Beginning and Intermediate Algebra - Wallace Math Courses ...

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$Solving Linear Equations - Distance, Rate and Time - Wallace Math ...$

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$Solving Linear Equations - Absolute Value - Wallace Math Courses$

Thus, from our previous example we have two solutions, x = 4 or x = − 3. Again, it is important to remember that the absolute value must be alone first before we consider the positive and negative possibilities. This is illustrated in below. Example 89. 2 − 4|2x + 3| = − 18 To get the absolute value alone we first need to get rid of the 2 by subtracting, then divide by − 4. Notice we cannot combine the 2 and − 4 becuase they are not like terms, the − 4 has the absolute value connected to it. Also notice we do not distribute the − 4 into the absolute value. This is because the numbers outside cannot be combined with the numbers inside the absolute value. Thus we get the absolute value alone in the following way: 2 − 4|2x+3| = − 18 Notice absolute value is not alone − 2 − 2 Subtract2from both sides − 4|2x+3| = − 20 Absolute value still not alone − 4 − 4 Divide both sides by − 4 |2x +3| =5 Absoloute value can be positive or negative 2x +3=5 or 2x + 3=−5 Two equations to solve Now we just solve these two remaining equations to find our solutions. 2x + 3=5 − 3 − 3 2x = 2 2 2 x = 1 We now have our two solutions, x = 1 and x = − 4. or 2x + 3=−5 − 3 − 3 2x = − 8 2 2 x = − 4 As we are solving absolute value equations it is important to be aware of special cases. Remember the result of an absolute value must always be positive. Notice what happens in the next example. 54
Example 90. 7+|2x − 5| =4 Notice absolute value is not alone − 7 − 7 Subtract 7 from both sides |2x − 5| = − 3 Result of absolute value is negative! Notice the absolute value equals a negative number! This is impossible with absolute value. When this occurs we say there is no solution or ∅. One other type of absolute value problem is when two absolute values are equal to eachother. We still will consider both the positive and negative result, the difference here will be that we will have to distribute a negative into the second absolute value for the negative possibility. Example 91. |2x − 7| = |4x + 6| Absolute value can be positive or negative 2x − 7 =4x +6 or 2x − 7=− (4x + 6) makesecondpartofsecondequationnegative Notice the first equation is the positive possibility and has no significant difference other than the missing absolute value bars. The second equation considers the negative possibility. For this reason we have a negative in front of the expression which will be distributed through the equation on the first step of solving. So we solve both these equations as follows: 2x − 7=4x + 6 − 2x − 2x − 7=2x +6 − 6 − 6 − 13 = 2x 2 2 − 13 = x 2 This gives us our two solutions, x = or − 13 2 2x − 7=−(4x +6) 2x − 7=−4x − 6 + 4x + 4x 6x − 7=−6 + 7 + 7 6x = 1 6 6 x = 1 6 or x = 1 6 . 55
Page 1 and 2:
Beginning and Intermediate Algebra
Page 3 and 4: Special thanks to: My beautiful wif
Page 5 and 6: Chapter 6: Factoring 6.1 Greatest C
Page 7 and 8: 0.1 Pre-Algebra - Integers Objectiv
Page 9 and 10: Multiplication and division of inte
Page 11 and 12: 45) (4)( − 6) Find each quotient.
