the inequality is the set ofall solutions of the inequality.For example, x 3 is a solution of x 2 6 (when we replace x with 3, we obtain the true statement3 2 6 ). However, it is not the solution set, because there are other solutions of the inequality, forexample x 1 ( 12 6). As expected,To Solve an InequalityTo solve an inequality is to find the set of solutions of theinequality or prove that it does not have a solution.96


Operations that can be performed on any inequality without changing its solutionThe following operations can be performed to any inequality without changing its solution.Any quantity can be added to, or subtracted from both sides of an inequality.For example, If x y then x 2 y 2If x y then x 2 y 2Both sides of an inequality can be multiplied, or divided by any nonzero quantity, but ifthe quantity is negative the direction of the inequality symbol has to be reversed.For example, If x y then 2x 2ybut 2x 2yIf x y thenx y x y but 2 2 2 2We must always perform the same operation on both sides of an inequality, just as we did in our workwith equations.Example 10.4 Knowing 6 3x, which of the following inequalities must also be true?a) 2 xb) 0 3x 6Solution:a) Since x appears without 3 on the right-hand side of the inequality 2 x,we conclude that the original inequality must have been divided by 3 on both sides.But any time you divide an inequality by a negative number, the inequality sign must bereversed. The resulting inequality would have been 2 x (not 2 x). Thus 2 xdoes not have to be true.b) Since 6 on the left-hand side of the inequality 6 3xis „replaced‟ withzero, we conclude that 6 must have been subtracted from both sides of the inequality.The resulting inequality would be 6 6 3x 6 , which after simplification gives us0 3x 6 . Thus, if 6 3xis true, 0 3x 6 must also be true.Solving linear inequalities in one variableThere are many types of inequalities, but in this course we will focus on linear inequalities only. Tosolve a linear inequality, we apply the same strategy as in the case of a linear equation. That is, we add,subtract, multiply or divide both sides of an inequality by suitable quantities with the goal of isolatingthe variable on one side of an inequality. However, we must remember the following rule.Any time we multiply or divide both sides of an inequality by a negative number,the direction of the inequality sign must be reversed.Consider the following example.97


■ Solve the inequality 4x 9 1. 4x 9 1Subtract 9 from both sides. 4x 9 9 19Perform the indicated operations (on each sideseparately). 4x 8Divide both sides by 4 and reverse thedirection of the inequality sign. 4x 8Cancel fractions. 4 4x 2The solution of the inequality 4x 9 1is x 2■This means that all numbers that are less than or equal to 2 make this inequality true. That is, if in theoriginal inequality x is replaced by any number less than or equal to 2, the resulting numericalstatement will always be true. It also means that any other number, i.e. any number that is greater than2, when substituted for x , would give us a false numerical statement.Observe that there are infinitely many solutions and that is why, instead of listing them, we must usethe appropriate inequality symbol in recording the result.Example 10.5 Solve the following inequalities and graph their solution.a) x 4b) 3x 7 2Solution:a) x 4Multiply both sides by 1, remember about reversing theinequality sign.( 1)(x ) ( 1)(4)Perform the indicated operations.x 4The graph of the solution:b) 3x 7 2Add 7 to both sides.3x 7 7 2 7 Simplify.3x 5Divide each side by 3.3x5 3 3Simplify.5x 3The graph of the solution:98


4 x x 2Example 10.6 Solve the following inequality 52 4Solution:4 x x 2To get rid of fractions, multiply both sides by 4 ( a 52 4common multiple of both denominators). 4 x x 2 4 54 Apply the Distributive Law to remove parentheses. 2 4 4 x x 24 4 54 Simplify (place the parentheses correctly)2 42(4 x ) ( x 2) 20 Again apply the Distributive Law to removeparentheses.8 2x x 2 20Collect like terms on each side of the inequalityseparately 3x 6 20Subtract 6 from both sides. 3x 6 6 20 6Simplify. 3x 14Divide each side by 3 and reverse the inequalitysign. 3x14Simplify. 3 314x 34 x x 214The solution of 5 is x .2 43Linear inequalities with no solution or solution consisting of all real numbersIt might happen that during the process of solving an inequality, the variable gets cancelled and theinequality reduces to a numerical statement (you can remember this occurring when solving someequations).■ Solve the following inequality x 3 x 5.x 3 x 5Subtract x from each sidex 3 x x 5 xSimplify3 5The statement 3 5 is always true, regardless of the value of x . Thus, the solution set of x 3 x 5consists of all real numbers. ■99


■ Solve the following inequality x 5 x 3.This time, after subtracting x from each side we obtain5 3This is a false statement, false for all values of x . There is no value of x that would make thisstatement true. Thus x 3 x 5 has no solutions. ■Example 10.6 Solve the following inequality. 3(2x 5) 6xSolution: 3(2x 5) 6xRemove parentheses. 6x 15 6xAdd 6 x to both sides. 6x 15 6x 6x 6xSimplify. 15 0Since the obtained inequality 15 0 is always true, we conclude that all real numbers are solutionsof the inequality 3(2x 5) 6x.Common mistakes and misconceptionsMistake 10.1DO NOT forget to reverse the inequality sign any time you multiply or divide both sides of aninequality by a negative number.Mistake 10.2When solving an inequality DO NOT change the inequality sign to an equation sign.Mistake 10.3DO NOT forget to write the final answer. If you are getting 3 5 that means there is nosolution. You must write that there is „no solution‟. Likewise, if you get something like 5>3,which is true regardless of the value(s) of the variable(s), that means all numbers are solutions.You must write “all real numbers”.<strong>Exercises</strong> with Answers (For answers see Appendix A)Ex.1 List two numbers satisfying x 5 . List two numbers satisfying a 5. Is the second questiondifferent from the first one? How about asking: List two numbers satisfying 5 x ?Ex.2 Which of the following statements has the same meaning as x 2.a) 2 xb) 2 xc) x 2100


d) x 2and x 2e) All numbers x that are at least 2 .f) All numbers x that are no more than 2 .Ex.3 Name three numbers that satisfy the condition x 1Ex.4Find a number that satisfies2x but does not satisfy32x .3Ex.5 Circle all numbers that satisfy the following inequality x 3 3, 2 , 1, 0, 1, 2, 3, 4, 5Ex.6 Determine which of the following numbers do not satisfy the inequality x 3 3, 2 , 1, 0, 1, 2, 3, 4, 5Ex.7 Determine which of the following numbers satisfies the inequality x 0. 6 .1, .666, 6, 0.6 .5999,02Ex.8Describe the following sets of numbers using inequality signs.a) All negative numbers x .b) All non-positive numbers x .c) All numbers x that are at least equal to 6.d) All numbers x that are at most equal to 6.e) All numbers x that are not more than 6Ex.9 Graph the following sets on a number line provided. Assume that all marks on the line areequally spaced.a) x 4b)2x 13c) 4 xd) x 2e)f)4x 35x 2101


Ex.10 Graph the following number sets on a number line.a) All numbers that are no less than 2b) All numbers no more than 4c) All non-negative numbersd) All numbers that are at most 1Ex.11 Using inequality symbols, describe the set that is graphed below.a)b)c)d)2Ex.12 Graph x and32also satisfies x .32x on one number line, and then find a number that satisfies32x and3Ex.13 Plot the points in part (a) and (b) on separate number lines. Then write an inequality for eachthat is satisfied by all points from the set.a) 3 , 4, 6b) 2,1,3Ex.14 Plot the points in part (a) and (b) on separate number lines. Then write an inequality for eachthat is not satisfied by any of these points.1 1a) 0 , 1, 4b) , 1 22 2102


Ex.15 Find an inequality that is satisfied by 1 but not by 3 (if it helps, you might plot the points).Ex.16 Find an inequality that is satisfied by 43 but not by 5 (if it helps, you might plot the points).Ex.17 The number 5 is a solution of which of the following inequalities. Determine your answerwithout solving the inequality. Show your work.a) x 2 4b) x 8 13c) x 3 1d) 3x 15Ex.18 The solution of an inequality is x 2Here are the answers given by students.Student A: x 2Student B: 2 xStudent C: x 2Student D: 2 xStudent E: x 3List all students who correctly solved the inequality.Ex.19 The solution of a given inequality is x 0.a) Is x 2 a solution of this inequality?b) Is x 2 the solution of this inequality? Why?c) How many solutions does this inequality have?d) List three solutions of the inequality.e) List three numbers that are not solutions of this inequality.f) If we write the solution of the inequality as 0 x , would that also be correct? Why?g) If we write the solution of the inequality as 0 x , would that also be correct? Why?h) Would it be right to say that the solution consists of all positive numbers?i) Would it be right to say that the solution consists of all non-negative numbers?Ex.20 Determine which of the following operations requires the change of inequality sign.a) Multiplying both sides of an inequality by 2 .b) Multiplying both sides of an inequality by 2.c) Adding 2 to both sides of an inequality.d) Subtracting 21 from both sides of an inequality.e) Dividing both sides of an inequality by 2 .Ex.21 Name the operation that must be performed on both sides of an inequality to isolate z on oneside. Determine if the operation requires the change of inequality sign (indicate it in writing), and thenperform the operation, reversing the inequality sign, if needed (for example, to isolate z in theinequality z 1 3, 1 must be added to both sides, the operation of adding 1 does not require thechange of sign, the resulting inequality is z 4).a) z 5 8b) 2 z 1c) 12 4z103


d) z 3e) z 13 Ex.22 Knowing x 10, which of the following inequalities must also be true?a) x 10 0b) 2x 20x3c) 5d) x 3210Ex.23 Knowing 2x 0, which of the following inequalities must also be true?a) x 2b) x 0c) x 2d) 2x 0Ex.24 Solve the following inequalities and graph their solutions.a) 2x 8b) x 1 xc) 0.3x 0. 6d) 3 2a11e) 6 x 4f) 2a1 3Ex.25 Solve the following inequalities. Each time you perform the operation on both sides ofinequality, name the operation together with the operand (for example write “adding 2 to both sides”,“dividing both sides by 3”, and so on).a) 4 3 ab) 3x1 63x xc) 15d) 5 44xx 1e) 1 1f) 144Ex.26 Solve the following inequalities.a) 3x 18 xb) 12 c) 2 3x 14d) x xe) x xf) 3a 4ag) 4x 17x 5h) x 1 6x 23i) x 12j) x 8 x 9k) 2 a 4a 5a 2l) 4x1 2( x 5)m) 3(x 2) 3x 21n) 3x1 3( x )3o) 3(2 5) 6x163 2x1 2 x x 2q) 5 y y 3r) 1x p) 242a 3s) 1 5a52323t) 3 23a 5 6a104


Lesson 11__________________________________________________________________________________Topics: Recognizing and matching patterns; Writing expressions in a prescribed way; Definition ofa linear equation; Factorization of the difference of squares.__________________________________________________________________________________This lesson is devoted to developing the ability to rewrite expressions according to a prescribed rule orpattern, as well as to match variables in the formula with appropriate parts of the expression.Matching expressions2If we are asked to remove parentheses in the expression, let us say ( x 3) , we might do it byremembering that raising an expression to the second power means to multiply it by itself two timesand thus222( x 3) ( x 3)( x 3) x 3x 3x 9 x 6x 9 .Alternatively, in this case, we could recall and use the equation that says that for any two numbers a2 222and b, ( a b) a 2ab b . To this end, we have to match the expression ( x 3) to the left-2hand side of the equation ( a b) and identify a and b in this representation. It might be helpful towrite the two expressions one under the other.2( a b)2( x 3)This makes it easier for us to see that a x and b 3 in this representation. Now, we rewrite theright side of the equation by replacing a with x and b with 3 in the right-hand side of the equation.22a 2ab b x2 2Thus ( x 3) x 6x 9 .2 2x332 x2 6x 9In order to use the formulas, we need to learn how to match expressions.2 xExample 11.1 The expression is written in the formz 1z 1represent a, b, and c, respectively.a cbc. What algebraic expressionsSolution:We matchwithto see that2 xz 1z 1a bc ca 2 , b x , and c z 1.105


