Chapter 7 - XYZ Custom Plus

Solving Equations7IntroductionImage © 2008 SanbornImage © 2008 DigitalGlobeChapter Outline7.1 The Distributive Propertyand Algebraic Expressions7.2 The Addition Property ofEquality7.3 The MultiplicationProperty of Equality7.4 Linear Equations in OneVariable7.5 Applications7.6 Evaluating Formulas7.7 Paired Data and theRectangular CoordinateSystemCentral Park in New York City was the first landscaped public park in the UnitedStates. More than 25 million people visit the park each year. Central Park is _1 2 mile wide and covers an area of 1.4 square miles. A person who jogs around theperimeter of the park will cover approximately 6.6 miles. Because the park canbe modeled with a rectangle, we can use these numbers to find the length of thepark. In fact, solving either of the two equations below will give us the length._1 2 x = 1.4 2x + 2 ⋅ 1 _= 6.62In this chapter, we will learn how to take the numbers and relationships givenin the paragraph above and translate them into equations like the ones above.Before we do that, we will learn how to solve these equations, and many othersas well.Comparing ParksCentral Park (New York)Stanley Park (Vancouver)843 acres1,000 acresRichmond Park (London)2,360 acresGriffith Park (Los Angeles)4,210 acresThe illustration here shows the area of Central Park compared to other prominentparks in large cities.409

Chapter PretestThe pretest below contains problems that are representative of the problems you will find in the chapter.Simplify.1. 4a − 1 − 5a + 8−a + 72. 2(5y − 6) + 4y14y − 12Solve each equation.3. a + 4 = −2−67. 2a + 1 = 5(a − 2) − 14. 5y + 9 − 4y = −7 + 11−55. _1 x = 3 15 6. −2x + 9 = 115−18. _x 3 − x _= 5 60449. Find the value of 3x − 4 when x = −2. −10 10. Is x = 5 a solution to the equation 6x − 28 = 1? No11. The sum of a number and 6 is −17. Find the number.−2312. If four times a number is decreased by 7, the result is25. Find the number. 8Getting Ready for Chapter 7The problems below review material covered previously that you need to know in order to be successful in Chapter 7.If you have any difficulty with the problems here, you need to go back and review before going on to Chapter 7.Simplify.1. −2 + 7 5 2. 180 − 45 135 3. −2 + (−4) −6 4. _5 8 + 3 _4 11 _ 85. (−4)(5) −20 6. 6 _−3 −2 7. 3 _2 (12) 18 8. − 5 _4 − 4 _5 19. − 5 _4 8 _15 − 2 _3 10. −4(−1) + 9 13 11. 1 _3 (15) + 2 7 12. 5 _(95 − 32) 35913. 3x + 7x 10x 14. −4(3x ) −12x 15. 4(x − 5) 4x − 20 16. 2 1 _2 x x17. Write in symbols: the sum of x and 2. x + 2 18. Find the perimeter. P = 8x3xx410Chapter 7 Solving Equations

The Distributive Propertyand Algebraic ExpressionsWe recall that the distributive property from Section 1.5 can be used to find thearea of a rectangle using two different methods.x 34x 347.1ObjectivesA Apply the distributive property toan expression.B Combine similar terms.C Find the value of an algebraicexpression.D Solve applications involvingcomplementary and supplementaryangles.Area = 4(x ) + 4(3) Area = 4(x + 3)= 4x + 12 = 4x + 12Since the areas are equal, the equation 4(x + 3) = 4(x) + 4(3) is a true statement.AThe Distributive PropertyExample 1Apply the distributive property to the expression:5(x + 3)Practice Problems1. Apply the distributive propertyto the expression 6(x + 4).SolutionDistributing the 5 over x and 3, we have5(x + 3) = 5(x) + 5(3) Distributive property= 5x + 15 MultiplicationRemember, 5x means “5 times x.”The distributive property can be applied to more complicated expressionsinvolving negative numbers.Example 2Multiply: −4(3x + 5)SolutionMultiplying both the 3x and the 5 by −4, we have2. Multiply: −3(2x + 4)−4(3x + 5) = −4(3x) + (−4)5 Distributive property= −12x + (−20) Multiplication= −12x − 20 Definition of subtractionNotice, first of all, that when we apply the distributive property here, we multiplythrough by −4. It is important to include the sign with the number when we usethe distributive property. Second, when we multiply −4 and 3x, the result is −12xbecause−4(3x) = (−4 ⋅ 3)x Associative property= −12x MultiplicationInstructor NoteA number of the problems shown inthe examples here have been solvedpreviously in the text. Even so, Ihave included them here to give studentsa good start in this chapter.Answers1. 6x + 24 2. −6x − 127.1 The Distributive Property and Algebraic Expressions411

412Chapter 7 Solving Equations3. Multiply: _1 (2x − 4)2Example 3Multiply: _1 (3x − 12)3Solution_1 3 (3x − 12) = 1 _3 (3x) − 1 _(12) Distributive property3= 1x − 12 _3 Simplify= x − 4 DivideWe can also use the distributive property to simplify expressions like 4x + 3x.Because multiplication is a commutative operation, we can rewrite the distributiveproperty like this:b ⋅ a + c ⋅ a = (b + c)aApplying the distributive property in this form to the expression 4x + 3x, wehave:4x + 3x = (4 + 3)x Distributive property= 7x AdditionNoteWe are using the wordterm in a differentsense here than wedid with fractions. (The terms of afraction are the numerator and thedenominator.)B Similar TermsRecall that expressions like 4x and 3x are called similar terms because the variableparts are the same. Some other examples of similar terms are 5y and −6y, and theterms 7a, −13a, _3 a. To simplify an algebraic expression (an expression that4involves both numbers and variables), we combine similar terms by applying thedistributive property. Table 1 reviews how we combine similar terms using thedistributive property.Table 1Original Apply Distribution SimplifiedExpression P property Expression4x + 3x = (4 + 3)x = 7x7a + a = (7 + 1)a = 8a−5x + 7x = (−5 + 7)x = 2x8y − y = (8 − 1)y = 7y−4a − 2a = (−4 − 2)a = −6a3x − 7x = (3 − 7)x = −4xAs you can see from the table, the distributive property can be applied to anycombination of positive and negative terms so long as they are similar terms.4. Simplify: 6x − 2 + 3x + 8Example 4Simplify: 5x − 2 + 3x + 7Solution We begin by changing subtraction to addition of the opposite andapplying the commutative property to rearrange the order of the terms. We wantsimilar terms to be written next to each other.5x − 2 + 3x + 7 = 5x + 3x + (−2) + 7 Commutative property= (5 + 3)x + (−2) + 7 Distributive property= 8x + 5 AdditionNotice that we take the negative sign in front of the 2 with the 2 when we rearrangeterms. How do we justify doing this?Answers3. x − 2 4. 9x + 6

7.1 The Distributive Property and Algebraic Expressions413Example 5Simplify: 3(4x + 5) + 6Solution We begin by distributing the 3 across the sum of 4x and 5. Then wecombine similar terms.5. Simplify: 2(4x + 3) + 73(4x + 5) + 6 = 12x + 15 + 6 Distributive property= 12x + 21 Add 15 and 6Example 6Simplify: 2(3x + 1) + 4(2x − 5)Solution Again, we apply the distributive property first; then we combine similarterms. Here is the solution showing only the essential steps:6. Simplify: 3(2x + 1) + 5(4x − 3)2(3x + 1) + 4(2x − 5) = 6x + 2 + 8x − 20 Distributive property= 14x − 18 Combine similar termsCThe Value of an Algebraic ExpressionAn expression such as 3x + 5 will take on different values depending on whatx is. If we were to let x equal 2, the expression 3x + 5 would become 11. On theother hand, if x is 10, the same expression has a value of 35:When x = 2 When x = 10the expression 3x + 5 the expression 3x + 5becomes 3(2) + 5 becomes 3(10) + 5= 6 + 5 = 30 + 5= 11 = 35ExamplesFind the value of each of the following expressions byreplacing the variable with the given number.Original Value of Value ofExpression the Variable the Expression7. 3x − 1 x = 2 3(2) − 1 = 6 − 1 = 58. 2x − 3 + 4x x = −1 2(−1) − 3 + 4(−1) = −2 − 3 + (−4) = −99. y 2 − 6y + 9 y = 4 4 2 − 6(4) + 9 = 16 − 24 + 9 = 17. Find the value of 4x − 7 whenx = 3.8. Find the value of2x − 5 + 6x when x = −2.9. Find the value ofy 2 − 10y + 25 when y = −2.Example 10Find the area of a 30-W solarpanel shown here with a length of 15 inches and awidth of 10 + 3x inches.Solution Previously we worked with area, so weknow that Area = (length) (width). Using the values forlength and width, we have:A = lwA = 15(10 + 3x) length = 15; width = 10 + 3x= 150 + 45x in 2 Distributive property15”10 + 3x10. Find the area of a 30-W solarpanel with a length of 25 cmand a width of 8 + 2x cm.Answers5. 8x + 13 6. 26x − 12 7. 58. −21 9. 49 10. 200 + 50x cm 2