Page 13 and 14: 9 ÷ 3 3 = 21 ÷ 3 7 Our Soultion T
Page 15 and 16: Example 21. 13 6 − 9 6 4 6 2 3 Sa
Page 17 and 18: Find each quotient. 37) − 2 ÷ 7
Page 19 and 20: the expression that is to be evalua
Page 21 and 22: Solve. 1) − 6 · 4( − 1) 3) 3+(
Page 23 and 24: It will be more common in our study
Page 25 and 26: 0.4 Practice - Properties of Algebr
Page 27 and 28: Chapter 1 : Solving Linear Equation
Page 29 and 30: Addition Problems To solve equation
Page 31 and 32: 8x = − 24 8 8 x = − 3 Division
Page 33 and 34: 1.2 Linear Equations - Two-Step Equ
Page 35 and 36: − 1 − 1 Divide both sides by
Page 37 and 38: 1.3 Solving Linear Equations - Gene
Page 39 and 40: Example 63. − 3x +9=6x − 27 Not
Page 41 and 42: 4[ − 2]+9=−15 +8(2) Finish mult
Page 43 and 44: 1.4 Solving Linear Equations - Frac
Page 45 and 46: Example 72. � � 3 5 4 x + = 3 2
Page 47 and 48: 1.5 Solving Linear Equations - Form
Page 49 and 50: Example 79. A = πr 2 +πrs for s S
Page 51 and 52: 1.5 Practice - Formulas Solve each
Page 53: |x| = 5 Absolute value can be posit
Page 57 and 58: 1.7 Solving Linear Equations - Vari
Page 59 and 60: Once we have found the constant of
Page 61 and 62: 1.7 Practice - Variation Write the
Page 63 and 64: sure. If the pressure of a certain
Page 65 and 66: Fifteen more than three times a num
Page 67 and 68: When we started with our first, sec
Page 69 and 70: 1.8 Practice - Number and Geometry
Page 71 and 72: 30. The perimeter of a college bask
Page 73 and 74: Our equation comes from the future
Page 75 and 76: 4x + 2=102 Solve the two − step e
Page 77 and 78: years ago, the bronze plaque was on
Page 79 and 80: 1.10 Solving Linear Equations - Dis
Page 81 and 82: Rate Time Distance Bob r + 2 3 Fred
Page 83 and 84: 2 =t Our solution fort, she catches
Page 85 and 86: Find the distance he rode. 8. A man
Page 87 and 88: The first plan is flying 25 mph slo
Page 89 and 90: 2.1 Graphing - Points and Lines Obj
Page 91 and 92: E B D F G C A Our Solution The main
Page 93 and 94: x y − 3 − 4 0 − 2 3 0 Our com
Page 95 and 96: 2.2 Graphing - Slope Objective: Fin
Page 97 and 98: When mathematicians began working w
Page 99 and 100: Example 131. Find the value of x be
Page 101 and 102: 9) 10) Find the slope of the line t
Page 103 and 104: y = − 2 x + 3 Our Solution 3 Iden
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2.3 Practice - Slope-Intercept Writ
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2.4 Graphing - Point-Slope Form Obj
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Find the equation of the line throu
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Write the point-slope form of the e
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m = 2 5 m = 2 5 Example 147. Slope
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2.5 Practice - Parallel and Perpend
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Chapter 3 : Inequalities 3.1 Solve
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than 4. If we have an expression su
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− 5 − 5 − 2x � 6 Divide bot
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Solve each inequality, graph each s
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As the graphs overlap, we take the
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3.2 Practice - Compound Inequalitie
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edition of a college algebra text!
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It is important to remember as we a
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Chapter 4 : Systems of Equations 4.