Example 11.2 The expression( 22 2x 2) ( y 1)3 is written in the form22 2( x p) ( y q) rWhat are the values of p, q, and r ?Solution:Compare22 2( x p) ( y q) r22with ( x 2) ( y 1) 3to find that p 2,q 1, and r 3.2Rewriting expressions in a prescribed formIt is quite often that in order to match expressions we must rewrite a given expression in aprescribed form.Example 11.3 Write the expressiona) x y as a sum of two expressions, that is in the form A B .b) x y as a difference of two expressions, that is in the form A B .Solution:a) The difference of any two expressions always can be written as a sumx y x ( y)b) The sum of any two expressions always can be written as a differencex y x ( y)To see that (a) and (b) are true, recall the rules for signed numbers x ( y) x y ,x ( y) x y . Now, interchange the left-hand side of each of the equations with itsright-hand one (it can always be done!), and we obtain exactly what we claimed in part(a) and (b) of the solutions.22Example 11.4 Write the expression ( x 2) ( y 1)16in the form22 2( x p) ( y q) rWhat are the values of p, q, and r ?Solution:22We rewrite ( x 2) ( y 1)16asCompare( xr22 2 p) ( y q) with22( x ( 2)) ( y ( 1))to find that p 2,q 1,and r 4 .22( x ( 2)) ( y ( 1))4242Example 11.5 Write the following expression in the formEach time identify A .2A , where A is any algebraic expression.106


a)281x b)0.09xy8x c)6Solution:a) Express 81 as a square of a number 9 2 81. We writeand thus A 9x.m n mn8 4b) Recall that a a , and so x x 222 2 281x 9 x (9x. Notice that the power of x insidethe parentheses multiplied by 2 must be equal to 8. Since 42 8 , x needs to be4raised to the fourth power. In the expression 2expression4x we identify A as A x .c) Write all factors that are not already written in this form, as a square of some2 6 30.09 0.3 , 22 must be equal to 6, thus we haveThe expression 0.3x3 y y y (the power of y inside the parentheses multiplied by0.09x6y2220.3 x3 2y3y ). As a result, we have22 0.3x3 . y is written in the form0.3xA .y2A , with3Example 11.6 Write the following expressions in the form Ax By , where A , B are any numbers.Identify A and B in your representation. x 8ya)b) 3x (2y x)c) x ( y x)4Solution:a) Recall that x 8y x 8y1 x 2y. Compare the obtained4 4 4 41 1expression x 2ywith Ax By to recognize that A , B 2 .44b) We simplify the expression (hoping to obtain the desired form of theexpression). 3x (2y x) 3x 2y x 2x 2y 2x ( 2)y . Notice how, in the laststep, we expressed subtraction as the addition. Comparing 2x ( 2)y to Ax Bygives us A 2 , B 2.c) We simplify the expression x ( y x) x y x y 0x ( 1)y .Notice that after the simplification the variable x has „disappeared‟ from theexpression. Since zero multiplied by any quantity is equal to zero, in such situation, wecan write 0 x to match the prescribed form. This „trick‟ is often used, and thus worthremembering. The comparison of 0x ( 1)y with Ax By gives us A 0,B 1.Example 11.7 Write the following equation in the form y mx b , where m and b are any numbers.In the expression you obtained, identify m and b .a) y 2x 4b) y x 0c) 2y 4x 3)2107


Solution:a) To match the form y mx b , subtraction of 4 must be expressed as anaddition. We rewrite y 2x 4 as y 2x ( 4); m 2, b 4.b) To match the form y mx b , we must isolate y on one side of the equation(which means solving the equation for y ). To this end, we add x on both sides and gety x , or equivalently, y 1x 0 ; We see that m 1, b 0 .c) We must solve the equation for y , and then rewrite the right-hand side insuch way that it matches mx b .2y 4x 3 Divide each side by 2. 4x 3y Write the fraction as a sum of two fractions.2 4x3y Simplify.2 23y 2x23As a result, m 2, b .2The definition of a linear equationThe ability to rewrite an expression in a prescribed form is needed to determine if a given equation islinear. We use the following definition.Linear equationin one variableA linear equation in one variable x is an equation that can bewritten in the formax+b = 0,where b is any real number, and a any real number differentfrom zero ( a ≠ 0 ).The equation 2x 3 0 is already in the form ax b 0.2x 3 0ax b 0In this representation a 2,b 3 , and thus 2x 3 0 is an example of a linear equation.To show that the equation 5 x 1 x is also an example of a linear equation, one must be able torewrite it in the prescribed form given in the definition above, namely ax b 0. To this end, wesubtract x from both sides of the equation 5 x 1 x x x , collect like terms, and obtain theequation 4x 1 0 that matches the form ax b 04x1 0ax b 0We can now see that a 4 and b 1, and we have matched to the pattern given in the definition.Therefore, we can conclude that 5 x 1 x is a linear equation in one variable.108


2Equations x 3 0 and 4 3 2x 5 are not linear equations. No algebraic operation would „remove‟ x3or x from the equation, and hence each of the equations can never be written to match the desiredform ax b 0.Example 11.8 Determine if the following equations are linear in one variable. If so, express them inthe form ax b 0, a 0 . Determine the values of a and b in your representation.x2a) 1 0 b) 7x 2 3xc) 2x(x 3) 2xd) 3x2 1 04Solution:1a) It is a linear equation since it can be rewritten it in the form x ( 1) 0;41 1 a 0 , b 1. Notice, that this representation is not unique. If we multiply4 4 xboth sides of the equation 1 0 by 4, we get x 4 0 . This can be rewritten as41x ( 4) 0 , and then a 1, b 4. Both answers would be correct. There are otherrepresentations possible.b) Subtract 3 x from both sides 7x 2 3x 0 . Collect like terms 4x 2 0.This matches the form ax b 0, with a 4 ( 4 0), b 2. Thus 7x 2 3xis alinear equation in one variable.222c) Remove parentheses 2x 6x 2x, subtract 2x from both sides222x 6x 2x 0, and simplify to obtain 6x 0. We rewrite 6x 0 as 6x 0 0(adding 0 does not change the value of the expression, but allows us to match the form).2Thus a 6 ( 6 0), b 0 in our representation, and 2x(x 3) 2xis a linearequation in one variable.2d) Since no operation can remove x from the equation, it is not a linearequation.Factoring with the use of the Difference of Squares FormulaConsider the expression ( a b)(a b). When we remove parentheses and simplify, we get22 2 2( a b)(a b) a ab ba b a bIt is called the Difference of Squares Formula.The Difference ofSquaresFormula2a b2 ( a b)(a b)Let us consider the expression 2 2x 9 . Since 9 3 , we can rewrite x 2 9 in its equivalent form,2 22 2x 3 . The expression x 3 matches the left-hand side of the Difference of Squares Formula with109


a x and b 3 , so now we can rewrite the right-hand side of the formula, substituting x in place of2 2a and 3 in place of b . We will write the formula directly above the expression x 3 to make iteasier to see how x must replace a and 3 must replace b on the right hand side.2 2a b ( a b)(a b)x2 3 2 ( x 3)( x 3)Notice, that by applying the Difference of Squares formula we were able to rewrite the expression2 2x 9 in a form of a product, namely x 9 ( x 3)( x 3). Recall that the process of rewriting anexpression as a product is called factorization (we used factorization (*) to simplify algebraicfractions). So far the only method of factorization we learned was a method of factoring a commonfactor (see, Lesson 5). The example above provides us with another method of factorization, a method2 2that can be applied to any expression that can be rewritten to match the form a b . Let ussummarize the above considerations.If we need to factor the expression x2 9 , we must notice that it can be rewritten as a difference of2 2two squares, x 3 and then the Difference of Squares formula can be applied to get the desiredfactorization2x 9 ( x 3)( x 3) ..Example 11.9 Using the formula2 2a b ( a b)(a b), factor the following expressions.2a) 25 xb) x2 192 4c) 36m 49n2d) m2 ( 1 3m)Solution:22 22a) The expression 25 x must be written in the form a b . Since 25 5 ,2 2we have 5 x . We recognize that a 5 and b x , and substitute those values in theright-hand side of the equation( a b)(a b) 5 x 5 x .The expression 2 is factored: 25 x 5 x5 x225 xb) Write the expression2x 2 1 in the form922a b2.. To this end, notice that1b , and31 1 1 2 1 2 1 2 , and thus x x . We recognize that a x ,9 3 39 3substitute those values in the right-hand side of the equation 1 1 ( a b)(a b) x x . 332 1 1 1 Thus x x x .9 33 __________________________(*) We should make sure that we understand that factorization of an expression does not change its “meaning”. Theobtained expression, although written in a different form, is still equal to the original one. In the example, we can convinceourselves that the two expressions are equal by removing parentheses in (x − 3) (x + 3) and collecting like terms.(x − 3) (x + 3) = xx +3x−3x−3∙3 = x² − 9.110


c) Since36 m22m (6 ) ,4 249n (7n)2, thus36mn2 42 2 2 49n (6m) (7 ) .2 2The left-hand side of the equation is written in the form a b with a 6mand2b 7n . Substituting those values in ( a b)(a b)gives us the desired factorization2 42236m 49n (6m 7n)(6m 7n) .2 2d) The expression is already written in the form a b , with a m andb 1 3m . We replace a and b in ( a b)(a b)with m and 1 3m, respectively, andsimplify.( a b)(a b) [ m (1 3m)][m (1 3m)] ( m 13m)(m 13m) (4m1)(1 2m)22. This gives us the desired factorization: m (1 3m) (4m1)(1 2m).<strong>Exercises</strong> with Answers (For answers see Appendix A)Ex.1The expression2 3 y is written in the form 42( y a) . What is the value of a?Ex.2 The following expressions are written in the form ax by . What are the values of a and b inthese representations?2a) 3x 4yb) 4x y3Ex.3 The expression 3 x is writtena) in the form p x . Determine the value of p in this representation.b) in the form p x . Determine the value of p in this representationEx.4The expression25x is written in the forma bx . What are the values of a and b?2Ex.5 The expression ( 7) x is written in the form a 2 x . What is the value of a?Ex.6 The expression 4x 2yis written in the form A 2B. What algebraic expression represents Aand B.2Ex.7 The following expressions are written in the form XY For each of them determine whatalgebraic expression represents X and Y .22a) 3ab b) 3(ab )cEx.8 The following expressions are written in the form . For each of expressions identifyb(without rewriting the expression) what algebraic expression represents c and b .99( x 1)xa)b)92y9111