414Chapter 7 Solving EquationsThe area of this solar panel is 150 + 45x square inches.facts from geometry AnglesAn angle is formed by two rays with the same endpoint. The common endpointis called the vertex of the angle, and the rays are called the sides of theangle.In Figure 1, angle θ (theta) is formed by the two rays OA and OB. The vertexof θ is O. Angle θ is also denoted as angle AOB, where the letter associatedwith the vertex is always the middle letter in the three letters used to denotethe angle.Degree Measure The angle formed by rotating a ray through one completerevolution about its endpoint (Figure 2) has a measure of 360 degrees,which we write as 360°.BOAOne complete revolution = 360Figure 1Figure 2One degree of angle measure, written 1°, is _1 of a complete rotation of a ray360about its endpoint; there are 360° in one full rotation. (The number 360 wasdecided upon by early civilizations because it was believed that the Earth wasat the center of the universe and the Sun would rotate once around the Earthevery 360 days.) Similarly, 180° is half of a complete rotation, and 90° is aquarter of a full rotation. Angles that measure 90° are called right angles, andangles that mea sure 180° are called straight angles. If an angle measuresbetween 0° and 90° it is called an acute angle, and an angle that measuresbetween 90° and 180° is an obtuse angle. Figure 3 illustrates further.90180Right angleStraight angleAcute angleObtuse angleDFigure 3Complementary Angles and Supplementary Angles If two anglesadd up to 90°, we call them complementary angles, and each is called thecomplement of the other. If two angles have a sum of 180°, we call themsupplementary angles, and each is called the supplement of the other. Figure4 illustrates the relationship between angles that are complementary andangles that are supplementary.Complementary angles: 90°Supplementary angles: 180°Figure 4

7.1 The Distributive Property and Algebraic Expressions415Example 11Find x in each of the following diagrams.a. b.x30°x45°Supplementary angles11. Find x in each of the followingdiagrams.a.x45°SolutionComplementary anglesWe use subtraction to find each angle.a. Because the two angles are complementary, we can find x bysubtracting 30° from 90°:x = 90° − 30° = 60°Complementary anglesb.x60°We say 30° and 60° are complementary angles. The complementof 30° is 60°.Supplementary anglesb. The two angles in the diagram are supplementary. To find x, wesubtract 45° from 180°:x = 180° − 45° = 135°We say 45° and 135° are supplementary angles. The supplementof 45° is 135°.U s i n gCalculatorsT e c h n o l o g yWhen we think of technology, we think of computers and calculators. However,some simpler devices are also in the category of technology, because they helpus do things that would be difficult to do without them. The protractor belowcan be used to draw and measure angles. In the diagram below, the protractoris being used to measure an angle of 120°. It can also be used to draw anglesof any size.4014013050120601107080100909010080110701206013050140401503016020120°15030160201701017010018018000 1 2 3 4 5 6 7 8 9 10 11If you have a protractor, use it to draw the following angles: 30°, 45°, 60°, 120°,135°, and 150°. Then imagine how you would draw these angles without aprotractor.Answer11. a. 45° b. 120°

416Chapter 7 Solving EquationsGetting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is the distributive property?2. What property allows 5(x + 3) to be rewritten as 5x + 5(3)?3. What property allows 3x + 4x to be rewritten as 7x?4. True or false? The expression 3x means 3 multiplied by x.

7.1 Problem Set417Problem Set 7.1A For review, use the distributive property to combine each of the following pairs of similar terms. [Examples 1–3]1. 2x + 8x10x2. 3x + 7x10x3. −4y + 5yy4. −3y + 10y7y5. 4a − a3a6. 9a − a8a7. 8(x + 2)8x + 168. 8(x − 2)8x − 169. 2(3a + 7)6a + 1410. 5(3a + 2)15a + 1011. _1 (3x + 6)3x + 212. _1 (2x + 4)2x + 2B Simplify the following expressions by combining similar terms. In some cases the order of the terms must be rearrangedfirst by using the commutative property. [Examples 4–6]13. 4x + 2x + 3 + 86x + 1114. 7x + 5x + 2 + 912x + 1115. 7x − 5x + 6 − 42x + 216. 10x − 7x + 9 − 63x + 317. −2a + a + 7 + 5−a + 1218. −8a + 3a + 12 + 1−5a + 1319. 6y − 2y − 5 + 14y − 420. 4y − 3y − 7 + 2y − 521. 4x + 2x − 8x + 4−2x + 422. 6x + 5x − 12x + 6−x + 623. 9x − x − 5 − 18x − 624. 2x − x − 3 − 8x − 1125. 2a + 4 + 3a + 55a + 926. 9a + 1 + 2a + 611a + 727. 3x + 2 − 4x + 1−x + 328. 7x + 5 − 2x + 65x + 1129. 12y + 3 + 5y17y + 330. 8y + 1 + 6y14y + 131. 4a − 3 − 5a + 2aa − 332. 6a − 4 − 2a + 6a10a − 4

418Chapter 7 Solving EquationsApply the distributive property to each expression and then simplify.33. 2(3x + 4) + 834. 2(5x + 1) + 1035. 5(2x − 3) + 436. 6(4x − 2) + 76x + 1610x + 1210x − 1124x − 537. 8(2y + 4) + 3y38. 2(5y + 1) + 2y39. 6(4y − 3) + 6y40. 5(2y − 6) + 4y19y + 3212y + 230y − 1814y − 3041. 2(x + 3) + 4(x + 2)42. 3(x + 1) + 2(x + 5)43. 3(2a + 4) + 7(3a − 1)44. 7(2a + 2) + 4(5a − 1)6x + 145x + 1327a + 534a + 10C Find the value of each of the following expressions when x = 5. [Examples 7–9]45. 2x + 41446. 3x + 21747. 7x − 82748. 8x − 93149. −4x + 1−1950. −3x + 7−851. −8 + 3x752. −7 + 2x3Find the value of each of the following expressions when a = −2.53. 2a + 5154. 3a + 4−255. −7a + 41856. −9a + 32157. −a + 101258. −a + 81059. −4 + 3a−1060. −6 + 5a−16

7.1 Problem Set419Find the value of each of the following expressions when x = 3. You may substitute 3 for x in each expression the way it iswritten, or you may simplify each expression first and then substitute 3 for x.61. 3x + 5x + 42862. 6x + 8x + 74963. 9x + x + 3 + 74064. 5x + 3x + 2 + 43065. 4x + 3 + 2x + 52666. 7x + 6 + 2x + 94267. 3x − 8 + 2x − 3468. 7x − 2 + 4x − 130Find the value of each of 12x − 3 for each of the following values of x.69. _1 2 370. _1 3 171. _1 4 072. _1 6 −173. _3 2 1574. _2 3 575. _3 4 676. _5 6 7Use the distributive property to write two equivalent expressions for the area of each figure.77. 6(x + 4) = 6x + 24 78. 7(x + 5) = 7x + 3567x 4x 5Write an expression for the perimeter of each figure.79. 4x + 4 80. 3x 210x + 2Squarex 1Rectangle 2x 181. 3x 110x − 4 82. 13x + 64x 1 4x 1ParallelogramTriangle2x 35x 4

420Chapter 7 Solving EquationsDApplying the Concepts83. Buildings This Google Earthimage shows the LeaningTower of Pisa. Mostbuildings stand at a rightangle, but the tower issinking on one side. Theangle of inclination is theangle between the verticaland the tower. If theangle between the towerand the ground is 85°what is the angle ofinclination? 5°Image © 2008 DigitalGlobe,Image NASA84. Geometry This Google Earthimage shows the Pentagon.The interior angles of a regularpentagon are all thesame and sum to 540°. Findthe size of each angle.108°Image © 2008 SanbornFind x in each figure and decide if the two angles are complementary or supplementary. [Example 11]85.86.xx35°145°; supplementary angles35°55°; complementary angles87.88.xx70°70°20°; complementary angles110°; supplementary angles89. Luke earns $12 per hour working as a math tutor.We can express the amount he earns each week forworking x hours with the expression 12x. Indicatewith a yes or no, which of the following could be oneof Luke’s paychecks. If you answer no, explain youranswer.a. $60 for working five hoursYesb. $100 for working nine hoursNo, he should earn $108 for working 9 hoursc. $80 for working seven hoursNo, he should earn $84 for working 7 hoursd. $168 for working 14 hoursYes90. Kelly earns $15 per hour working as a graphic designer.We can express the amount she earns each week forworking x hours with the expression 15x. Indicate with ayes or no which of the following could be one of Kelly’spaychecks. If you answer no, explain your answer.a. $75 for working five hoursYesb. $125 for working nine hoursNo, she should earn $135 for working 9 hoursc. $90 for working six hoursYesd. $500 for working 35 hoursNo, she should earn $525 for working 35 hours

7.1 Problem Set42191. Temperature and Altitude On a certain day, the temperatureon the ground is 72 degrees Fahrenheit, and thetemperature at an altitude of A feet above the groundis found from the expression 72 − _A . Find the temperatureat the following altitudes.300a. 12,000 feet b. 15,000 feet c. 27,000 feet32°F 22°F −18°F92. Perimeter of a Rectangle As you know, the expression2l + 2w gives the perimeter of a rectangle with length land width w. The garden below has a width of 3_1 2 feetand a length of 8 feet. What is the length of the fencethat surrounds the garden?23 ft3.5 ftA72˚F8 ft93. Cost of Bottled Water A water bottling company charges$7 per month for their water dispenser and $2 for eachgallon of water delivered. If you have g gallons of waterdelivered in a month, then the expression 7 + 2g givesthe amount of your bill for that month. Find the monthlybill for each of the following deliveries.a. 10 gallons b. 20 gallons$27 $4794. Cellular Phone Rates A cellular phone company charges$35 per month plus 25 cents for each minute, or fractionof a minute, that you use one of their cellular phones.3500 + 25tThe expression _ gives the amount of money,100in dollars, you will pay for using one of their phones fort minutes a month. Find the monthly bill for using one oftheir phones:a. 20 minutes in a month b. 40 minutes in a month$40 $45MONTHLY BILL234 5th StreetGlendora, CA 91740Water dispenserGallons of water18DUE 07/23/10$7.00$16.00$23.00Cell Phone CompanyGrover Beach, CAAugust 2010 DUE 08/15/10Monthly Access per Phone:Charges:$0.25/minute150$35.00$12.50$47.50Getting Ready for the Next SectionAdd.95. 4 + (−4)096. 2 + (−2)097. −2 + (−4)−698. −2 + (−5)−799. −5 + 2−3100. −3 + 129101. 5 _8 + 3 _4 102. _5 6 + 2 _3 _103. − 38 _ 3 2 0_ 114 + 3 _4 104. − 2 _3 + 2 _3 0