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ested in the point that is a soluti
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that parallel lines have the same s
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4.2 Systems of Equations - Substitu
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tion. This process is described and
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World View Note: French mathematici
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31) 2x + y = 2 3x + 7y = 14 33) x +
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6x +9y = 6 New second equation −
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Just as with graphing and substutio
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4.4 Systems of Equations - Three Va
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just the ordered pair (x, y). In th
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15x − 12y +9z = − 12 5x − 4y
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23) 3x + 3y − 2z = 13 6x + 2y −
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are given to use. This means someti
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− 0.5 − 0.5 c=17 We have c, num
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− 0.06x − 0.06y = − 240 Add e
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14) A bank contains 27 coins in dim
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4.6 Systems of Equations - Mixture
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Amount Part Total Start 40 3 120 Ad
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4(30 −n)+2.5n=105 Substitute into
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16) A certain grade of milk contain
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175
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5.1 Polynomials - Exponent Properti
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A quicker method to arrive at the s
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Example 210. Example 211. 7a 3 (2a
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5.2 Polynomials - Negative Exponent
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As we simplified our fraction we to
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5.2 Practice - Negative Exponents S
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Example 223. Convert 3.21 × 10 5 t
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5.3 Practice - Scientific Notation
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World View Note: Ada Lovelace in 18
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21) (3+b 4 )+(7+2b + b 4 ) 22) (1+6
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Just as we distribute a monomial th
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Example 243. (2x − 5)(4x 2 − 7x
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5.6 Polynomials - Multiply Special
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Be very careful when we are squarin
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5.7 Polynomials - Divide Polynomial
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3. Change the sign of the terms and
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Example 261. 2x 3 + 42 − 4x x + 3
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Chapter 6 : Factoring 6.1 Greatest
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ers using mental math, then we take
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6.1 Practice - Greatest Common Fact
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Example 273. 10ab + 15b + 4a+6 Spli
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example the binomials are (a + b) a
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6.3 Factoring - Trinomials where a
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As the past few examples illustrate
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6.3 Practice - Trinomials where a =
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Example 296. 10x 2 − 27x +5 Multi
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6.5 Factoring - Factoring Special P
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Example 306. 4x 2 + 20xy + 25y 2 Mu
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6.5 Practice - Factoring Special Pr
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Example 313. Example 314. Example 3
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6.7 Factoring - Solve by Factoring
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0=(2x +3)(2x − 3) One factor must
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6.7 Practice - Solve by Factoring S
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7.1 Rational Expressions - Reduce R
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However, if there is more than just
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41) 8m+16 20m − 12 43) 2x2 − 10
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Example 333. 25x2 24y4 · 9y8 55x7
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Simplify each expression. 1) 8x2 9
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7.3 Rational Expressions - Least Co
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As the above example illustrates, w
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7.4 Rational Expressions - Add & Su
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13 12 Our Solution The same process
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7.4 Practice - Add and Subtract Add
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2(12) 1(12) − 3 4 5(12) 1(12) + 6
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LCD =(x + 3)(x − 3) Multiply each
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23) 25) 27) 29) x − 4+ 9 2x +3 x
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(20)(9)=6x Multiply 180 = 6x Divide
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3.5 6 The man hasashadow of 3.5 fee
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33) Kali reduced the size of a pain
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LCD will be more involved. We will
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x=5 or − 2 Our Solution World Vie
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7.8 Rational Expressions - Dimensio
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� 435 1 �� 1 lbs = 16 435
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To focus on the process of conversi
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7.8 Practice - Dimensional Analysis
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Chapter 8 : Radicals 8.1 Square Roo
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√ process is being able to transl
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Simplify. √ 1) 245 √ 3) 36 √
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Example 384. 3√ 54 We are working
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8.3 Radicals - Adding Radicals Obje
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Simiplify 1) 2 5 √ + 2 5 √ + 2
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√ √ √ √ 4 50 + 6 30 − 8 3
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The previous example could have bee
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8.5 Radicals - Rationalize Denomina
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2x2 √ √ 10 − 6x 3x Our Soluti
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Example 408. 2 5 √ − 3 7 √ 5
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37) 5 2 √ + 3 √ 5+5 2 √ 38) 3
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1 (3) 4 Evaluate exponent 1 81 Our