Ex.9 For each of the following expressions (1-6) indicate if they match A, B, C, D, E or F. Each timeidentify a and b .3 33(1) x 1(A) ( a b)(2)2( x 1)(B)2a 2(3) (8 x)(648x x )(C) ( a b)(a b)(4)(5)( 3 y322x 5 )(D) ( a b)(a ab b )( 3 y2 2x) (5 )(E)(6) ( 3 y)(3 y)(F)3a bb232( a b)Ex.10 Among the expressions below identify those that are written in the form A B and those thatare in the form A B , where A and B are any expressions except 0. Those that are in the formA B rewrite as A B , those that are in the form A B rewrite as A B (In other words, rewritesums as differences and differences as sums).a) 5 ( n)b) 5 nc) 5 ( n)d) 5 n3Ex.11 Rewrite the following expressions in the form ax b , where a and b are any numbers.Identify a and b in your representation.a) x 3 3b) 2x3 33x4x3 3c) 1 d)2233 2xe)f) (2x)3 422Ex.12 Write the following expressions in the form a , where a is any algebraic expression or anumber. In each case state what a is equal to.a) 36 b) 4009c) 0 . 16d) 49e)22b25y f) 1002g) 0.49c 4h) X6i) 4x2j) 81xy3Ex.13 Write the following expressions in the form a , where a is any algebraic expression or anumber. In each case state what a is equal to.a) 1b) 278c) 0.027 d) 125e)3 zf)364x8112


3 x6g)h) y8159xi) 1000x j)38yEx.14 Write the expressions4a) a6b) a12c) a24x in the following forms.7Ex.15 Write the following expressions in the form A . Identify A in your representation.714a) xb) x21yc) x 14 y 21d)70zmEx.16 Write the following expressions in the form a , where a is any algebraic expression and m is apositive integer different from 1. Identify a and m in your representation.a) x 2 x 37 7b) s (tv)c)e)b (b3 ) 2d)3 x y z x yzf)BC33264x2 2Ex.17 Write the following equations in the form ax by 0 , where a and b are any numbers.Identify a and b in your representation.a) 2 2 2 2 2x y 0b) 3x y2x 2c) ( y 6) 622 23x ye) 04d) x2 0f)22 8x 5y x 2Ex.18 Write in the form Ax By Cz , where A, B ,C are any numbers. Identify A, B , and C .a) x 3 z yb) 3 (2x y) z2 43x 2y zc)d) x y4x 3yz 2e) x 2zf) z43 5Ex.19 Write the following equations to match the form y mx b , where m and b are any numbers.In each case determine the value of m and b .a) y 3x 2b) y x2113


114xc) y 5d) y 3x 2 0e) 3y 6x1f) y x x 5g) 3 y 3 y 4x 2h) 422 22Ex.20 Recall the formula for the square of the sum ( a b) a 2ab b . The following2expressions are written in the form ( a b) . For each such expression, identify a and b, and thensubstitute their values in a2a) ( 3 b)b)c)( 2x 3y) y22 52222 2ab b . Simplify.Ex.21 Determine if the following equations are linear equations in one variable. If so, express them inthe form ax b 0, where b is any real number, a is any real number except zero, and x is unknown.Determine the values of a, and b in your representation.xa) 3x 2 9 0b) 1 03x 8c) x 0d) 0. 18e) x ( x 1) x2 2f) 3 x 7 xx 1g) ( x 2) 0. 5h) 2x 5 2Ex.22 The equation 4x 2 0 is a linear equation written in the form ax b 0. Record the valuesof a and b in this representation. Obtain an equivalent equation by multiplying both side of theequation by3 . What are the values of a and b in the new representation?Ex.23 Using the formula for the difference of two squares a 2 b2 ( a b)(a b), factor thefollowing expressions. Simplify your answer, if possible.a) 2 2x 1b) 4 9x222 2c) y 100ad) 25s 64te)4 12 2x f) x y 0. 2592a22g) 36h) m ( 2m1)2b222 2i) ( x 1) (3x 5)j) ( 2a 3) 9a3 322Ex.24 The following formula is true a b ( a b)(a ab b ). Factor each of the followingexpressions using the above formula. To this end- rewrite each expression to match the left-hand side of the equation


- identify the value of a and b in your representation22- replace a and b in ( a b)(a ab b ) with their representation- simplify, if possiblea) 3 3x 64b) 18y3 31c) 8x 27yd) x6 27Ex.25 The following formulas are true2 2a b ( a b)(a b)a3322 b ( a b)(a ab b ) .3 322a b ( a b)(a ab b )Factor each of the following expressions using one of the above formulas. To this end, you must firstmatch each expression with one of the above formulas, then identify the value of a and b in yourrepresentation, and finally replace a and b in the right-hand side of the used formula, Please, simplifyyour answer..3 322a) 8x yb) 36x 0.01yc)3 3x 27yd)3 3 364a b c2 22 2Ex.26 Calculate 10.7 9.3 using a b ( a b)(a b)(Show how you matched to the givenidentity in order to arrive at your answer).Ex.27 The following formula is true (you are not asked to check it, although you certainly can; youhave enough knowledge to do so)4 4 3 2 2 3 4( a b) a 4ab 6ab 4ab b4Use the above formula to calculate 11 . Hint: Use the fact that 11 10 1to write4form ( a b) .411 to match the115


APPENDIX A:Lesson 1ANSWERS TO EXERCISES21 3x 2 , y , bc ( 2a 1)3 are examples of algebraic expressions. x , y,a,b,c are examples of variables2but also examples of algebraic expressions . Variables represent unknown numbers. If we know the value of x , wecan evaluate 3x 2 , and as a result we get a number.2 a) “a squared” or “a raised to the second power.” b) “a cubed” or “a raised to the third power.” c) “a raised to thetwelfth power.” d) ”2 to the m” or ” 2 raised to the m-th power.” e) “minus y” or “the opposite of y” f) “theproduct of c and d” or “c times d” g) “a minus b” h) “two-fifth times x” or “two-fifth x”3 a) 7 n b) −5 k m c) “there is no multiplication performed” d) − x (−y) e) 3 x f) 2 x−y z+w (−t)24 a) multiplication b) division c) exponentiation d) division e) subtraction f) multiplication5 a) Any time two operation signs are next to each other, parentheses are needed. b) If parentheses areremoved, only m would be raised to the fourth power. c) Any time two operation signs are next to eachother, parentheses are needed, even if the multiplication sign is not explicitly displayed d) If parenthesesare removed, only a would be raised to the fourth power.each other, parentheses are needed, even if the multiplication sign is not explicitly displayedf) Any time two operation signs are next to each other, parentheses are needed6 a) s b) s c) st d) t e) xxf)y7 a) x b) −x c) multiplication8 a) 1∙x = x; b) −1∙x =−x c) 0∙x = 0x 3e) Any time two operation signs are next tog) st h) s i) y s9 a) ( x)b) c) x 3 y d) x 3 y y 10 a)1y b)32y c) y 5 d) v y e) y f) y 3 g) y x h) xy i) 2 y2 4 a11 a) a ( b)b) a ( b)c) a( b)d) −C f) −(−C) g) h) v( t)( p)i)12 a) x 10b)2x3c Bc)3x13dt14 ma15 1 ( ) 12 2bh16 a) 3 5 8 b) 3 2 1c)31171 can not be evaluated with 0xd)j)100xm( x) k)x y m be) x 2 f) 30 x g) x 3 h) x 53 d) 4 3 12 e) 96 33 2 f) 2x , because the denominator of a fraction can not be 0.1If x 0,can be evaluated:1 1 1 , but if x 5 thenx 5x 5 0 5 5x example could be: cannot be evaluated with y 4. (answers vary)3y 418 a) 3 0 0 b) 0 2 2c) undefined d)0 0 e)719 a) 3 2 9b) 2 3 8c) 2 2 4152 0 31 1 is undefined. Another5 5 023f) undefined116Answers to <strong>Exercises</strong>


20 a) A 2b) A 221 a) 6 8 2b) 10 6 16c) 4 6 2 d) 6 6 0 e) 2 6 6 222 a) 2 ( 2) 0 b) 2 ( 2) 4 c) 2 ( 2) 0 d) 5 ( 2) 4 1e) 6 ( 2)10 ( 2) 423 a) 3(10) 30 b) 5(10) 50c) 200 20d)10 5102e)1 5 10 0.5 or 2f) 10 4 10, 00024 a) 1000(12)12,000b)12 26c) 5(12) 60d)61212“not possible” e) 24 ( 12) 2 f) ( 12) 14425 a)7 or12 b)13c)8 d)13 e)8 or22 f)11 or2 33 3 15 21 12 3 3 3 39 1626 a) b) c)3 110 35 5 513 29 9 e) or 220 10 10427 a) b) 245c) d) e)1 177 38 2 217928 a) b) 19c)16 203d)21e)4f) 029 a) 3.41 b) 34. 81 c) 0.05 d) −8 e) 0.06 f) −530 a) −3.9 b) −2.1 c) 2 d) 360 e) −0.0006 f) −0.21631 a) −1.58 b) −1.1 c) −10 d) 0.3 e) 2032 a)4 115 15514c) 1. 28331 13a) b)15 14c) 0 . 88434 a) b)3 39 5 5c) 0. 00235940a) b)1219c) 2036 a) 10,000, 000 b) 16 c)1 d)80 .00001 or1100,00037 1 3 3 1 3 1 2 1a) 2 x 2 1b) x c) 2 4 4 2 8 2 438 t 1if t 1, t ( 1)1if t 1; t may be positive or negative depending on the value of tLesson 21 a) x ( m 2n) b) ( m 2n)c) ( m 2n)7d) m 2n 3ae) ( k 2 3k1)f) 4 ( 4x y)2 a) 3 x y b) 4(a b)c)f)3( 9x ) g)3 a) 2( 7)a bc3h) xy 6( x) d) z ( y 3) or ( y 3)zi) ( M 3)j)21x b) ( x c)c) x 5 d) c 9x34g) x 3 6 h) y ( x 4)i) ( y x)j) ( x 5) e)3 7( x) y3 x e)x k) 2 39x121( x) l)f) 4( x)2 xAnswers to <strong>Exercises</strong> 117


49C 3255 2(L+W)62mc75a) a b exponentiation b)c)8 x exponentiation d)5( a b) addition8( x) opposite of xa be)csubtraction f) a bcdivisiong) 4 7ymultiplication h) 3 a bdivision8 a) Multiply x by 4 and then subtract y. b) Add a and 3 and then divide the result by x.c) Add x and 3 and then multiply the result by y. d) Divide s by t and then add 2 to the result.e) Square x and then multiply by 3. f) Multiply x by 3 and then square the result.g) Add a and c and then raise the result to the 4 th power. h) Raise c to the 4 th power and then add the result to a.9 a) parentheses are needed b) c−3−a c) 3a+x d) parentheses are needed e ) parentheses are neededf) c d g) parentheses are needed8h) yx i) parentheses are needed j) parentheses are neededa10 a) 2 b) 2 c) 2 d) Yes, because we performed the same operations. e) −211 a) 235 11b) 4 33 5 c) cannot be performed3 3d) ( 3) 9e) 3 2 93 3f) 04 3g) 3 3 2712 a) 2 4 -16 b) ( 2) 16 c) ( 4) 16 d) 4 2 1613 a) (−(−1))+( −1)=0 b) (−(−1)) −( −1)=2 c) (−(−1))( − ( −1))=1 d) (−1) 2 =1 e) − (−1) 2 =−114 a)1 2b) 121121212215 a) 0.39 b) −5 c) −1 d) −27316 b) c) and f) We would have 0 in the denominator in these cases.17 a) −19 b) 60 c) −34 d) −35 e) −5052c) 1 2 1 d) 2 418 a) 1 48 10 9b) 1 4 10 1c) 4 1 372 8 5 8 5 5 8 202d) 1 4 27 8 e) 4 1 1 cannot be performed .f) 4 3 8 5 40 5 8 8 5 10419 a) 1 22 b) 2 4 16 c)2 1 2 4 16 381 3 81 3 d) 3 3 1 81 1 2 33 320 a) −1 b) 64 c)1 1 d)9 921 a) − 2.2 b) −2.122 a) −0.1 b) 0.1 c) −0.223 a) −2 b) −2e) 1 f) −11 3 13 4324 a) −5 b) 0 c) 1.3 d) 1 e) 2 f)2 2 15 1525 a) 4 b) 7 c) 0.5 d) 7 e) 3 f) 17 2 514 3 326 a) 10 b) −50 c) −0.05 435 1 2 1 2 41 8 e)61241 2 3 3 1 2 3 3 49118Answers to <strong>Exercises</strong>