422Chapter 7 Solving EquationsSimplify.105. x + 0x106. y + 0y107. y + 4 − 6y − 2108. y + 6 − 2y + 4Maintaining Your SkillsGive the opposite of each number.109. 9−9110. 12−12111. −66112. −55Problems 113–118 review material we covered in Chapter 1. Match each statement on the left with the property that justifiesit on the right.113. 2(6 + 5) = 2(6) + 2(5)a114. 3 + (4 + 1) = (3 + 4) + 1b115. x + 5 = 5 + xc116. (a + 3) + 2 = a + (3 + 2)b117. (x + 5) + 1 = 1 + (x + 5)c118. (a + 4) + 2 = (4 + 2) + ada. Distributive propertyb. Associative propertyc. Commutative propertyd. Commutative and associative propertiesPerform the indicated operation.119. − 5 _4 8 _15 _− 2 3 _− 8 5 18_120. − 4 3 _6 5 121. 12 ÷ _2 3 122. 6 ÷ _3 5 10123. 2 _3 − 3 _4 124. _3_− 1 12 5 − 5 _8 − 1 _40

Introduction . . .The Addition Property of EqualityPreviously we defined complementary angles as two angles whose sum is 90°. IfA and B are complementary angles, then7.2ObjectivesA Identify a solution to an equation.B Use the addition property ofequality to solve linear equations.A + B = 90°ABComplementary anglesIf we know that A = 30°, then we can substitute 30° for A in the formula above toobtain the equation30° + B = 90°In this section we will learn how to solve equations like this one that involveaddition and subtraction with one variable.In general, solving an equation involves finding all replacements for the variablethat make the equation a true statement.A Solutions to EquationsDefinitionA solution for an equation is a number that when used in place of thevariable makes the equation a true statement.NoteAlthough an equationmay have many solutions,the equationswe work with in the first part ofthis chapter will always have asingle solution.For example, the equation x + 3 = 7 has as its solution the number 4, becausereplacing x with 4 in the equation gives a true statement:When x = 4the equation x + 3 = 7becomes 4 + 3 = 7or 7 = 7 A true statementExample 1Is x = 5 the solution to the equation 3x + 2 = 17?Solution To see if it is, we replace x with 5 in the equation and find out if theresult is a true statement:Practice Problems1. Show that x = 3 is the solutionto the equation 5x − 4 = 11.When x = 5the equation 3x + 2 = 17becomes 3(5) + 2 = 1715 + 2 = 1717 = 17 A true statementBecause the result is a true statement, we can conclude that x = 5 is the solutionto 3x + 2 = 17.Answer1. See solutions section.7.2 The Addition Property of Equality423

424Chapter 7 Solving Equations2. Is a = −3 the solution to theequation 6a − 3 = 2a + 4?Example 2Is a = −2 the solution to the equation 7a + 4 = 3a − 2?Solution When a = −2the equation 7a + 4 = 3a − 2becomes 7(−2) + 4 = 3(−2) − 2−14 + 4 = −6 − 2−10 = −8A false statementBecause the result is a false statement, we must conclude that a = −2 is not thesolution to the equation 7a + 4 = 3a − 2.BAddition Property of EqualityInstructor NoteAs you know, the addition propertyof equality can be extended to includesubtracting the same numberfrom both sides of an equation. Idon’t mention this in class, nor doI work examples using subtraction,because I have found that my studentswill make fewer mistakes ifthey think in terms of addition. However,in the next section we coverthe multiplication property of equality,and we do extend that propertyto include dividing both sides of anequation by the same number.We want to develop a process for solving equations with one variable. The mostimportant property needed for solving the equations in this section is called theaddition property of equality. The formal definition looks like this:Addition Property of EqualityLet A, B, and C represent algebraic expressions.If A = Bthen A + C = B + CIn words: Adding the same quantity to both sides of an equation never changesthe solution to the equation.This property is extremely useful in solving equations. Our goal in solving equationsis to isolate the variable on one side of the equation. We want to end upwith an equation of the formx = a numberTo do so we use the addition property of equality. Remember to follow this basicrule of algebra: Whatever is done to one side of an equation must be done to theother side in order to preserve the equality.3. Solve for x: x + 5 = −2NoteWith some of theequations in this section,you will be ableto see the solution just by lookingat the equation. But it is importantthat you show all the steps usedto solve the equations anyway. Theequations you come across in thefuture will not be as easy to solve,so you should learn the stepsinvolved very well.Example 3Solve for x: x + 4 = −2Solution We want to isolate x on one side of the equation. If we add −4 toboth sides, the left side will be x + 4 + (−4), which is x + 0 or just x.x + 4 = −2x + 4 + (−4) = −2 + (−4) Add −4 to both sidesx + 0 = −6Additionx = −6x + 0 = xThe solution is −6. We can check it if we want to by replacing x with −6 in theoriginal equation:Whenx = −6the equation x + 4 = −2becomes −6 + 4 = −2−2 = −2 A true statementAnswers2. No 3. −7

7.2 The Addition Property of Equality425Example 4Solve for a: a − 3 = 5Solution a − 3 = 5a − 3 + 3 = 5 + 3 Add 3 to both sidesa + 0 = 8Additiona = 8a + 0 = a4. Solve for a: a − 2 = 7The solution to a − 3 = 5 is a = 8.Example 5Solve for y: y + 4 − 6 = 7 − 1Solution Before we apply the addition property of equality, we must simplifyeach side of the equation as much as possible:5. Solve for y: y + 6 − 2 = 8 − 9y + 4 − 6 = 7 − 1y − 2 = 6y − 2 + 2 = 6 + 2y + 0 = 8y = 8Simplify each sideAdd 2 to both sidesAdditiony + 0 = yExample 6Solve for x: 3x − 2 − 2x = 4 − 9SolutionSimplifying each side as much as possible, we have6. Solve for x: 5x − 3 − 4x = 4 − 73x − 2 − 2x = 4 − 9x − 2 = −5x − 2 + 2 = −5 + 2x + 0 = −3x = −33x − 2x = xAdd 2 to both sidesAdditionx + 0 = xExample 7Solve for x: −3 − 6 = x + 4Solution The variable appears on the right side of the equation in this problem.This makes no difference; we can isolate x on either side of the equation. Wecan leave it on the right side if we like:7. Solve for x: −5 − 7 = x + 2−3 − 6 = x + 4−9 = x + 4−9 + (−4) = x + 4 + (−4)−13 = x + 0−13 = xSimplify the left sideAdd −4 to both sidesAdditionx + 0 = xThe statement −13 = x is equivalent to the statement x = −13. In either case thesolution to our equation is −13.Example 8Solve: a − 3 _4 = 5 _8 SolutionTo isolate a we add _3 to each side:4a − 3 _4 = 5 _8 a − 3 _4 + 3 _4 = 5 _8 + 3 _4 a = 11 _8 When solving equations we will leave answers like _11 as improper fractions,8rather than change them to mixed numbers.8. Solve: a − 2 _3 = 5 _6 Instructor NoteI have my students write solutionsto equations as improper fractionsrather than as mixed numbers.Answers4. 9 5. −5 6. 0 7. −148. _3 2

426Chapter 7 Solving Equations9. Solve: 5(3a − 4) − 14a = 25Example 9Solve: 4(2a − 3) − 7a = 2 − 5.Solution We must begin by applying the distributive property to separateterms on the left side of the equation. Following that, we combine similar termsand then apply the addition property of equality.4(2a − 3) − 7a = 2 − 5 Original equation8a − 12 − 7a = 2 − 5 Distributive propertya − 12 = −3 Simplify each sidea − 12 + 12 = −3 + 12 Add 12 to each sidea = 9AdditionA Note on SubtractionAlthough the addition property of equality is stated for addition only, we can subtractthe same number from both sides of an equation as well. Because subtractionis defined as addition of the opposite, subtracting the same quantity fromboth sides of an equation will not change the solution. If we were to solve theequation in Example 3 using subtraction instead of addition, the steps would looklike this:x + 4 = −2x + 4 − 4 = −2 − 4x = −6Original equationSubtract 4 from each sideSubtractionIn my experience teaching algebra, I find that students make fewer mistakes ifthey think in terms of addition rather than subtraction. So, you are probably betteroff if you continue to use the addition property just the way we have used it in theexamples in this section. But, if you are curious as to whether you can subtractthe same number from both sides of an equation, the answer is yes.Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences. An answer of true or false shouldbe accompanied by a sentence explaining why the answer is true orfalse.1. What is a solution to an equation?2. True or false? According to the addition property of equality, adding thesame value to both sides of an equation will never change the solutionto the equation.3. Show that x = 5 is a solution to the equation 3x + 2 = 17 without solvingthe equation.4. True or false? The equations below have the same solution.Equation 1: 7x + 5 = 19Equation 2: 7x + 5 + 3 = 19 + 3Answer9. 45

7.2 Problem Set427Problem Set 7.2A Check to see if the number to the right of each of the following equations is the solution to the equation. [Examples 1, 2]1. 2x + 1 = 5; 2Yes2. 4x + 3 = 7; 1Yes3. 3x + 4 = 19; 5Yes4. 3x + 8 = 14; 2Yes5. 2x − 4 = 2; 4No6. 5x − 6 = 9; 3Yes7. 2x + 1 = 3x + 3; −2Yes8. 4x + 5 = 2x − 1; −6No9. x − 4 = 2x + 1; −4No10. x − 8 = 3x + 2; −5YesB Solve each equation. [Examples 3, 4, 8]11. x + 2 = 8612. x + 3 = 5213. x − 4 = 71114. x − 6 = 2815. a + 9 = −6−1516. a + 3 = −1−417. x − 5 = −4118. x − 8 = −3519. y − 3 = −6−320. y − 5 = −1421. a + 1 _3 = −2 _3 −122. a + 1 _4 = −3 _4 −123. x − _3 5 = 4 _5 24. x − _7 8 = 3 _8 25. y + 7.3 = −2.7_7 5 _ 5 4 −1026. y + 8.2 = −2.8−11