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8.6 Practice - Rational Exponents W
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Example 420. ab2 3√ a2 4√ b Rew
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8.7 Practice - Radicals of Mixed In
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Example 426. Example 427. i 35 Divi
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√ Dividing with complex numbers a
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Simplify. 1) 3 − ( − 8+4i) 3) (
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Chapter 9 : Quadratics 9.1 Solving
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Example 444. + 1 +1 Add1to both sid
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2 − 2=0 Subtract 0=0 True! It wor
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Solve. √ 1) 2x + 3 √ 3) 6x −
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x4 4√ = ± 4√ 16 Simplify roots
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5√ 4x +1 = ± 3 Clear radical by
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9.3 Quadratics - Complete the Squar
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x2 + 5 25 x+ 3 36 � x + 5 �2 6
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Example 464. − 7 3 � 3 3 + 3 3
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9.4 Quadratics - Quadratic Formula
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x = Example 468. √ 30 ± 900 + 11
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9.4 Practice - Quadratic Formula So
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12x 2 − 17x + 6=0 Our Solution If
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9.5 Practice - Build Quadratics fro
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Example 479. a −2 − a −1 −
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12 = (x − 3) 2 or 13 =(x − 3) 2
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9.7 Quadratics - Rectangles Objecti
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Example 486. 2 Our Solution The len
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4(x 2 + 10x − 24) =0 Factor trino
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that require 24 square meters for p
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1(12x) 3 + 1(12x) x 1 1 5 + = 3 x 1
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6x +6(x − 1)=5x(x − 1) Reduce e
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16) A sink can be filled from the f
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These simultaneous product equation
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9.10 Quadratics - Revenue and Dista
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96n(n − 2) 96n(n − 2) +4n(n −
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upstream, the current will pull aga
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12) A pilot flying at a constant ra
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The above method to graph a parabol
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It is important to remember the gra
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Chapter 10 : Functions 10.1 Functio
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Example 504. Which of the following
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2x − 3 � 0 Set up an inequality
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Solve. 10.1 Practice - Function Not
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10.2 Functions - Operations on Func
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(x − 5)(x +1) (x − 5) Divide ou
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We can also evaluate a composition
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23) f(x) = x 2 − 5x g(x) = x + 5
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10.3 Functions - Inverse Functions
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(3x +6)(3 − 4x) 3 − 4x (4x+8)(3
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10.3 Practice - Inverse Functions S
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Example 530. 8 3x = 32 Rewrite 8 as
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10.4 Practice - Exponential Functio
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log3x =7 Identify base, 3, answer,
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10.5 Practice - Logarithmic Functio
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A = 25000(1.01625) 20 Evaluate expo
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The variable e is a constant simila
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j. All of the above compounded cont
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Using the diagram at right, find ea
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10.7 Practice - Trigonometric Funct
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21) 23) 25) 27) A B A 5 38 ◦ x x
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37) 39) A A 37.1 ◦ C x x C 13.1 4
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θ 17 From angle θ the given sides
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function to find the missing angle
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Find the measure of each angle indi
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27) 29) 14 A A θ C 15 7 C 14 θ B
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437
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7) 5 4 8) 4 3 9) 3 2 10) 8 3 11) 5
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54) 60v − 7 55) − 3x + 8x 2 56)
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45) 12 46) All real numbers 47) No
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15) wx 3 = 1458 16) h = 1.5 j 17) a
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1) B(4, − 3) C(1, 2) D( − 1, 4)
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1) y = 2x +5 2) y = − 6x +4 3) y
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23) x =4 24) y − 4 = 7 (x − 1)
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25) 5�x
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1) ( − 2, 4) 2) (2,4) 3) No solut
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27) 56, 144 28) 1.5, 3.5 29) 30 30)
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42) 1.2 × 10 6 5.4 1) 3 2) 7 3)
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5) 2x 3 + 4x 2 + x 2 6) 5p3 4 +4p2
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7) (b +8)(b + 4) 8) (b − 10)(b
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11) n(n − 1) 12) (5x +3)(x − 5)
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5) 3x2 2 6) 5p 2 7) 5m 8) 7 10 9) r
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7) 8) 5 24r 7x + 3y x 2 y 2 9) 15t+
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21) 0, 5 22) − 2, 5 3 23) 4,7 24)
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42) − 18xz 4x3yz3 � 4 8.3 1) 6
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16 3 √ + 4 5 √ 43 18) 19) 2 √
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8.8 1) 11 − 4i 2) − 4i 3) − 3
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√ √ 33) 4+i 39, 4 −i 39 34)
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9) ± 2, ± 4 10) 2,3, − 1 ±i 3
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1) 2) 3) 4) 5) 6) (-2,0) (0,-8) (-1
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11) 100 12) − 74 13) 1 5 14) 27 1
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10.5 1) 9 2 = 81 2) b −16 = a 3)
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21) 51.3 ◦ 22) 45 ◦ 23) 56.4
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