27 a) 0 b) −628 a) addition, 7+x b) multiplication c) multiplication , 12x d) addition , 7xe) exponentiation, x∙8 f) subtraction 1x29 a) (3 )x ( 4)x 81 b) 5 c) 9 3 270.2 x Lesson 31 In the expression x 2yterms.4 2 ,24x and22 a) 3 , x b) ab, cd c) , 2y, 122 y are called factors . In the expression 4x 2 2y,xy2d) (2 b), , z3 All these expressions are equal (equivalent) because of the commutative property of addition.4 a) m n 1, n m 1; they are not equivalent. b) True24x and2 y are called5 a) Terms: 2 m, z ; 2m z z 2mb) Terms: x , 2 ; x 2 2 x c) Terms: 3c, 2 ; 3c 2 2 3c33 32 y22 2x, 222 y y x 2 2x22 2e) Terms: c( d f ), y ; c(d f ) y y c(d f )2 s 22 s 2 s 22 ( x y), ; ( x y) ( x y)33 3x mn 2 mn 2 b) 3(2a 3b) 4x 4x 3(2a 3b)d) Terms:f) Terms:6 a) x7 a) Terms: x 2 3 232 3 32 , x x ; x x x x x x x x x (answers vary)3b) Terms: , 2 ,x2 2 3x3x2 3x2 a bc ; a 2bc a 2bc 2bc a (answers vary)22 228 (1) -- (C), (2) -- (E), (3) --(A) , (4) --(B), (5)-- (D9 a) 2 a ; factors: 2, a b) 3 ( a b); factors: 3,( a b)c)2 3x ; factors: 3, x,2 d) 4 ( x y)(b c); factors: 4,( x y),( b c)yymn b) 57 7( 5)c) cd d( c)d) c( a d) ( a d)(c)vst stv svt b) v ( x y)t vt(x y) ( x y)vt ( x y)tv .10 a) nm11 a) tsv12 a) All these expressions are equal (equivalent) because of the commutative property of multiplicationb) 3AB 312 6; BA 3 213 1when A 1 and B 2 , thus they are not equivalent,13 They are all equivalent because of the commutative property of addition and multiplication142 5 2 5 a 6 a 6 aa) b) c)a 222b c b cd) 2227 7 7 3 3 3 a b a b a b ab c ab c ab c1522m n7m n5m 2nA B 2Ca)b)c)d)444c 2x164 7 27m n 3 m 3 ta)b)c)54xs 1172 x 2 2 2 ( a 2b)1 1a) x b) 2 c) x d) 2 e) x f) ( a 2b)3y 3y 3 3183m 3(m)a( b) a( b)( s 4)t4( a)a) b)c)d)e)f)n n44n n 1Answers to <strong>Exercises</strong> 119xy


3(m n)a(x1)g)h)2tx19s s s sThe opposite to is or equivalently , . All students were right.t tt t202a 2a2a2a c (2a c)2a ca) b) b b b 2d2d2d212x 4y2x y 2x 7y3x y 2x 3y 12na)b)c)d)e)f)333 ttk t3(a b) 2( cd 1) ( 2 a ) 3( m n)g)h)txy22 a)x3xx 3x1a b 1 b) 3 c) 3xd) ( a b)y yy y y yy y23 a) 4 a b) 32ac) 2 a d) 64 x e) not possible f) not possible g) 2 x h) 2 xmi) 32j) x2k) xy l) not possible m) 2 x n) 2 x86ybdo) x p) q) 1 x r)5xz 3acs) 0.06xyzt) 4x24 Yes, both are equivalent.25 (x+y)(1+a) and x+y(1+a) are not equivalent. In (x+y)(1+a) the entire expression of (x+y) is multiplied with (1+a) andin x+y(1+a) only y is multiplied with (1+a).26 a) a 2 b c c 2b a b) x x 44 xy 1c) xy4 4aa 2 2e)x y x yab b f) z c c z g) xyz yz( x)h) a 3 3 32 227m m , ,1 1m (1)are equivalent to m.28 m (n)and n m are equivalent to m n :29 a b c d and d b c a are equivalent to a c b d .30 a 1 2aa, a , , are equivalent to a .b b 2b bb31 All of the expressions are equal to32x 4x x, , and2 8 233 8a 3, 3 a 8,345x except656xare equivalent to x 23 8a2 are equivalent to 3 8a21 3ab b 3a1(3a b), ,, and (3a b)are equivalent to66 6 66m n 1 11 n m are equivalent to35m n, m n,( m n) and .3 3 33 33 336 Both John and Mary are right.37 ( )4 4x 1, x 1. Since 1 1 the expressions are not equivalent.2 238 y 53a b .6x , ( x y)2 1, when x 1, y 2 . Since 5 1 the expressions are not equivalent.n ( 1)m 11 m 1, when m 1 b) ( 1)m 1,1 m 1, when m 339 m−n+p=−2, m− (n+p)= −4,when m=2, n=5 5 ,and p=1. Since −2 −4 the expressions are not equivalent.40 a) , m 1 m 1, when m 5 d) ( 1)m 1,1 m 1, when m 7c) ( 1) 1,120Answers to <strong>Exercises</strong>


Lesson 4e) No, we cannot. Even if the expressions have the same answers in a-d, we cannot conclude that the expression willalways be equivalent. f) (−1) m =1, −1 m =−1, when m=2. Yes, we can, they are not equivalent. It is enough to findone set of values of variables for which two expressions are not equal to determine that they are not equivalent.1 coefficient, exponent or power, the base.2 a) first b) zero3 a) b b) ab c) de d) ─a e) a f) 2x y 43 a4 a)2 2 2 b) a c) d) ( 5a ) e) 3 75 base exponent coefficienta) x 4 3b) x m −16 a)3cc) x 323d) a bc2 −1e)xym 1f) x y714g)3x z37w4h) ab 51256 b)i) m+ n 2 +m j)34z c)3 a3 4 d) x5 y 2 e)34k k k k) z z zn n314z28a f)4 a a f)l) x y ( z)3z z z5( x) g)10 x2 2x y xy g)m)xx243 x3( a b) h) n) 2a 3 o) p) q)r) b x3 s) m n)m n( w 2v) m ( m p n)7 a) (−4) (−4) (−4) (−4) (−4) b) −4∙4∙4∙4∙4 c) (−m) (−m) (−m) d) −m∙ m∙ me) ( 2a )(2a)(2a)f) 2 a a ag) ( a b)(a b)h) a b b8 a) 1 b) 3 c) x d) a e) ab f) 1 g) ab 1h) 193a) x4b) ( x) , necessary7c) xd)3, necessarye)( b3a 2 ) , necessary f)3a 2bg)3a )2( 2t3 ) 4( 3 x)2( t) ( n m p)m( bc , necessary h)a ( 2b) 2x y 4, necessary10 a) 2,000,000 b) 8,000,000The answers are different, since the order of operations is different. In part b, we must first complete operations withinparentheses.11 a) To multiply exponential expressions with the same bases one needs to add the exponents.b) To divide exponential expressions with the same bases one needs to subtract their exponentsc) To raise an exponential expression to another power one needs to multiply exponents.12 a)i)23n b)169a j)14s c)12x k)6x d)1 a23 l)28b e)3x m)216x f)326a n)902m g) b h) 115s o)12 5t p)x 1218x3Answers to <strong>Exercises</strong> 121


52233x1 2 x5 3 1513 a) m ( 2m) 2mb) x orc) ( 2a ) 8a318x6 614 a) a 2b) 2 4 3a 32c) a 8d) a2 41 11 115 a) m 8 1b) m c) m 28d) 2m57 22 23 32416 a) x 49b) 49c) d) 0. 00499588417 a) B nc: 1 b) B nc: 1 c) B nc: 1 d) B nc: 13348 1 2 118 a) 8x b) −32x c) 6x d) 12x e) a3 f) x21g) 64xh) a4 3261 6i) a4256 2 6 xj) 2x k) 27x l) 1 m) v n) 9a o) x p) 25 0. 1619 a)20 a)1222 39 3xx) 27b)4a20 16a2( xx 10 y 6 b)g) 1 h)x 3 y 310c) 3ab d)1 x 3 i) 5 3162 y21 a)( xy7 ) 55 32 x yb)3y422 a) 223ax j)c)3 722( a ) (a) a62 4 a3a b e) f)416b5144(m n)k) ( a b)l)23 2 2 7( 3ab ) b 9abmn b) m 2 4c) ( n)2 14 y2 2 2,525y 2, y y 5,2 2 2 y , yy 55m d) 1 x5 y 5a8 bc 22420a3,a32020,(a31220)8,12a a2038,4(a )2035254 x2 3y ,( 2 x2 3y ) ,2 3 2) y ( 2)4xy)2 x( x( , 4 ( xy)2 x262 b6 2 3a c ,2 a2 3 2 3a b c ,112(abc ) ,2 37cb a146,3 32(a c) 2 b27 a) Ψ=8 b) Ψ= 10 c) Ψ=400 d) Ψ=12 e) Ψ=14 f) Ψ=1028 a) 15 b) 3 2 = 9 c) −12 d) 0.5 2 = 0.25 e) 4 2 = 16 f) −2 3 = −8Lesson 51 Based on the Commutative Law of Multiplication: ( a b)c c(a b), and from here we can apply the DistributiveLaw: c( a b) ca cb .2 a) 2L+2R b) R − Rx c) P+Prt d) R 2 s−r 2 s e) −3−xy f) x 3 −7zx g) 2ac+c 2 −c 6 h) a−a 2 +23 Based on the Commutative Law of Addition xb yb xc yc xb xc yb yc , so the two answers areequivalent. Some other ways (answers vary):122Answers to <strong>Exercises</strong>