428Chapter 7 Solving EquationsB Simplify each side of the following equations before applying the addition property. [Examples 5–7]27. x + 4 − 7 = 3 − 10−428. x + 6 − 2 = 5 − 12−1129. x − 6 + 4 = −3 − 2−330. x − 8 + 2 = −7 − 1−231. 3 − 5 = a − 4232. 2 − 6 = a − 1−333. 3a + 7 − 2a = 1−634. 5a + 6 − 4a = 4−235. 6a − 2 − 5a = −9 + 1−636. 7a − 6 − 6a = −3 + 1437. 8 − 5 = 3x − 2x + 4−138. 10 − 6 = 8x − 7x + 6−2B The following equations contain parentheses. Apply the distributive property to remove the parentheses, then simplifyeach side before using the addition property of equality. [Example 9]39. 2(x + 3) − x = 4−240. 5(x + 1) − 4x = 2−341. −3(x − 4) + 4x = 3 − 7−1642. −2(x − 5) + 3x = 4 − 9−1543. 5(2a + 1) − 9a = 8 − 6−344. 4(2a − 1) − 7a = 9 − 5845. −(x + 3) + 2x − 1 = 61046. −(x − 7) + 2x − 8 = 45Find the value of x for each of the figures, given the perimeter.47. P = 36 x = 4 48. P = 30 x = 1310 105xx 121249. P = 16 x = 12 50. P = 60 x = 245 510x26x 6

7.2 Problem Set429Applying the ConceptsTemperature The chart shows the temperatures for some of the world’shottest places. To convert from Celsius to Kelvin we use the formulay = x + 273, where y is the temperature in Kelvin and x is the temperaturein Celsius. Use the formula to answer Questions 51 and 52.51. The hottest temperature in Al’Aziziyah was 331 Kelvin. Convertthis to Celsius.58° Celsius52. The hottest temperature in Kebili, Tunisia, was 328 Kelvin.Convert this to Celsius.55° CelsiusHeating Up137˚F Al’Aziziyah, Libya134˚F Greenland Ranch, Death Valley, United States131˚F Ghudamis, Libya131˚F Kebili, Tunisia130˚F Tombouctou, MaliSource: Aneki.com16014012010080604053. Geometry Two angles are complementary angles. Ifone of the angles is 23°, then solving the equationx + 23° = 90° will give you the other angle. Solve theequation.67°54. Geometry Two angles are supplementary angles. Ifone of the angles is 23°, then solving the equationx + 23° = 180° will give you the other angle. Solve theequation.157°ABComplementary angles55. Theater Tickets The El Portal Center for the Arts inNorth Hollywood, California, holds a maximum of 400people. The two balconies hold 86 and 89 people each;the rest of the seats are at the stage level. Solving theequation x + 86 + 89 = 400 will give you the numberof seats on the stage level.a. Solve the equation for x.225b. If tickets on the stage level are $30 each, and ticketsin either balcony are $25 each, what is the maximumamount of money the theater can bring in fora show?$11,12556. Geometry The sum of the angles in the triangle on theswing set is 180°. Use this fact to write an equation containingx. Then solve the equation.x + 67° + 67° = 180°; 46°x67° 67°El PortalCENTER FOR THE ARTSTICKETStage LevelSeats$30 00TICKETBalconySeats$25 00

430Chapter 7 Solving EquationsGetting Ready for the Next SectionFind the reciprocal of each number.57. 458. 359. _1_1 4 _ 1 3 2 260. _1 3 361. _2 3 62. _3 5 _ 3 2 _ 5 3 Multiply.63. 2 ⋅ 1 _2 64. 1 _4 ⋅ 465. − 1 _3 (−3)66. − 1 _4 (−4)111167. 3 _2 2 _3 68. 5 _3 3 _5 69. − 5 _4 − 4 _5 70. − 4 _3 − 3 _4 1111Simplify.71. 1 ⋅ x72. 1 ⋅ a73. 4x − 11 + 3x74. 2x − 11 + 3xxa7x − 115x − 11Maintaining Your Skills75. 3 _2 + 5 _10 276. _1 3 + 4 _12 77. _2 7 + 1 _14 78. _3 8 + 1 _16 _ 2 3 _ 514 _ 716 79. _1 3 − 2 _5 80. _3 4 − 3 _7 81. _1 6 − 4 _3 82. _2_− 1 15 _ 928 _− 7 6 5 − 5 _10 − 1 _10 Translating Translate each of the following into an equation, and then solve the equation.83. The sum of x and 12 is 30.Equation: x + 12 = 30; solution: 1884. The difference of x and 12 is 30.Equation: x − 12 = 30; solution: 4285. The difference of 8 and 5 is equal to the sum of x and 7.Equation: 8 − 5 = x + 7; solution: −486. The sum of 8 and 5 is equal to the difference of x and 7.Equation: 8 + 5 = x − 7; solution: 20

The Multiplication Property of EqualityIn this section we will continue to solve equations in one variable. We will againuse the addition property of equality, but we will also use another property—themultiplication property of equality—to solve the equations in this section. We willstate the multiplication property of equality and then see how it is used by lookingat some examples.The most popular Internet video download of all time was a Star Wars movietrailer. The video was compressed so it would be small enough for people todownload over the Internet. In movie theaters, a film plays at 24 frames per second.Over the Internet, that number is sometimes cut in half, to 12 frames persecond, to make the file size smaller.We can use the equation 240 = _x to find the number of total frames, x, in a12240-second movie clip that plays at 12 frames per second.7.3ObjectivesA Use the multiplication property ofequality to solve equations.AMultiplication Property of EqualityMultiplication Property of EqualityLet A, B, and C represent algebraic expressions, with C not equal to 0.IfthenA = BAC = BCIn words: Multiplying both sides of an equation by the same nonzero quantitynever changes the solution to the equation.Now, because division is defined as multiplication by the reciprocal, we are alsofree to divide both sides of an equation by the same nonzero quantity and alwaysbe sure we have not changed the solution to the equation.Example 1Solve for x: _1 2 x = 3Solution Our goal here is the same as it was in Section 4.2. We want to isolatex (that is, 1x) on one side of the equation. We have _1 x on the left side. If we2multiply both sides by 2, we will have 1x on the left side. Here is how it looks:Practice Problems1. Solve for x: 1 _3 x = 5 1 _2 x = 32 1 _2 x = 2(3) Multiply both sides by 2x = 6MultiplicationTo see why 2 1 _2 x is equivalent to x, we use the associative property:2 1 _2 x = 2 ⋅ 1 _2 x Associative property= 1 ⋅ x 2 ⋅ 1 _2 = 1Instructor NoteAs you can see, we do allow for dividingboth sides of an equation bythe same number. Example 5, alongwith the note in the margin next toit, justifies using division to solveequations. However, you may wantto give some further explanation inclass.= x 1 ⋅ x = xAlthough we will not show this step when solving problems, it is implied.7.3 The Multiplication Property of EqualityAnswer1. 15431

432Chapter 7 Solving Equations2. Solve for a: 1 _5 a + 3 = 7Example 2Solve for a: 1 _3 a + 2 = 7SolutionWe begin by adding −2 to both sides to get _1 a by itself. We then3multiply by 3 to solve for a. 1 _3 a + 2 = 7_1 a + 2 + (−2) = 7 + (−2) Add −2 to both sides31_a = 5 Addition33 ⋅ _1 a = 3 ⋅ 53Multiply both sides by 3a = 15 MultiplicationWe can check our solution to see that it is correct:When a = 15the equationbecomes 1 _3 a + 2 = 7 1 _3 (15) + 2 = 75 + 2 = 77 = 7 A true statement3. Solve for y: 3 _5 y = 6Example 3Solve for y: 2 _Solutiony = 123In this case we multiply each side of the equation by the reciprocalof _2 3 , which is 3 _2 . _2 y = 123 3 _2 2 _3 y = 3 _2 (12)y = 18The solution checks because _2 of 18 is 12.3Instructor NoteWhen I work a problem similarto the one shown in Example 3, Ialways show a few extra steps in orderto point out how the associativeproperty is used to simplify the leftside of the equation to just y.Note The reciprocal of a negative number is also a negative number. Remember,reciprocals are two numbers that have a product of 1. Since 1 is a positive number,any two numbers we multiply to get 1 must both have the same sign. Hereare some negative numbers and their reciprocals:The reciprocal of −2 is − 1 _2 .The reciprocal of −7 is − 1 _7 ._The reciprocal of − 1 is −3.3The reciprocal of − 3 _4 is −4 _3 .Answers2. 20 3. 10The reciprocal of − 9 _5 is −5 _9 .

7.3 The Multiplication Property of Equality433Example 4Solve for x: − 4 _5 x = 8 _15 SolutionThe reciprocal of − 4 _5 is −5 _4 .4. Solve for x: − 3 _4 x = 6 _5 − 4 _5 x = 8 _15 − 5 _4 − 4 _5 x = − 5 _4 8 _15 x = − 2 _3 Many times, it is convenient to divide both sides by a nonzero number to solvean equation, as the next example shows.Example 5Solve for x: 4x = −20Solution If we divide both sides by 4, the left side will be just x, which is whatwe want. It is okay to divide both sides by 4 because division by 4 is equivalent tomultiplication by _1 , and the multiplication property of equality states that we can4multiply both sides by any number so long as it isn’t 0.4x = −20_4x = −20_ Divide both sides by 44 4x = −5 DivisionBecause 4x means “4 times x,” the factors in the numerator of _4x are 4 and x.4Because the factor 4 is common to the numerator and the denominator, we divideit out to get just x.5. Solve for x: 6x = −42NoteIf we multiply eachside by 1_ 4, the solutionlooks like this: 1 _4 (4x) = 1 _4 (−20) 1 _4 ⋅ 4 x = −51x = −5x = −5Example 6Solve for x: −3x + 7 = −5SolutionWe begin by adding −7 to both sides to reduce the left side to −3x.−3x + 7 = −5−3x + 7 + (−7) = −5 + (−7) Add −7 to both sides−3x = −12Addition_ −3x= −12_−3 −3 Divide both sides by −3x = 4Division6. Solve for x: −5x + 6 = −14With more complicated equations we simplify each side separately beforeapplying the addition or multiplication properties of equality. The examples belowillustrate.Example 7Solve for x: 5x − 8x + 3 = 4 − 10Solutionusual.We combine similar terms to simplify each side and then solve as5x − 8x + 3 = 4 − 10−3x + 3 = −6−3x + 3 + (−3) = −6 + (−3)−3x = −9_ −3x= −9_−3 −3 x = 3Simplify each sideAdd −3 to both sidesAdditionDivide both sides by −3Division7. Solve for x: 3x − 7x + 5 = 3 − 18Answers_4. − 8 5. −7 6. 4 7. 55

434Chapter 7 Solving Equations8. Solve for x:−5 + 4 = 2x − 11 + 3xExample 8Solve for x: −8 + 11 = 4x − 11 + 3xSolutionWe begin by simplifying each side separately.−8 + 11 = 4x − 11 + 3x3 = 7x − 11 Simplify both sides3 + 11 = 7x − 11 + 11 Add 11 to both sides14 = 7x Addition_14 = 7x_ Divide both sides by 77 72 = x or x = 2Again, it makes no difference which side of the equation x ends up on, so long asit is just one x.Common MistakesBefore we end this section, we should mention a very common mistake madeby students when they first begin to solve equations. It involves trying to subtractaway the number in front of the variable—like this:7x = 217x − 7 = 21 − 7 Add −7 to both sidesx = 14 m88888 MistakeThe mistake is not in trying to subtract 7 from both sides of the equation. Themistake occurs when we say 7x − 7 = x. It just isn’t true. We can add and subtractonly similar terms. The terms 7x and 7 are not similar, because one containsx and the other doesn’t. The correct way to do the problem is like this:7x = 21_7x = 21_ Divide both sides by 77 7x = 3 DivisionGetting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. True or false? Multiplying both sides of an equation by the same nonzeroquantity will never change the solution to the equation.2. If we were to multiply the right side of an equation by 2, then the leftside should be multiplied by .3. Dividing both sides of the equation 4x = −20 by 4 is the same as multiplyingboth sides by what number?Answer8. 2

7.3 Problem Set435Problem Set 7.3AUse the multiplication property of equality to solve each of the following equations. In each case, show all the steps.[Examples 1, 3–5]1. 1 _4 x = 22. 1 _3 x = 73. _1 x = −324. _1 x = −65821−6−305. − 1 _3 x = 26. − 1 _3 x = 5_7. − 1 x = −16_8. − 1 x = −42−6−15689. _3 y = 1241610. _2 y = 1832711. 3a = 481612. 2a = 2814_13. − 3 5 x = 9 _10 _14. − 4 5 x = − _ 815 15. 5x = −35−7_− 3 2 _ 2 3 16. 7x = −35−517. −8y = 64−818. −9y = 27−319. −7x = −42620. −6x = −427

436Chapter 7 Solving EquationsA Using the addition property of equality first, solve each of the following equations. [Examples 2, 6]21. 3x − 1 = 5222. 2x + 4 = 6123. −4a + 3 = −9324. −5a + 10 = 50−825. 6x − 5 = 1926. 7x − 5 = 3027. _1 a + 3 = −5328. _1 a + 2 = −7245−24−1829. − 1 _4 a + 5 = 230. − 1 _5 a + 3 = 731. 2x − 4 = −2032. 3x − 5 = −2612−20−8−733. 2 _3 x − 4 = 634. 3 _4 x − 2 = 735. −11a + 4 = −2936. −12a + 1 = −4715123437. −3y − 2 = 1−138. −2y − 8 = 2−539. −2x − 5 = −7140. −3x − 6 = −3610A Simplify each side of the following equations first, then solve. [Examples 7, 8]41. 2x + 3x − 5 = 7 + 3342. 4x + 5x − 8 = 6 + 4243. 4x − 7 + 2x = 9 − 10144. 5x − 6 + 3x = −6 − 8−145. 3a + 2a + a = 7 − 13−146. 8a − 6a + a = 8 − 14−247. 5x + 4x + 3x = 4 − 848. 4x + 8x − 2x = 15 − 10_− 1 3 _ 1 2 49. 5 − 18 = 3y − 2y + 1−1450. 7 − 16 = 4y − 3y + 2−11

7.3 Problem Set437Find the value of x for each of the figures, given the perimeter.51. P = 72 x = 9 52. P = 96 x = 82x3x53. P = 80 x = 8 54. P = 64 x = 43x5x2x3xApplying the Concepts55. Cars The chart shows the fastest cars in America. Toconvert miles per hour to feet per second we use theformula y = _15 x where x is the car’s speed in feet per22second and y is the speed in miles per hour. Find thespeed of the Ford GT in feet per second. Round to thenearest tenth.300.7 ft/secReady for the RacesFord GT 205 mphEvans 487 210 mphSaleen S7 Twin Turbo 260 mphSSC Ultimate Aero 273 mph56. Mountains The map shows the heights of the tallestmountains in the world. To convert the heights of themountains into miles, we use the formula y = 5,280x,where y is in feet and x is in miles. Find the height of K 2in miles. Round to the nearest tenth of a mile.5.3 milesThe Greatest HeightsPAKISTANNEPALINDIASource: Forrester Research, 2005K2 28,238 ftMount Everest 29,035 ftKangchenjunga 28,208 ftCHINASource: Forbes.com

438Chapter 7 Solving Equations57. MP 3 Players Southwest Electronics tracked the numberof MP 3 players it sold each month for a year. The storemanager found that when he raised the price of theMP 3 players just slightly, sales went down. He usedthe equation 60 = −2x + 130 to determine the price xhe needs to charge if he wants to sell 60 MP 3 players amonth. Solve this equation.$35.0058. Part-time Tuition Costs Many two-year colleges havea large number of students who take courses on apart-time basis. Students pay a charge for each credithour taken plus an activity fee. Suppose the equation$1960 = $175x + $35 can be used to determine the numberof credit hours a student is taking during the upcomingsemester. Solve this equation.11 credit hours59. Super Bowl XLII According to Nielsen Media Research,the New York Giants’ victory over the New EnglandPatriots in Super Bowl XLII was the most watchedSuper Bowl ever, with 3 million more viewers thanthe previous record for Super Bowl XXX in 1996. Theequation 192,000,000 = 2x − 3,000,000 shows that thetotal number of viewers for both Super Bowl gameswas 192 million. Solve for x to determine how manyviewers watched Super Bowl XLII.97,500,000 or 97.5 million viewers60. Blending Gasoline In an attempt to save money at thegas pump, customers will combine two different octanegasolines to get a blend that is slightly higher in octanethan regular gas but not as expensive as premium gas.The equation 14x + 120 − 6x = 200 can be used to findout how many gallons of one octane are needed. Solvethis equation.10 gallonsOCTANE1OCTANEBLENDOCTANE214x + 120 − 6x = 200192,000,000 viewers totalMaintaining Your SkillsTranslations Translate each sentence below into an equation, then solve the equation.61. The sum of 2x and 5 is 19.2x + 5 = 19; x = 762. The sum of 8 and 3x is 2.8 + 3x = 2; x = −263. The difference of 5x and 6 is −9.64. The difference of 9 and 6x is 21._5x − 6 = −9; x = − 3 5 9 − 6x = 21; x = −2Getting Ready for the Next SectionApply the distributive property to each of the following expressions.65. 2(3a − 8)66. 4(2a − 5)67. −3(5x − 1)68. −2(7x − 3)6a − 168a − 20−15x + 3−14x + 6Simplify each of the following expressions as much as possible.69. 3(y − 5) + 670. 5(y + 3) + 771. 6(2x − 1) + 4x72. 8(3x − 2) + 4x3y − 95y + 2216x − 628x − 16

Introduction . . .The Rhind Papyrus is an ancientEgyptian document, created around1650 bc, that contains some mathematicalriddles. One problem on the RhindPapyrus asked the reader to find a quantitysuch that when it is added to onefourthof itself the sum is 15. The equationthat describes this situation isLinear Equations in One Variablex + _1 x = 154As you can see, this equation containsa fraction. One of the topics we will discuss in this section is how to solve equationsthat contain fractions.In this chapter we have been solving what are called linear equations in onevariable. They are equations that contain only one variable, and that variable isalways raised to the first power and never appears in a denominator. Here aresome examples of linear equations in one variable:3x + 2 = 17, 7a + 4 = 3a − 2, 2(3y − 5) = 6Because of the work we have done in the first three sections of this chapter, weare now able to solve any linear equation in one variable. The steps outlinedbelow can be used as a guide to solving these equations.ASolving Linear Equations with One VariableStrategyExample 1Solve: 3(x + 2) = −9SolutionSolving a Linear Equation with One VariableStep 1: Simplify each side of the equation as much as possible. This step isdone using the commutative, associative, and distributive properties.Step 2: Use the addition property of equality to get all variable terms on oneside of the equation and all constant terms on the other, and thencombine like terms. A variable term is any term that contains the variable.A constant term is any term that contains only a number.Step 3: Use the multiplication property of equality to get the variable by itselfon one side of the equation.Step 4: Check your solution in the original equation if you think it is necessary.We begin by applying the distributive property to the left side:Step 1Step 2Step 3{{{Bridgeman Art Library/Getty Images3(x + 2) = −93x + 6 = −93x + 6 + (−6) = −9 + (−6)3x = −15Distributive propertyAdd −6 to both sidesAddition_3x = −15_3 3Divide both sides by 3x = −5Division7.4 Linear Equations in One Variable7.4ObjectivesA Solve linear equations with onevariable.B Solve linear equations involvingfractions and decimals.NoteOnce you have somepractice at solvingequations, these stepswill seem almost automatic. Untilthat time, it is a good idea to payclose attention to these steps.Instructor NoteUsing division on the last step insolving equations is always shownin the examples in this section. However,if I were working Example 1 inclass, when I reached the last stepI would say, “We can multiply bothsides by 1/3 or we can divide bothsides by 3,” just to remind them thatthe two processes are equivalent.Practice Problems1. Solve: 4(x + 3) = −8Answer1. −5439