( x y)(b c) xb xc yc yb yc yb xb xc yc xc yb xb xc xb yc yb55 46 53 2 2 24 a) F 10b) 18y 24yc) x 2x d) 2c 22de) a a b a b94 2 3 2 54 2 3f) 2xy xy x y g) xz yz xw yw h) 10a 4b a a b753 23 2i) ac bc gc ad bd gd j) x y x y xy y x x x 1k) 2xy 4y 3x 6 l) 2ac c 2abc bc 4a 2 4ab 2b225 a) ( x y)5 5x 5y b) ( 4x 1) 4x1c) a( c 1) ac a33 3 e) ( x 1)(y 2) xy y 2x 2d) 2y(a 2b d) 2ya 4yb 2yd2 42 4f) ( x x 2x) x x 2x6 a) 3a 3b 3b 3a( a b)3 a 3b3 b3a 3(answers vary)b) 2 z yz yz 2z z(2 y) z 2 zy 2z zy (answers vary)7 . mn mp( n p)m pm nm mn p mp nm8 ( 1)( x y z)( x y z)(1) 1(x y z)( x y z)1 x y z x y z x y z ( x z y)9 a) 2 b) 3 c) m10 a) 5(x y)b) 7(1 7a)c) ( 3 2c)2e) 11 t(14t)211 a) (2 a )c d) y ( 8xy 5 1)2 f) 2(lw lh wh)g) x (4x 5x 5)xy b) xy( x y)c) xy ( a b 1)12 a) 1(3 x) b) 1( b 1)13 a) 5 (2 3a)a b) 2e) d2 (8 ad)a c) x y z 1 a 2 13 211t t 4 c) 5 (3x1) 2x d) 4y5 (2xy1)c f) 3 2 x1x y( 13xy)g) ( 2 a)h) (1 s)ya b2 ( x 2 y 2zb) 1 1 3x 5c) )10.6(6x 10y1d) (5a 6b) 5 62 (2x 2 17311f) 0.1(20a 3b1)g) (11xy 2x ) h) 0.02(20m 50n)113( a b)(6 x b) ( a b)[4 3( a b)]14 a) )e) )15 a) )16 a) 2xy((1y 2xy)173 3 44 2 242 4 b) a b ( b 5a1)c) 17xy ( x y 2xy 3)d) m n(1 5n m )34 5e) 4ac( 4 2c 3cd) 3 a(1a a 18 a) 2(2 )2222 f) 7ab (5b 2ab 3)g) ( x 2y)(2 z ) h) ( c d)[1 4a(c d)]x 4 2 1 1 1 b) x 1 c) 2x d) 4x 2 x x 2 x 4 19 a) Numerator: One term, t. (can be viewed as two factors, 1 and t) Denominator: two terms: 2t (with factors 2 and t) and─ty (with factors ─1, t and y). Therefore t is a common factor of ALL terms (numerator and denominator.) We can dividethe numerator and the denominator by t.b) Numerator: two terms, x (with factors 1 and x) and xy (with factors x and y) Denominator: one term, 2ax (with factors2, a and x). ALL terms in the numerator AND in the denominator have a common factor, x. We can therefore divide boththe numerator and denominator by x.Answers to <strong>Exercises</strong> 123


c) Numerator: one term, 3ab (with three factors 3, a and b). Denominator: two terms, ab (with factors a and b) and ─ 22(with factors ─1 and a ). ALL terms in the numerator AND in the denominator have a common factor, a. We cantherefore divide both the numerator AND the denominator by a.20 x is NOT a factor in the denominator, but it can be viewed as a factor in the numerator. We can NOT cancel x, because itis not a factor in the denominator.217x IS a factor in BOTH the numerator and the denominator, therefore we can cancel x. The result isxy22 a IS a factor in BOTH the numerator and denominator, thus we can cancel it. The result: a x .23 ab IS a factor in the denominator, but NOT in the denominator, therefore we cannot cancel it.1 c24 a) ; 3xy b)3 4f)25 a)h)m)Lesson 6a( b c)21x y132 x1y x3; a g) 20x2 ;b)y z3; 2ab c)12b;2a d)xy 3 ; 5y e)43(a b)51 x h) c( b e); b i) ; ( a b)j) not possible7c) 4 5xd)i) not possible j)n) 111y1x2 o) 2 3u1e)v p)113xf); 5 x2 4 3a b 4ba g) 1k) not possible l) x 3y 2z13x 3 6xyza1 b) and d)2 Yes.3 No. (All three are unlike)42 26x y 2x 2 y5 xy0.3yx2563 ab225(ab ) 2 a 22b 77y 3 x 2 xy 3 x 2x 3 y2 y 2 xxbayyxxy78 a) x2 5 3a b a b 2 a5 b3 3( ab ) b3 22(ab ) a2 b) not possible c) 0 d) not possible e) st or 2tsh) 6 hmv i)3 z3 5xy j) m 3 n9 k) not possible l)a3 b 52 f)2 22ab9 a) ( 3 4)x xb) 1 2 1 x x c) 2 3 1 x x 3 7 21 1122 22d) ( 0.3 0.5) x 0.2x 7 2 53 1 2 3 1e) x x x 9 5 45 5 3 10 610 a) x b) 5ac) x d) x e) 7ab11 a) 3x 8x 2x 8y 13x 8yb) 5a 7a 3b 2b 12a 5bc) 2ab 4ba 3ab 2 1 3ab1 f) x f) not possible g) not possible h) 1.2x 2 y22ac g) not possible124Answers to <strong>Exercises</strong>


13 412 a) j b) 2 a c) 0 .7zd) 2 7me) 2x x f) 3 x y g) x 131 22 3 2 3h) y x i) cd d c j) 4ba 5ab k) 4yx 4 11 7 7xl) ab 52 338 813 Student A and C.14 x y a) 4 b)9 c) 2. 75215 x ; a) 4 b) 10 c) .3816 a) 18a 2 b) 27a 2 c) 4y 6 d) 2t 10e) x 1 f) 1.3a 0. 2 g) 2x 1h)541 29q 48 i) a b j) 10x 9yk) 2a 5dl) a 2 3 m) 3xy 3514 2n) 5bc abc 2ao) b p) 6x 3x 2 q) 12x 2 6 r) 2.5m 3. 9732 222 217 a) The student claiming that ( A B) A 2AB B was right. b) ( A B) A 2AB B18 a) x 8 4x4 4 b) 9 2 22 1 1 2 29x 6x1c) a 2a 8 d) 3b bc c e) c x2 235 3 2 2 4 8 2a 2ab a b 2bh) x 4xi) 3b 2cj) 13x 5y9 32 x 5x l) 5a 2 12a 4 m) 3 x 2 x n) 2 2 2 m 3m 2 o) 2k 63yx ( 2xy) 5b) ( 3x1) ( 4x 2) x13 2f) 2x 5x x 2 g)k) 819 a) xy333c) ( 4 2a) (4a 2) 8a 2a 2k p) 0a d) ( a 2 3b) ( 2b a) 2a 5b 22e)1 2 16a 2a 2a2 a f) 35 15 15g) mnk 4nmk ( 3mn) 3mnk 3mn224 2 2( 2x y)(3y x ) 2x 7xy 3y2 2 h) (3x 2) 9x12x 42 22 b) 4a 4abj) ( 5x 2) 2( 4x1) 3x 4( 2abi)203 1 3a 3b;a) 9 b) or 12 2c) 5.421 −3xy a) −9 b)1 c) 0.32Lesson 711 a) 20 b) 27c) 322 a) 6 b) −36 c) 4 d) −93 a) −4 b) −4 c) 424 a) −14 b) c) 2 d) 4722245 a) b) c) d)77 7 496 a) −7 b) 7 c) −7 d) −7 e) −77 a) 1 b) −1 c) −2 d) −1Answers to <strong>Exercises</strong> 125


18 a) 9 b) 3 c) 3 d) 3e) 6 f) 279 a)1 1 b)3 3c)1 d)910 a) 1 b) 2 c) 3911 a) −3 b) −0.4 c) −0.1212 a) −12 b) −3 c) −2 d) −22213 a) b) 32c) 9114 a) 0. 2 b) 231c) 0.05 or 201 e) 3 f)3d) 415 a) x z 1b) z 3166xa) 3x2172 3x3 2 7 27a) or x b) x c) x32 2 2 2 d) x182a) x2b) 25x2c) x2d) 25x2e) x 2x 1f)219 a) 1 2x x b)d) 1022x 2 2 9 2x c) x x x d) 0444x b) 3x 2 2x1c) 6x 2 4x 2 d) 3x2 5x 29x 2 6x f) 3x 2 7x 2 g) 8x 1h) 2x2 2x 32s 3 735 2 b) 2s c) 10s d) s20 a) 3e) 121 a) s22 x4 3m mb)4 34s 2s23 a)244a) 10s10 12b) 8s 8s4c) s252a) 26s2b) 4m2c) 4m264 44a) ( 2b) 16bb) 16b c) equal d) 04 4 4 4 4e) b 4b 6b 4b b 0 f) equal27 4 x28 a) 3mmb) 2mc) 63d) m e) m f) m29 a) 3yb)30 ( 6x 2)(4x 2)Lesson 82 3y c) y d) 5y1 One can solve an equation but not an algebraic expression_. If the left hand side of an equation is equal to the right handside of the equation for x 7 , then 7 is called a _solution_. The solutions_ of an equation are all values of variablesthat make the equation true. The statement that contains two quantities separated by an equal sign is called an _equation .A _solution__ always makes the _equation_ true.2 Equations: b) 5x 2 c) x 2 36 e) x 4 2x3 Both Tom’s and Mary’s answers are correct, because x=3 is equivalent to both 3=x and −x=−3.4 False. 7 is not a solution of 2(x + 1) − x = 7.5 None of the numbers is a solution of x 4 16 . The number 2 and −2 are solutions of x 4 =16.6 a) No b) Yes136x126Answers to <strong>Exercises</strong>


7 a) Yes b) No c) Yes8 a) No b) Yes c) No d) Yes9 a) Yes b) Yes c) No10 a) algebraic expression b) equation; −3 is a solution c) equation; −3 is not a solutiond) equation; −3 is not a solution e) algebraic expression f) equation; −3 is a solution11 a) algebraic expression b) equation; 0.6 is a solution c) equation; 0.6 is a solutiond) algebraic expression e) equation; 0.6 is a solution f) equation; 0.6 is not a solution12 a) equation; −1 b) algebraic expression c) equation; 0 or 1 d) equation; 1 e) algebraic expression f) equation;13 a) −2 b) 3 (answers vary; any number different from −2)14 a) 5 b) 5 (answers vary; any number different from −5.)15 a) a solution b) a solution c) a solution d) not a solution e) not a solution f) a solution16 For example, −2, −1, 0, 1 (answers vary; every number is a solution of this equation)17 We can only divide by a variable if we assume it is not 0. It is better if we always try to avoid dividing by a variable or byan algebraic expression. A better way to solve this equation is: x = 2x; x −2x = 2x−2x; x = 018 a) x= −3 b) x = 6 c) x= 5 d) x = −20 e) −50 f) x= −3 g) x= −2 h) x= −819 a) x=−4 b) x= 3 c) x= −34d) x 5e) x= −18 f) x= −52012 523a) x b) x c) x= 0 d) x e) x 5 778f) x= −121 f) x= 6 g) x= −1 c) x= −8 h) x= −1 e) x= 6 f) x= 122 a) no solution b) x= 0 (exactly one solution) c) all real numbers d) no solution e) all real numbers23 a) x= −2 b) a= −60 c) x= 7 d) x=0 e) a= 0 f) x= 5 g) y= 222 1121h) y i) x= 0 j) x= −6 k) x l) a= 10 m) B 6 356n) x= 3 o) all real numbersLesson 9p) x= −2 q)1 a) x 18b)2 a) no solution b)3 a) x 0 b)4 a) x 8b)5 a) no solution b)13x r) no solution s) no solution t) no solution u)25x c) 15 162 1x c) 5 4g) x 0 h)10211x 5c) x 02 11x c) x 7 212a c) a 77x 1 2x b) x abx b) x b ac bx acx a bx d)a d)x i) all real numbers j)6 No. 5yis not an equation, so we can not solve it.7 a) 148 a) 59 a) x 6 b)10 a) x 2 b)8x 5e) 21 64 13x k) 9 8y f)x e) no solution f)x l)75x v) x= 21x 61x 49x 2Answers to <strong>Exercises</strong> 127