440Chapter 7 Solving EquationsThis general method of solving linear equations involves using the two propertiesdeveloped in Sections 7.2 and 7.3. We can add any number to both sides ofan equation or multiply (or divide) both sides by the same nonzero number andalways be sure we have not changed the solution to the equation. The equationsmay change in form, but the solution to the equation stays the same. Lookingback to Example 1, we can see that each equation looks a little different from thepreceding one. What is interesting, and useful, is that each of the equations saysthe same thing about x. They all say that x is −5. The last equation, of course, isthe easiest to read. That is why our goal is to end up with x isolated on one sideof the equation.2. Solve: 6a + 7 = 4a − 3Example 2Solve: 4a + 5 = 2a − 7Solution Neither side can be simplified any further. What we have to do is getthe variable terms (4a and 2a) on the same side of the equation. We can eliminatethe variable term from the right side by adding −2a to both sides:4a + 5 = 2a − 7{4a + (−2a) + 5 = 2a + (−2a) − 7 Add −2a to both sidesStep 2 2a + 5 = −7 Addition2a + 5 + (−5) = −7 + (−5) Add −5 to both sides2a = −12AdditionStep 3{_2a = −12_2 2Divide by 2a = −6Division3. Solve: 5(x − 2) + 3 = −12Example 3Solve: 2(x − 4) + 5 = −11Solutionx − 4:We begin by applying the distributive property to multiply 2 andStep 1Step 2Step 3{{{2(x − 4) + 5 = −112x − 8 + 5 = −11 Distributive property2x − 3 = −11 Addition2x − 3 + 3 = −11 + 3 Add 3 to both sides2x = −8 Addition_2x = −8_2 2Divide by 2x = −4 Division4. Solve: 3(4x − 5) + 6 = 3x + 9Answers2. −5 3. −1 4. 2Example 4Solve: 5(2x − 4) + 3 = 4x − 5Solution We apply the distributive property to multiply 5 and 2x − 4. We thencombine similar terms and solve as usual:Step 1Step 2Step 3{5(2x − 4) + 3 = 4x − 510x − 20 + 3 = 4x − 5Distributive property10x − 17 = 4x − 5Simplify the left side10x + (−4x) − 17 = 4x + (−4x) − 5 Add −4x to both sides{6x − 17 = −5Addition6x − 17 + 17 = −5 + 17 Add 17 to both sides6x = 12Addition{_6x = 12_6 6Divide by 6x = 2Division

7.4 Linear Equations in One Variable441B Equations Involving FractionsWe will now solve some equations that involve fractions. Because integers areusually easier to work with than fractions, we will begin each problem by clearingthe equation we are trying to solve of all fractions. To do this, we will use themultiplication property of equality to multiply each side of the equation by theLCD for all fractions appearing in the equation. Here is an example.Example 5Solve the equation x _2 + x _= 8.6Solution The LCD for the fractions _x 2 and x _is 6. It has the property that both 26and 6 divide it evenly. Therefore, if we multiply both sides of the equation by 6, wewill be left with an equation that does not involve fractions.6 x _2 + x _6 = 6(8) Multiply each side by 66 x _2 + 6 x _6 = 6(8) Apply the distributive property3x + x = 48 Multiplication4x = 48 Combine similar termsx = 12 Divide each side by 45. Solve: x _3 + x _6 = 9Instructor NoteExamples 5–7 are different from theprevious examples in that we usethe multiplication property first toclear the equations of fractions. Isometimes take two days to coverthis section and wait until the secondday to cover these problems.We could check our solution by substituting 12 for x in the original equation. If wedo so, the result is a true statement. The solution is 12.As you can see from Example 5, the most important step in solving an equationthat involves fractions is the first step. In that first step we multiply both sides ofthe equation by the LCD for all the fractions in the equation. After we have doneso, the equation is clear of fractions because the LCD has the property that all thedenominators divide it evenly.Example 6Solve the equation 2x + _1 2 = 3 _4 .Solution This time the LCD is 4. We begin by multiplying both sides of theequation by 4 to clear the equation of fractions.6. Solve: 3x + 1 _4 = 5 _8 4 2x + 1 _2 = 4 3 _4 Multiply each side by the LCD, 44(2x) + 4 1 _2 = 4 3 _4 Apply the distributive property8x + 2 = 38x = 1MultiplicationAdd −2 to each sideExample 7Solve for x: 3 _x = _1 Divide each side by 88x + 2 = 1 _. (Assume x is not 0.)2Solution This time the LCD is 2x. Following the steps we used in Examples 5and 6, we have2x 3 _x + 2 = 2x 1 _2 Multiply through by the LCD, 2x2x 3 _x + 2x(2) = 2x 1 _2 Distributive property6 + 4x = x Multiplication6 = −3x Add −4x to each side−2 = xDivide each side by −37. Solve: _4 + 3 = 11_x 5 Instructor NoteThe equation shown in Example 7is not a linear equation in one variable.Sometimes I point this out tomy classes. If I were doing this kindof problem in my algebra classes,I would mention the possibility ofobtaining extraneous solutions.However, I never mention this fact inprealgebra or basic math.Answers5. 18 6. _1 7. −58

442Chapter 7 Solving EquationsEquations Containing Decimals8. Solve: _1 x − 2.4 = 8.35Example 8Solve: 1 _Solutionmultiply each side by 2.x − 3.78 = 2.522We begin by adding 3.78 to each side of the equation. Then we_1 x − 3.78 = 2.522_1 x − 3.78 + 3.78 = 2.52 + 3.78 Add 3.78 to each side21_x = 6.3022 1 _2 x = 2(6.30) Multiply each side by 2x = 12.69. Solve: 7a − 0.18 = 2a + 0.77Example 9Solve: 5a − 0.42 = −3a + 0.98Solution We can isolate a on the left side of the equation by adding 3a toeach side.5a + 3a − 0.42 = −3a + 3a + 0.98 Add 3a to each side8a − 0.42 = 0.988a − 0.42 + 0.42 = 0.98 + 0.42 Add 0.42 to each side8a = 1.40_8a = 1.40_ Divide each side by 88 8a = 0.175Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. Apply the distributive property to the expression 3(x + 4).2. Write the equation that results when −4a is added to both sides of theequation below.6a + 9 = 4a − 33. Solve the equation 2x + _1 2 = 3 _4 by first adding −1 _to each side.2Compare your answer with the solution to the equation shown inExample 6.Answers8. 53.5 9. 0.19

7.4 Problem Set443Problem Set 7.4A Solve each equation using the methods shown in this section. [Examples 1–4]1. 5(x + 1) = 2032. 4(x + 2) = 2443. 6(x − 3) = −624. 7(x − 2) = −715. 2x + 4 = 3x + 7−36. 5x + 3 = 2x + (−3)−27. 7y − 3 = 4y − 15−48. 3y + 5 = 9y + 89. 12x + 3 = −2x + 17_− 1 2 110. 15x + 1 = −4x + 20111. 6x − 8 = −x − 8012. 7x − 5 = −x − 5013. 7(a − 1) + 4 = 11214. 3(a − 2) + 1 = 4315. 8(x + 5) − 6 = 18−216. 7(x + 8) − 4 = 10−617. 2(3x − 6) + 1 = 7318. 5(2x − 4) + 8 = 38519. 10(y + 1) + 4 = 3y + 7−120. 12(y + 2) + 5 = 2y − 1−321. 4(x − 6) + 1 = 2x − 9722. 7(x − 4) + 3 = 5x − 9823. 2(3x + 1) = 4(x − 1)−324. 7(x − 8) = 2(x − 13)625. 3a + 4 = 2(a − 5) + 15126. 10a + 3 = 4(a − 1) + 1−127. 9x − 6 = −3(x + 2) − 24−228. 8x − 10 = −4(x + 3) + 2029. 3x − 5 = 11 + 2(x − 6)430. 5x − 7 = −7 + 2(x + 3)2

444Chapter 7 Solving EquationsB Solve each equation by first finding the LCD for the fractions in the equation and then multiplying both sides of theequation by it. (Assume x is not 0 in Problems 39–46.) [Examples 5–7]31. x _3 + x _6 = 51032. x _2 − x _4 = 31233. x _5 − x = 4−534. x _3 + x = 8635. 3x + _1 2 = 1 _4 36. 3x − _1 3 = 1 _6 37. _x 3 + 1 _2 = −1 _2 _− 1 12 _ 1 6 −338. x _2 + 4 _3 = −2 _3 −439. 4 _x = 1 _5 2040. 2 _3 = 6 _x 941. 3 _x + 1 = 2 _x −142. 4 _x + 3 = 1 _x −143. 3 _x − 2 _x = 1 _5 544. 7 _x + 1 _x = 2445. 1 _x − 1 _2 = −1 _4 446. 3 _x − 4 _5 = −1 _5 5Solve each equation.47. 4x − 4.7 = 3.52.0548. 2x + 3.8 = −7.7−5.7549. 0.02 + 5y = −0.3−0.06450. 0.8 + 10y = −0.7−0.1551. _1 x − 2.99 = 1.02312.0352. _1 x + 2.87 = −3.017−41.1653. 7n − 0.32 = 5n + 0.560.4454. 6n + 0.88 = 2n − 0.77−0.412555. 3a + 4.6 = 7a + 5.3−0.17556. 2a − 3.3 = 7a − 5.20.3857. 0.5x + 0.1(x + 20) = 3.2258. 0.1x + 0.5(x + 8) = 75Find the value of x for each of the figures, given the perimeter.59. P = 36 x = 10 60. P = 30 x = 5xxx2x 3x 62x 261. P = 16 x = 5 62. P = 60 x = 10xxx2x 62x 4x 1

Introduction . . .ApplicationsAs you begin reading through the examples in this section, you may find yourselfasking why some of these problems seem so contrived. The title of the sectionis “Applications,” but many of the problems here don’t seem to have much to dowith real life. You are right about that. Example 5 is what we refer to as an “ageproblem.” Realistically, it is not the kind of problem you would expect to find if youchoose a career in which you use algebra. However, solving age problems is goodpractice for someone with little experience with application problems, because thesolution process has a form that can be applied to all similar age problems.To begin this section we list the steps used in solving application problems. Wecall this strategy the Blueprint for Problem Solving. It is an outline that will overlaythe solution process we use on all application problems.7.5ObjectivesA Set up and solve number problemsusing linear equations.B Set up and solve geometryproblems using linear equations.C Set up and solve age problemsusing linear equations.Blueprint for Problem SolvingStep 1: Read the problem, and then mentally list the items that are knownand the items that are unknown.Step 2: Assign a variable to one of the unknown items. (In most casesthis will amount to letting x equal the item that is asked for in theproblem.) Then translate the other information in the problem toexpressions involving the variable.Step 3: Reread the problem, and then write an equation, using the itemsand variables listed in Steps 1 and 2, that describes the situation.Step 4: Solve the equation found in Step 3.Step 5: Write your answer using a complete sentence.Step 6: Reread the problem, and check your solution with the originalwords in the problem.There are a number of substeps within each of the steps in our blueprint. Forinstance, with Steps 1 and 2 it is always a good idea to draw a diagram or pictureif it helps you to visualize the relationship between the items in the problem.It is important for you to remember that solving application problems is moreof an art than a science. Be flexible. No one strategy works all of the time. Try tostay away from looking for the “one way” to set up and solve a problem. Think ofthe blueprint for problem solving as guidelines that will help you organize yourapproach to these problems, rather than as a set of rules.ANumber ProblemsExample 1The sum of a number and 2 is 8. Find the number.Solution Using our blueprint for problem solving as an outline, we solve theproblem as follows:Step 1 Read the problem, and then mentally list the items that are known andthe items that are unknown.Practice Problems1. The sum of a number and 3 is10. Find the number.Known items: The numbers 2 and 8Unknown item:The number in question7.5 Applications447

448Chapter 7 Solving EquationsStep 2 Assign a variable to one of the unknown items. Then translate the otherinformation in the problem to expressions involving the variable.Let x = the number asked for in the problemThen “The sum of a number and 2” translates to x + 2.Step 3 Reread the problem, and then write an equation, using the items and variableslisted in Steps 1 and 2, that describes the situation.With all word problems, the word “is” translates to = .The sum of x and 2 is 8.x + 2 = 8Step 4 Solve the equation found in Step 3.x + 2 = 8x + 2 + (−2) = 8 + (−2)x = 6Add −2 to each sideStep 5 Write your answer using a complete sentence.The number is 6.Step 6 Reread the problem, and check your solution with the original words inthe problem.The sum of 6 and 2 is 8.A true statementTo help with other problems of the type shown in Example 1, here are somecommon English words and phrases and their mathematical translations.EnglishAlgebraThe sum of a and ba + bThe difference of a and ba − bThe product of a and ba ⋅ bThe quotient of a and ba_b Of ⋅ (multiply)Is = (equals)A numberx4 more than x x + 44 times x 4x4 less than x x − 4You may find some examples and problems in this section and the problem setthat follows that you can solve without using algebra or our blueprint. It is veryimportant that you solve those problems using the methods we are showing here.The purpose behind these problems is to give you experience using the blueprintas a guide to solving problems written in words. Your answers are much lessimportant than the work that you show in obtaining your answer.2. If 4 is added to the sum of twicea number and three times thenumber, the result is 34. Findthe number.Answer1. The number is 7.Example 2If 5 is added to the sum of twice a number and three timesthe number, the result is 25. Find the number.SolutionStep 1 Read and list.Known items:Unknown item:The numbers 5 and 25, twice a number, andthree times a numberThe number in question

7.5 Applications449Step 2 Assign a variable and translate the information.Let x = the number asked for in the problem.Then “The sum of twice a number and three times the number”translates to 2x + 3x.Step 3 Reread and write an equation.5 added to the sum of twice a number is 25{and three times the number{888887n8888877n88888n88888nStep 4 Solve the equation.5 + 2x + 3x = 255 + 2x + 3x = 25Step 5 Write your answer.5x + 5 = 255x + 5 + (−5) = 25 + (−5)5x = 20Simplify the left sideAdd −5 to both sidesAddition_ 5x= 20_ Divide by 55 5x = 4The number is 4.Step 6 Reread and check.Twice 4 is 8, and three times 4 is 12. Their sum is 8 + 12 = 20. Fiveadded to this is 25. Therefore, 5 added to the sum of twice 4 andthree times 4 is 25.BGeometry ProblemsExample 3The length of a rectangle is three times the width. Theperimeter is 72 centimeters. Find the width and the length.SolutionStep 1 Read and list.3. The length of a rectangle istwice the width. The perimeteris 42 centimeters. Find thelength and the width.Known items:Unknown items:The length is three times the width.The perimeter is 72 centimeters.The length and the widthStep 2 Assign a variable, and translate the information. We let x = the width. Becausethe length is three times the width, the length must be 3x. A picturewill help.Rectanglex (width)3x (length)Figure 1Answer2. The number is 6.

450Chapter 7 Solving EquationsStep 3 Reread and write an equation. Because the perimeter is the sum of thesides, it must be x + x + 3x + 3x (the sum of the four sides). But the perimeteris also given as 72 centimeters. Hence,x + x + 3x + 3x = 72Step 4 Solve the equation.x + x + 3x + 3x = 728x = 72x = 9Step 5 Write your answer. The width, x, is 9 centimeters. The length, 3x, must be27 centimeters.Step 6 Reread and check. From the diagram below, we see that these solutionscheck:Perimeter is 72Length = 3 × Width9 + 9 + 27 + 27 = 72 27 = 3 ⋅ 9279927Figure 2Next we review some facts about triangles that we introduced in a previouschapter.facts from geometry Labeling Triangles and the Sum of the Anglesin a TriangleOne way to label the important parts of a triangle is to label the vertices withcapital letters and the sides with small letters, as shown in Figure 3.BcaAbFigure 3CIn Figure 3, notice that side a is opposite vertex A, side b is opposite vertex B,and side c is opposite vertex C. Also, because each vertex is the vertex of oneof the angles of the triangle, we refer to the three interior angles as A, B, and C.In any triangle, the sum of the interior angles is 180°. For the triangle shownin Figure 3, the relationship is writtenAnswer3. The width is 7 cm, and thelength is 14 cm.A + B + C = 180°

7.5 Applications451Example 4The angles in a triangle are such that one angle is twicethe smallest angle, while the third angle is three times as large as the smallestangle. Find the measure of all three angles.SolutionStep 1 Read and list.4. The angles in a triangle are suchthat one angle is three times thesmallest angle, while the largestangle is five times the smallestangle. Find the measure of allthree angles.Known items:Unknown items:The sum of all three angles is 180°; one angle istwice the smallest angle; and the largest angleis three times the smallest angle.The measure of each angleStep 2 Assign a variable and translate information. Let x be the smallest angle,then 2x will be the measure of another angle, and 3x will be the measureof the largest angle.Step 3 Reread and write an equation. When working with geometric objects,drawing a generic diagram will sometimes help us visualize what it isthat we are asked to find. In Figure 4, we draw a triangle with angles A,B, and C.CbaAcBFigure 4We can let the value of A = x, the value of B = 2x, and the value of C = 3x. Weknow that the sum of angles A, B, and C will be 180°, so our equation becomesx + 2x + 3x = 180°Step 4 Solve the equation.x + 2x + 3x = 180°6x = 180°x = 30°Step 5 Write the answer.The smallest angle A measures 30°.Angle B measures 2x, or 2(30°) = 60°.Angle C measures 3x, or 3(30°) = 90°.Step 6 Reread and check. The angles must add to 180°:A + B + C = 180°30° + 60° + 90° = 180°180° = 180° Our answers checkAnswer4. The angles are 20°, 60°, and100°.