11 a) x 1 b)12 a) x abcx a b42 b 3b2b) b a c c) a d) a e) u a f) b a cc xy t 2 1x ( x 1)i) y x j) A 1x 1x 1k) s 1s s 3tv or v n)331m 2nx o) 12 22n mxxd b) x y( d e)c) y yd eb t t b bb bb) a or a 1 c) t or t tt1 a a 1h) xm) t13 a) e14 a) at g)x l)x p)15 x= a 9 = −116 n = −m = 417 a)2Ah bb) h 4 inches18P 2L Pa) W or W L2 2b) W 1 c) W 2 inchesLesson 103m 2n5bya 2x3b3m 2b b 1b1 For example x= 3 or x= −2 (answers vary). The same numbers could be solutions for the other inequality. Both xx state the same condition (are equivalent)2 a) and d)3 For example x= −1 or x= −3 or x= −10 (answers vary.)42x 35. 3, 2 , 1, 0, 1, 2, 3, 4, 56. 3, 2 , 17 −.666, −6, −0.68 a) x 0 b) x 0 c) x 6 d) x 6 e) x 69 , 0, 1, 2, 3, 4, 5b)c)d)e)f)128Answers to <strong>Exercises</strong>


10a)b)c)d)11 a) x 1b) x 4c) x 1d) x 1122x is the value that satisfies both inequalities.313 a) x 3 or x 7 (answers vary)b) x 2 or x 3 (answers vary)14 a) x 4 or x 0 (answers vary)b)1 2 or x 2x (answers vary)15 x 0 (answers vary.)16 x 4 (answers vary)17 a) −5+2 < 4 and b) 5+8 13 ; −5 is not a solution of (c) and (d) −3 − 3d) divide or multiply by -1; sign changes; z < 3 e) multiply by -3; sign changes; z < − 322 a) and d)23 b) and d)24 a) x


c) x 2d) a 7e) x 10f) a 225 a) “add 3 to both sides”, “divide each side by −1”; a > −7130b) “subtract 1 from both sides”, “divide each side by 3”;5x 3c) “multiply each side by 4”, “divide each side by 3”; x < −20d) “multiply each side by 4”, “divide each side by −1”; x 20e) “subtract 1 from both sides”, “multiply each side by 4”; x 8f) “multiply each side by 4”, “subtract 1 from both sides”; x 526 a) x 6b) x 2c) x 4d) all real numbers e) no solution f)h)37Lesson 111x i)2 3n) no solution o) no solution p)3a 42 a) a 3 and b 4 b)3 a) p 3b) p 34 a 2 and b 55 a 76 A 4 x and B y7 a) X a and Y b3 59 2134x q) x 18 r) y 0 s) 481x j) all real numbers k) all real numbers l)a 4 and b 3 b) X 3 and Y ab8 a) c x 1 and b 2 b)9 (1)-E; x and b 123Answers to <strong>Exercises</strong>a g)4x 3x m) all real numbersa t) no solution9c x and b ya (2)-F; a x and b 1(3)-D; a 8 and b xa 3xand b 5y(5)-B; a 3xand b 5y(6)-C; a 3 and b yA ; 5 ( n) 5 n b) form of A B ; 5 n 5 ( n)A ; 5 ( n) 5 n d) form of A B ; 5 n 5 ( n)a 1 and b b) a 2 and b 3c) 1 1 x 3 1; a and b 122332 3 3 33x ; a 2 and b e) x ; a 1and b f) 8x 3 4 a 8,b 422 22(4)-A;10 a) form of Bc) form of B11 a) 3d)


12 a) 6 2 ; 6a b) 20 2 ; a 20 c) 0.42 3; a 0. 4 d)3 2 ; 7 2 b a e) (5y); a 5yf) 71022 223 234 24(0.7c); a 0. 7 h) ( X ) ; a X i) ( 2x ) ; a 2xj) ( 9xy ) ; a 9xyg) c a b) 3 3 ; a 3 c) 0.33 ; 0. 3313 a) ( 1) ; 114 a)3f) (4x); a 4xg)3a d) 2 3 2 ; 5 3a e) ( z); a z55 x x 2 3 23 33 x ; a h) ( y ) ; a y i) ( 10x ) ; a 10xj) 2 22y 6 464 642 122( x ) ; a x b) ( x ) ; a x c) ( x ) ; a x2 722 3 7 2 3 yb) ( x ) ; A x c) ( x y ) ; A x y d) z57x ; a x;m b) ( stv ) ; a stv;m 7 c) b7 ; a b;m 7715 a) ( x); A x16 a) 5343107;A 32b; a 105x; a 2y B B x y x y2d) ; a ; m 3 e) ; a ; m 4 f) (8x); a 8x;m 2 C C z z2222x ( 2)y 0; a 1;b b) 3x ( 1)y 0; a 3; b 1c) 1 21( 1)22 2x y 0; a ; b 1d) 1x 0y 0; a 1;b 022e)3 2 1 3 1x y2 0; a ; b or 3x 2 y 2 0 if you were to multiply both sides by 4, then a 3,b 14 4 4 4222 2f) 10x ( 5)y 0; a 10;b 5or 10x 5y 0; a 10;b 517 a) 218 a)c) 1 31 3( 1) x y z ; A 1;B ; C b) 6x ( 3)y z;A 6;B 3;C 1 4 24 23 1 1 3 1 1x y z ; A ; B ; C d) 1x ( 1)y 0z;A 1;B 1;C 04 2 4 4 2 43 33 3 1 1 x y 2z;A ; B ; C f) 0 x 0 y z ; A 0; B 0; C 4 44 4 15 151 1y 3x ( 2);m 3; b b) y 1x 0; m 1;b 0 c) y x 0; m ; b 05 5 1 1y 3x 2; m 3;b e) y 2 x ;m 2; b 33y 0x ( 5);m 0; b g) y 0x 0; m 0; b 0 h) y 4x ( 6);m 4; b 6e) 219 a) 220 a)d) 2f) 5b)c)a a 2223;b b;3 2(3) b b 9 6b b22 222x;b 3y;(2x) 2(2x)(3y) (3y) 4x12xy 9y22 2 2 2 2 2 2 4 4 2 4a y ; b ; ( y ) 2( y ) y y 55 5 5 2511x ( 1) 0; a ; b c) 1x 0 0; a 1;b 03311x 0.9 0; a ; b 0. or x 7.2 0; a 1;b 7. 2 e) x 2 0 , a 1,b 28821 a) not linear b) 1d) 9yz310Answers to <strong>Exercises</strong> 131


x or 4x ( 7) 0; a 4; b 7g) x ( 2.5) 0; a 1;b 2.5x or 18x 5 0; a 18;b 5x where a 12;b 6x b) ( )(2 3x)c) ( y 10a)(y 10a)d) ( 5s 8t)(5s 8t)f) 4 7 0; a 4;b 7h) 18 ( 5) 0; a 18;b 522 a 4;b 2 ; 12 6 023 a) ( 1)(x 1)2 3x e) 1 2 12 a a x x f) ( xy 0.5)( xy 0.5)g) 6 6 33 b b h) [ m (2m1)][m (2m1)] ( m1)(3m1)i) [( x 1) (3x 5)][( x 1) (3x 5)] ( 2x 4)(4x 6)j) [( 2a 3) 3a][(2a 3) 3a] ( a 3)(5a 3)3 3224 a) 4 ; a x;b 4; ( x 4)( x 4x16)3 32x b) 1 (2y); a 1;b 2y;(1 2y)(1 2y 4y)3 322c) (2x) (3y); a 2x;b 3y;(2x 3y)(4x 6xy 9y)32 3 1 2 1 2 1 4 1 2 1 d) ( x ) ; a x ; b ; x x x 33 33 9 22(2x y)(4x 2xy yb) ( 6x 0.1y)(6x 0.1y)25 a) )22c) ( x 3y)(x 3xy 9y)22 2 d) (4a bc)(16a 4abc b c )2 226 10.7 9.3 (10.7 9.3)(10.7 9.3) 1.4 20 284 432 23 427 (10 1) 10 41016101 4101114641132Answers to <strong>Exercises</strong>


APPENDIX B:SAMPLE TESTS_____________________________________________________________________________________Sample Test 1 (Lesson 1-3)_____________________________________________________________________________________1. Write the following statements as algebraicexpressions. Remember to place parentheses whereneeded (please, place them only when needed).a) two seventh of xyb) the product of x and y 2c) xy raised to fourth powerd) Subtract A from B, and then multiply by Ce) take the opposite to x , and then raise it tothe third power___________________________________________2. By placing the minus sign differently, write thea bexpression in two additional equivalentcways. Use parentheses when needed.___________________________________________3. Let m 2. Rewrite the expression replacing thevariable with its value (Remember to placeparentheses ) and then evaluate, if possible.Otherwise write “undefined”.41a) m b) 5 m c) m d)2 m___________________________________________4. Let a 0.1,b 0.3 . Rewrite theexpression replacing the variable with its value(remember to place parentheses ) and thenevaluate, if possible. Otherwise, write3 16. Let x , y . Rewrite the expression4 5replacing the variable with its value (rememberto place parentheses ) and then evaluate, ifpossible. Otherwise write “undefined”.xa) 8x 5yb) 5 xy c) y___________________________________________7. Determine which of the following expressions areequal to xx xx( 1) 1x12x11___________________________________________8. Write the expressions 1 x y as a single3 3fraction. .__________________________________________9. Fill in the blanks to make a true statement.Aa) A___________y2a b 2ab) __________c c3ac) a ______________4b 2 ___________________________________________“undefined”. a) a b b) 100 ab c)10. Perform all numerical operations that area possible. If no simplification is possible, write “not_______________________________________ possible”. a) 4 2xb) ( 4x)( 2)5. Write an algebraic expression representing the2 3x Ac) ( 2x) d) e) 12 xopposite of (do not remove parentheses):93 B_____________________________________________________________________________________Sample Test 2 (Lesson 1-3)_____________________________________________________________________________________x 32. Write the following statements as algebraic1. The expression cannot be evaluated for expressions: x 2a) x raised to the 4which of the following values of x? List all.th power, then subtracted from y.b) the sum of 5 and three-forth of a number.a) x 2 b) x 2c) x 0 d) x 3c) the opposite of b 4d) 6 less than the square of a number.Sample Tests 133


3. Let m = ─3. Evaluate the following expressions if _______________________________________possible (otherwise, write “undefined”): 5a2 m 5 m 18. Write in three different ways.a) m b) c)d) 3 2m2b2 m 3______________________________________________________________________________________ 9. Determine which of the following expressions are24. Evaluate uv if possible (otherwise, write2aequivalent to2 1“undefined”), when u and v .b3 31 2a(1 1a2, 2a b, 2a, (1 1a) b,_______________________________________ b b bbm ___________________________________________5. Evaluate n if possible (otherwise, writep10. Determine which of the following expressions areequivalent to“undefined”), when m 0.1,n 0.02, p 1.mn k :k mn nm k mn ( k)m(n k) k mn___________________________________________________________6. Let F be a variable representing the temperature. __________________________________________Use F to write the following statement as an algebraic 11. Replace with expressions such that theexpression:five-ninth times the difference of F and 32.resulting statement is true. Use parentheses whenneeded.___________________________________________a) xyz yz7. Perform all numerical operations if possible (ifAperforming a numerical operation is not possible,b) A 7 write “not possible”).10b2 2x3 9ya b a) b) 2 3cc) x (3) d) c) ( a b) 53 5c ce) 2x 3 (2 5) 4nf) 24 ( 5)_____________________________________________________________________________________Sample Test 3 (Lesson 1-3)_____________________________________________________________________________________1. Use the letter x to represent a number and write thefollowing statements as algebraic expressions.a) Half of a number is subtracted from 7.b) Twice a number is added to 5, and then theresult is multiplied by 3.c) The number is tripled, and then squared.___________________________________________2. Write the opposite (do not simplify).x ya) 2x + 3y b)x y_________________________________________3. Name the operation that should be performed first.23 x y b) x yz c) 4x d) 4x 2a) ___________________________________________4. If possible (otherwise, write “undefined”), evaluate─3xy when a) x = 6 and y = 253b) x 6 and y 4 c) x = 0.1 and y = 0.2.5. If possible (otherwise, write “undefined”), evaluatex when a) x = 6 and y = 2y53b) x 6 and y 4 c) x = 0.1 and y = 0.2.___________________________________________6. If possible (otherwise, write “undefined”), evaluatexwhen1 xa) x 0 b) x 1 c) x 1___________________________________________x y7. State which expressions are equivalent to2x ya) x y 2 b) x y2 c) 2 2 y x 1d)e) ( x y)2 2134Sample Tests


8. Replace X with expressions such that the resultingstatement is true. Use parentheses when needed.a) x 2 y 3z 3z 2y Xa Xb) 3 by ym n nc) X .v v9. Perform all possible numerical operations. If noneis possible, write “not possible”a) 5 6 2xb) 6(2x)c) 6 2xd) 2 53 6x e) 2 5 x3 6___________________________________________2210. Show that ( x) is not equivalent to x by2evaluating both expressions when x 3_____________________________________________________________________________________Sample Test 1 (Lesson 4-6)_____________________________________________________________________________________1. Write the following statements as an algebraicexpression using parentheses where appropriate, andthen remove parentheses:a) The product of A and B A 3b) The opposite of x 2_______________________________________35 32. Factor 2a from the expression 8ab 10a.___________________________________________2 4 23. Factor from the expression a 7 7 7_______________________________________4. If possible, add (or subtract) the followingexpressions. If not possible, write “not possible”.12a) 2hk kh b) ab ab22x 1c) mn nmn d) x3 6__________________________________________5. Collect like terms in 4yx y 3xy 8y 7y,1and then evaluate it when x 2,y .4_______________________________________6. Remove parentheses and then collect like terms1a) 7b (3b 6)b) 4x ( x 2)32c) ( 3x b)d) 2x2 (2x1)(x 2)7. Simplify, if possible (assume that all denominatorsare different from zero). If not possible, write “notpossible”.ax b xy 3a)b)axx2( a b)2c c)d) 2xyb a xyx 2e)2 x_______________________________________8. Write as a single exponential expression. For eachexpression, identify the numerical coefficient.5( a)7 2 3a) b) x ( 2x) 32abc)33a2ab___________________________________________9. Evaluate the following expressions:4520a) b) 3a432___________________________________________10. Find such X that the following is true:Xa) 4 7 2 1 b) 3X 1 9 5_____________________________________________________________________________________Sample Test 2 (Lesson 4-6)_____________________________________________________________________________________Sample Tests 135


1. Write the following statements as algebraic 6. Subtract 2 a b from 3a 2band simplify.expressions using parentheses if appropriate, and then ___________________________________________remove parentheses. Simplify, if possible:7. Remove parentheses. Simplify.a) The product of 2a 3 and 3a 3 4a 1 b) 2 times the opposite of 3y 9a) 7x 23x b) 2b (1 b) 4 c) 3 subtracted from x, then squared.______________________________________________________________________________________ 8. Simplify if possible. If not possible, clearly say so:2. Factor b from the expression: 9b 2 5b63a 2b3 x 2ya___________________________________________ a)b) c)d)237 32b 3a6b 3 x 2ya3. Factor 5x from the expression: 15x 5x______________________________________________________________________________________4. Collect like terms:9. Write as a single exponential expression.a) 3ab 4ba 5ab22a) 48253 x a2x( 3x2 ) x b)b) 3x 4x 5 3x 64 c) 34a x x 2a______________________________________________________________________________________5. If possible, add the following expressions,otherwise write “unlike terms”.21 2 1 22210. Factor 4 2 8from the expression: y 2 y a) ab b ab) mn (mn)3 3 9 33 22 22c) x y ( 4xy)_____________________________________________________________________________________Sample Test 3 (Lesson 4-6)_____________________________________________________________________________________1. Circle all expressions equivalent tox2 y 6xy 6 3x xy 23x y 2 3 2 2y x y___________________________________________2. Write as a one exponential expression.43x2 9a) b) a23(2x)a c) 3m(3m)___________________________________________3. Replace with a number so the following isequal: a) 9 7 3 b) 25 8 5___________________________________________4. Remove parentheses and identify numericalcoefficient:33 22 4 3 7ba) ( 2x )( 3x) x b) ( 2a b ) c) ( 5b)210b___________________________________________5. Write each statement as an algebraic expressionusing parentheses, and then rewrite withoutparentheses and collect like terms.a) opposite of 3x 5yb) product of x 2 5 and 2x 3c) sum of 5x 4yand 2x 7d) difference of 2x 2 7x 3 and 5x2 8x 66. Factor3 2a) x from the expression 5x 4x 7x.b) 3xfrom the expression 15x 2 3xc) a z from the expression 4a zxa z 2cd) 1 from the expression 2d___________________________________________7. Simplify, if possible. If not possible then write "not6xyxy xzpossible". a)b)9yx4x5x 3yx 4c)d) 3y 5xx 62758. Evaluate the following expression: .255___________________________________________9. Collect like terms, and then evaluate, whena 3 .a) 2a 7a (4 8a)b) (3 a)2 6ac) 6a 9 ( a 3)2136Sample Tests


_____________________________________________________________________________________Sample Test 1 (Lesson 7-9)_____________________________________________________________________________________1. Evaluate the expressions, if x 1,anda)2xyb) yx3y 4___________________________________________2. Evaluate a 2 bc if12a) a 2,bcb) a c 0.4,b 0. 54_______________________________________3. Let x 3m.Express the following5 xexpression in terms of m and simplify:3 x_______________________________________4. Write the following expression 2 x y z interms of m , if z 2x3mand y m.Simplify your answer._______________________________________25. Is x a solution of the following31 34equations? a) b) x2 x 293. Evaluate 2a (3b c)ifa) a 3,b 2,c 2 b) a 2,c 3b 0. 6_______________________________________7. If P x3and Q 2x 4 , find such x ,Pthat a) P Q b) Q 12_______________________________________8. Solve the following equations:1a) b 2 b) 5 2x1c) 2 x 2x 436y 4d) ( x4) 2xe) 3y 22_______________________________________9. Solve the following equations for x . Any timeyou must perform the operation of division,assume that the devisor is different from zero.Simplify whenever possible:xa) 2bb) ax b 2bc) ax a2 x ab___________________________________________10. If l d 8, evaluatea) 2 l d b) 2l 2dc) l d 2_____________________________________________________________________________________Sample Test 2 (Lesson 7-9)_____________________________________________________________________________________1. Evaluate the following expressions, if abc2 2 2abc 2 a)b) a b c4______________________________________x22. Evaluate x 2 3 ifa) x 4b) x 2 1______________________________________3. Write the following expressions in terms of a , ifx y a and xy a 1. Simplify your answer.x ya) xy b)3 32 2x yx y4. Determine whether the followingmathematical sentences represent an equation oran algebraic expression. In the case of anequation, circle the right-hand side of theequation. 3(x 2)3x4,15x , x 25x 7 8 , 2 3 5______________________________________5. Let m 2yand n 3y. Find y so that thefollowing is true.a) m 1 n b) m nSample Tests 137


6. Solve the following equations:8. Is y 0a solution of the following equations:a) 0.6y 0.4 b) 3x121a) 02c) a 4 d) 2 x 2x 4y5 7 1e) 2(2c 1) 4c 2f)x xb) 3y2 y 03 y5 3 3c) 1y 3______________________________________________________________________________7. Solve the following equations for d (assumethat all devisors are different from zero).9. Which of the following numbers are solutionsSimplify your answer whenever possible.of x 2 9 ?a) d c c2 2cba) 3b) ac) db2 b2 bb) 3d_____________________________________________________________________________________Sample Test 3 (Lesson 7-9)_____________________________________________________________________________________31. If A B , evaluate:82a)b) B A c) ABA B_______________________________________2. Evaluate the following expressions, if1x y 0.2 and 0.6:v1x ya) x yb)vv_______________________________________6. Which of the following are solutions ofxy 24 ?a) x 6,y 41b) x 2, y 48c) x 0,y 24_______________________________________7. Solve the following equations.a) x 3 27 b) 2x 7xc) 3(x1) 13x3xd) 2x 15 e) x 12_______________________________________8. Let A 3 x,B x , and C x. Find x so23. Is x 2a solution of x x 2 ? Explain howA Byou arrived at your answer.that the following is true: C 2 3______________________________________________________________________________24. Rewrite the expression 2a b in terms of x. 9. Solve for x . Any time you must perform theWrite your answer without parentheses. operation of division, assume that the devisor isSimplify. a) a x 2,b x 1different from zero. Simplify your answer, ifb) a 3x,b 2xpossiblex_______________________________________ a) ax b b) d c) axacx23d5. Write the following expression 3m(m 2n)interms of t , if m n t . Simplify your answer._____________________________________________________________________________________Sample Test 1 (Lesson 10-11)_____________________________________________________________________________________138Sample Tests


1. Describe the following set of numbers using 5. Write the expression x 7 in the form: x p ,inequality sign and then graph the set of on the where x is unknown and p any number. Thennumber line: all numbers x at least equal to 1.identify p in your representation______________________________________________________________________________2. Which of the following numbers:37 10 96. Write the following expressions in the form A , 1, 0, 2.99, , ,where A is any algebraic expression. Identify A in4 3 33ysatisfy the inequality: x 3?your representation a)b) x3 y 6_______________________________________8_______________________________________3. Solve the following inequalities.1 7. Write the following expressions in the forma) 3x 2 b) 3(x 1) 4xc) 3 2x3 24y ax bx c , where a, b,and c are any_______________________________________ numbers. Then identify a, b,and c ?3x 32 x3 24. The following expression is written in a) y 2x b) 2y 2x 4x 744x p_______________________________________the form . Identify p in this representation.28. Factor the following expressions:a) x 2 1001 4b) y25_____________________________________________________________________________________Sample Test 2 (Lesson 10-11)_____________________________________________________________________________________1. The number 4 satisfies which of the following 5. Write the following expressions in the form ofstatements? a) x 2 b) x 4 c) x 4na , where a and n is an algebraic expression_______________________________________ and then identify a and n:2. Using inequality symbols, describe the set5a) 2 23 3b) ( 2zy) ( x)that is graphed below.a)_______________________________________6. Write the following equations in the formax by c, where a , b,c are any real numbers.b)Identify a, b, and c in your representation.a) 2x 3y 5 b) 3y 7x1c) x 0______________________________________________________________________________ 7. Determine if the following equations are linear3. Solve the following inequalities.equations in one variable. If so, express it in thea) 5x 2 13b) 5 4 x c) 2x 8 form ax b 0, where a is any real number x 6d) 3e) 4x12 4x 3except 0, b is any real number, and x is5unknown. Then identify a and b. If not a linear_______________________________________ equation, then state: not a linear equation44. Write the following expression as a difference2x8x x 4a) x 1 0 b) 1 0 c) xof two expressions, that is in the form A B ,38where A and B are any expressions except zero. _______________________________________Then identify A and Ba) a 3bb)m nkSample Tests 139


8. Factor the following expressions:22a) 1 ab) ( 2x1) x29. Using the formula3 3223a b ( a b)(a ab b ), factor 1 27x._____________________________________________________________________________________Sample Test 3 (Lesson 10-11)_____________________________________________________________________________________1. Graph the following sets on the number line:a) x 4b) all numbers that are at least 1_______________________________________2. Solve the following inequalities:a) x 5b) 3x 3 4x 6c) 0 d) ( 2x 1) 4 2x53 2e) x 12 3_______________________________________3. If x 1, what inequality is true forxa) x 2b) 2_______________________________________4. The expression3(2a 2) ( 2) is written in2the form: 3A B . What algebraic expressionrepresents A and B?_______________________________________35. Write the following expressions in the form x ,where x is any algebraic expression. Identify x in318your representation: a) 8y b) y6. Write the following equations in the form3az b , where a and b are any numbers. AlsoIdentify a and b in your representation.a) z 3 3z3 16z3 3b) 13_______________________________________7. Factor the following expressions.22a) 0.25 9xb) ( 3a 1) (2a)4c) 1y_______________________________________8. Determine if the following equations are linearequations in one variable. If so, express it in theform ax b 0, where a is any real numberexcept 0, b is any real number, and x isunknown. Then identify the values of a and bxa) 5x 7 8 b) 14 c) 2x 4 3x_______________________________________9. Using the formula3 322a b ( a b)(a ab b ) ,3 3factor x y 10002140Sample Tests


APPENDIX C:SAMPLE TESTS SOLUTIONS_____________________________________________________________________________________Sample Test 1 (Lesson 1-3) Solutions_____________________________________________________________________________________1. a) 2xy b) x y 2c) xy 43 156. a) 5 b) c) 74 4d) C(B −A) e) x 37. all except 1 − 2xa b ( a b)a b8. 2x y2. c c c33. a) 2 b) −10 c) −16 d) undefined1 b 39. a) b) c)4. a) 0.4 b) −3 c) 0.9y c 45. A B 10. a) −4+2x b) 8x c) 2x 2 d)x3e) −4x_____________________________________________________________________________________Sample Test 2 (Lesson 1-3) Solutions_____________________________________________________________________________________1. x 22.4 3xa) y x b) 5 c) ( b 4)d) x2 643. a) 9 b) 1 c) undefined d) 94. 2275. 0.08596. F 327. a) −2b b) not possible c) x −9d) 6 x 3y e) 6x 3 −4 f) −10 4 n58.5 a a 1 5apossible ways: , 5 , 2 b 2b2 b9. all except: (1+1a) b10. nm−k, mn+(−k), −k+mn11. a) replace with ( x)b) replace with 7c) replace with 1_____________________________________________________________________________________Sample Test 3 (Lesson 1-3) Solutions_____________________________________________________________________________________x1. a) 7 b) 3(2x+5) c) (3x) 28. a) X xb) X 3abc) mX 2v2. a) −(−2x+3y) b) x y9. a) −5−4x b) −12x c) 3xx yd) 2 5 x3. a) addition of x and y b) division of x by y3 6e) 1 x 6c) exponentiation of x d) exponentiation of −4x244. a) 36 b) 15c) −0.0610. If x then ( x )2 and3985. a) 3 b) 10 c) −0.596. a) 0 b) undefined c) 127. all are equivalent except (a): x y 24 9 x2 x thus 22 xSample Tests Solutions 141


_____________________________________________________________________________________Sample Test 1 (Lessons 4-6) Solutions_____________________________________________________________________________________b) x 2x 21. a) A BA3ABA 2 3A7. a) not possible b) xy3x y3 c) 2 abba 23 22. 2a( 4ab 5)d) 2c23. 7 2a1xy xy2 x 2 1(2 x)2cy e) 12 x 2 x28. a) a ; coefficient: −13xb) 8 x 13 ;coefficient 84. a) hk b) not possible c) 0 d)226c) 4a , coefficient 45. xy ; after evaluating:1 xy 29. a) 2456. a) 7b 1 3 3b62 43 22 4 b) 3a 0 3137bb28b210. a) 4 7 2 X ; X 14b) 4xx24xx25x2c) 3xb 2 9x 2 6bxb 21X b)3 1 5 9 ;X10d) −3x+2_____________________________________________________________________________________Sample Test 2 (Lessons 4-6) Solutions_____________________________________________________________________________________31. a) (2a 3)( 3a 4a)5. a)1 ab 2 b) unlike terms c) 215xy24 2 3 6a 8a9a12a6b) 2((3y 9)) 2(3y 9) 6y186. ( 3a 2b) (2a b) 5a 3b2 2c) ( x 3) x 6x 917. a) x b) b 12. b9b5213 443. 5x(3x1)8. a) 1 b) c) not possible d) a2b 14. a) 3ab 4ba 5ab 4ab1567b) 3x4x 2 53x 2 6 7x 2 9. a) 6x b) x c) 2a3x11 10. 23 2y2 1 3 y4 _____________________________________________________________________________________Sample Test 3 (Lessons 4-6) Solutions_____________________________________________________________________________________1. xy 6 3x 23xy 2 3 2 222x y y x yd) (2x 7x 3) (5x8x 6)3 3x2 x 922. a) x b)622a c) 9m246. a) x (5x 4x 7)b) 3x(5x1)3 a) Ψ=14 b) Ψ=16c66 12c) ( a z)(4 x)d) ( 2)4. a) 6x ; coefficient: 6 b) 8a b ;coefficient: 8d7 27c) b coefficient: 2 1227. a) b) ( y z)c) 1 d) not possible3 45. a) 3x5y 3x5y5 2723 2b) ( x 5)(2x 3) 2x 3x10x158.5c) 5x4y 25 52 252x7 7x4y79 a) a 4 when a 3, a 4 1142Sample Tests Solutions


c)2 a when 3a ,2 a = 92b) 9 a when a 3, 9 a2 18____________________________________________________________________________________Sample Test 1 (Lessons 7-9) Solutions_____________________________________________________________________________________1. a) 9 b) 4 16 32. a) 1 b) −0.23. 8 mmor 8 m 14. 4m5. a) yes, it is a solution b) 32 is not a solution1 77. a) x b) x 3 558. a) b b) x 2 c) x 23d) no solution e) all real numbers2 3b19. a) x 2b b) x c) x a 1 a6. a) −2 b) 3.410. a) 16 b) 16 c) −64_____________________________________________________________________________________Sample Test 2 (Lessons 7-9) Solutions_____________________________________________________________________________________11. a) b) 422 16. a) y b) x c) a 102. a) 11 b) 43 32 2a 3 a 2a1d) x 2 e) all real numbers f) 153. a)b)3ab b 14. 3(x 2) 3x 4 ; equation7. a) d c2 c b) d c) d 1a b5x ; expressionx 28. a) no, not a solution b) yes, it is a solution5x 7 8 ; equation 2+3 = 5 equationc) yes, it is a solution5. a) y 1 b) y 09. a) no, not a solution b) no, not a solution_____________________________________________________________________________________Sample Test 3 (Lessons 7-9 ) Solutions_____________________________________________________________________________________163 3c) it is a solution set1. a) b) c)38 87. a) x 30 b) x 0 c) no solution2. a) −0.4 b) −0.12d) x 2e) x 23. no, not a solution8. x 04. a) x 2 2 3 b) 6x 4xb3 a9. a) x b) x 3d c) x 25. a) 3taa c6. a) it is a solution set b) it is a solution set_____________________________________________________________________________________Sample Test 1 (Lessons 10-11) Solutions_____________________________________________________________________________________3. a) x 2 1. x ≥ − 13b) x 3 c)2.10 9 x ≥ 3 and satisfy this inequalityy6. a)3 3 8 y 3 ; A y 2 24. p = 35. x 7 x ( 7); p 7x 11 8Sample Tests Solutions 143


3 ;b) x 3 y 6 xy 2 A xy 28. a) x 2 100 x 10x 107. a) 1 11y x3 ( 2) x 0 ; a , b 2, c 0b)4425 y 4 1 5 x 2 1 5 x 2 b) 7y x3 2x2 ; a 1, b 2, c 7 22_____________________________________________________________________________________Sample Test 2 (Lessons 10-11) Solutions_____________________________________________________________________________________1. b and c12. a) x < 2 b) x 4b) x ( 1) 0 ; a linear equation with33. a) x < 3 b) x 9 c) x 41d) x ≤ 9 e) all real numbersa , b 14. a) a 3b a ( 3b); A = a B 3b3c) x 0 0 ; a linear equation withm n m n m nb) ; A , B a 1, b 0,c 0 (note: other resultsk k k k kfor a are possible but b 0,c 0 for all65 a) 2 25 2 ; a 2 and n 6results)3b) 2xyz ; a 2xyz,n 38. a) 1 a 2 (1 a)(1 a)2 26. a) 2x + (-3)y = 5 so: a 2,b 3,c 5b) 2x 1 x (2x1x)(2x1x)b) either: 7x 3y 7so ( x 1)(3x1)a 7, b 3,c 7or 7x 3y 7 so 9. 127x3 1 3 3x 3a 7, b 3,c 72 1 3x13x 9xc) 1x 0y 0 a 1,b 0, c 07. a) not a linear equation_____________________________________________________________________________________Sample Test 3 (Lessons 10-11) Solutions____________________________________________________________________________________1. a)b) 6z 3 0;a 6 , b 07. a) 0.25 9x 2 (0.5 3x)(0.5 3x)b) x 12 2b) 3a1 2a ( a 1)(5a1)2. a) x 5b)x 7 3d) no solution e)3. a) x – 2 says that x 1x1b) saysthat x 224. A a 2, B 23 35. a) 8y 2y ; x 2yyy18 66b) 3; x yx 10 9c) x ≥ 66. a) 2z 3 1;a 2,b 1242c) 1y 1 y 2 221 y 1 y 1 y 1 y1 y8. a) 5x – 7 = 8 can become 5x – 15 = 0;linear equation a = 5 and b 15xb) 14 can become x 4 0 ; linearequation with a 1,b 4c) 2x 4 3xcan become x 8 0 ;linear equation a 1, b 83 32 2x y 1000 xy 10 x y 10xy1009. 2144Sample Tests Solutions

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