452Chapter 7 Solving EquationsCAge Problem5. Joyce is 21 years older than herson Travis. In six years the sumof their ages will be 49. How oldare they now?Example 5Jo Ann is 22 years older than her daughter Stacey. In sixyears the sum of their ages will be 42. How old are they now?SolutionStep 1 Read and list:Known items:Unknown items:Jo Ann is 22 years older than Stacey. Six yearsfrom now their ages will add to 42.Their ages nowStep 2 Assign a variable and translate the information. Let x = Stacey’s age now.Because Jo Ann is 22 years older than Stacey, her age is x + 22.Step 3 Reread and write an equation. As an aid in writing the equation we usethe following table:nowIn Six yearsStacey x x + 6Jo Ann x + 22 x + 28Their ages in six yearswill be their ages nowplus 6Because the sum of their ages six years from now is 42, we write the equation as(x + 6) + (x + 28) = 42h hStacey’s Jo Ann’sage in age in6 years 6 yearsStep 4 Solve the equation.x + 6 + x + 28 = 422x + 34 = 422x = 8x = 4Step 5 Write your answer. Stacey is now 4 years old, and Jo Ann is 4 + 22 = 26years old.Step 6 Reread and check. To check, we see that in six years, Stacey will be 10,and Jo Ann will be 32. The sum of 10 and 32 is 42, which checks.Car Rental Problem6. If a car were rented from thecompany in Example 6 for 2days and the total charge was$41, how many miles was thecar driven?Example 6A car rental company charges $11 per day and 16 centsper mile for their cars. If a car were rented for 1 day and the charge was $25.40,how many miles was the car driven?SolutionStep 1 Read and list.Answer5. Travis is 8; Joyce is 29.Known items:Unknown items:Charges are $11 per day and 16 cents per mile.Car is rented for 1 day. Total charge is $25.40.How many miles the car was driven

7.5 Applications453Step 2 Assign a variable and translate information. If we let x = the number ofmiles driven, then the charge for the number of miles driven will be0.16x, the cost per mile times the number of miles.Step 3 Reread and write an equation. To find the total cost to rent the car, we add11 to 0.16x. Here is the equation that describes the situation:$11 perday16 cents+ = Total costper mile11 + 0.16x = 25.40Step 4 Solve the equation. To solve the equation, we add −11 to each side andthen divide each side by 0.16.11 + (−11) + 0.16x = 25.40 + (−11) Add −11 to each side0.16x = 14.40_ 0.16x= 14.40_0.16 0.16Divide each side by 0.16x = 90 14.40 ÷ 0.16 = 90Step 5 Write the answer. The car was driven 90 miles.Step 6 Reread and check. The charge for 1 day is $11. The 90 milesadds 90($0.16) = $14.40 to the 1-day charge. The total is$11 + $14.40 = $25.40, which checks with the total charge given in theproblem.Coin Problem{{{Example 7Diane has $1.60 in dimes and nickels. If she has 7 moredimes than nickels, how many of each coin does she have?SolutionStep 1 Read and list.7. Amy has $1.75 in dimes andquarters. If she has 7 moredimes than quarters, how manyof each coin does she have?Known items:Unknown items:We have dimes and nickels. There are 7 moredimes than nickels, and the total value of thecoins is $1.60.How many of each type of coin Diane hasStep 2 Assign a variable and translate information. If we let x = the number ofnickels, then the number of dimes must be x + 7, because Diane has 7more dimes than nickels. Because each nickel is worth 5 cents, theamount of money she has in nickels is 0.05x. Similarly, because eachdime is worth 10 cents, the amount of money she has in dimes is0.10(x + 7). Here is a table that summarizes what we have so far:nickelsDimesNumber of x x + 7Value of 0.05x 0.10(x + 7)Answer6. The car was driven 118.75 miles.

454Chapter 7 Solving EquationsStep 3 Reread and write an equation. Because the total value of all the coins is$1.60, the equation that describes this situation isAmount of money Amount of money Total amount+ =in nickelsin dimes of money0.05x + 0.10(x + 7) = 1.60Step 4 Solve the equation. This time, let’s show only the essential steps in thesolution.{{0.05x + 0.10x + 0.70 = 1.60 Distributive property{0.15x + 0.70 = 1.60 Add 0.05x and 0.10x to get 0.15x0.15x = 0.90 Add −0.70 to each sidex = 6 Divide each side by 0.15Step 5 Write the answer. Because x = 6, Diane has 6 nickels. To find the numberof dimes, we add 7 to the number of nickels (she has 7 more dimes thannickels). The number of dimes is 6 + 7 = 13.Step 6 Reread and check. Here is a check of our results.6 nickels are worth 6($0.05) = $0.3013 dimes are worth 13($0.10) = $1.30The total value is$1.60Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is the first step in solving a word problem?2. Write a mathematical expression equivalent to the phrase “the sum of xand ten.”3. Write a mathematical expression equivalent to the phrase “twice thesum of a number and ten.”4. Suppose the length of a rectangle is three times the width. If we let xrepresent the width of the rectangle, what expression do we use to representthe length?Answer7. There are 3 quarters and 10dimes.

7.5 Problem Set455Problem Set 7.5Write each of the following English phrases in symbols using the variable x.1. The sum of x and 3x + 32. The difference of x and 2x − 23. The sum of twice x and 12x + 14. The sum of three times x and 43x + 45. Five x decreased by 65x − 66. Twice the sum of x and 52(x + 5)7. Three times the sum of x and 13(x + 1)8. Four times the sum of twice x and 14(2x + 1)9. Five times the sum of three x and 45(3x + 4)10. Three x added to the sum of twice x and 1(2x + 1) + 3xUse the six steps in the “Blueprint for Problem Solving” to solve the following word problems. You may recognize thesolution to some of them by just reading the problem. In all cases, be sure to assign a variable and write the equation usedto describe the problem. Write your answer using a complete sentence.A Number Problems [Examples 1, 2]11. The sum of a number and 3 is 5. Find the number.The number is 2.12. If 2 is subtracted from a number, the result is 4. Find thenumber.The number is 6.13. The sum of twice a number and 1 is −3. Find thenumber.The number is −2.14. If three times a number is increased by 4, the result is−8. Find the number.The number is −4.15. When 6 is subtracted from five times a number, theresult is 9. Find the number.The number is 3.16. Twice the sum of a number and 5 is 4. Find the number.The number is −3.17. Three times the sum of a number and 1 is 18. Find thenumber.The number is 5.18. Four times the sum of twice a number and 6 is −8. Findthe number.The number is −4.19. Five times the sum of three times a number and 4 is−10. Find the number.The number is −2.20. If the sum of three times a number and two times thesame number is increased by 1, the result is 16. Find thenumber.The number is 3.

456Chapter 7 Solving EquationsB Geometry Problems [Examples 3, 4]21. The length of a rectangle is twice its width. The perimeteris 30 meters. Find the length and the width.The length is 10 m and the width is 5 m.22. The width of a rectangle is 3 feet less than its length. Ifthe perimeter is 22 feet, what is the width?The width is 4 ft.23. The perimeter of a square is 32 centimeters. What isthe length of one side?The length of one side is 8 cm.25. Two angles in a triangle are equal, and their sum isequal to the third angle in the triangle. What are themeasures of each of the three interior angles?The measures of the angles are 45°, 45°, and 90°.27. The smallest angle in a triangle is _1 as large as the3largest angle. The third angle is twice the smallestangle. Find the three angles.The angles are 30°, 60°, and 90°.24. Two sides of a triangle are equal in length, and the thirdside is 10 inches. If the perimeter is 26 inches, how longare the two equal sides?The two equal sides are each 8 in. long.26. One angle in a triangle measures twice the smallestangle, while the largest angle is six times the smallestangle. Find the measures of all three angles.The measures of the angles are 20°, 40°, and 120°.28. One angle in a triangle is half the largest angle, but threetimes the smallest. Find all three angles.The angles are 18°, 54°, and 108°.C Age Problems [Example 5]29. Pat is 20 years older than his son Patrick. In 2 years,the sum of their ages will be 90. How old are theynow? Patrick is 33 years old, and Pat is 53 years old.30. Diane is 23 years older than her daughter Amy. In 5years, the sum of their ages will be 91. How old are theynow? Amy is 29 years old, and Diane is 52 years old.nowIn 2 YearsnowIn 5 YearsPatrickxAmyxPatDiane31. Dale is 4 years older than Sue. Five years ago the sumof their ages was 64. How old are they now?Sue is 35 years old, and Dale is 39 years old.32. Pat is 2 years younger than his wife, Wynn. Ten years agothe sum of their ages was 48. How old are they now?Wynn is 35 years old, and Pat is 33 years old.Renting a Car [Example 6]33. A car rental company charges $10 a day and 16 centsper mile for their cars. If a car were rented for 1 dayfor a total charge of $23.92, how many miles was itdriven?87 mi34. A car rental company charges $12 a day and 18 centsper mile to rent their cars. If the total charge for a1-day rental were $33.78, how many miles was the cardriven?121 mi35. A rental company charges $9 per day and 15 cents amile for their cars. If a car were rented for 2 days for atotal charge of $40.05, how many miles was it driven?147 mi36. A car rental company charges $11 a day and 18 centsper mile to rent their cars. If the total charge for a2-day rental were $61.60, how many miles was it driven?220 mi

7.5 Problem Set457Coin Problems [Example 7]37. Mary has $2.20 in dimes and nickels. If she has 10more dimes than nickels, how many of each coin doesshe have?8 nickels, 18 dimes38. Bob has $1.65 in dimes and nickels. If he has 9 morenickels than dimes, how many of each coin does hehave?17 nickels, 8 dimes39. Suppose you have $9.60 in dimes and quarters. Howmany of each coin do you have if you have twice asmany quarters as dimes?16 dimes, 32 quarters40. A collection of dimes and quarters has a total value of$2.75. If there are 3 times as many dimes as quarters,how many of each coin is in the collection?5 quarters, 15 dimesMiscellaneous Problems41. Magic Square The sum of the numbers in each row,each column, and each diagonal of the square below is15. Use this fact, along with the information in the firstcolumn of the square, to write an equation containingthe variable x, then solve the equation to find x. Next,write and solve equations that will give you y and z.x = 8, y = 6, z = 9x 1 y3 5 74 z 242. Magic Square The sum of the numbers in each row, eachcolumn, and each diagonal of the square below is 3. Usethis fact, along with the information in the second row ofthe square, to write an equation containing the variable a,then solve the equation to find a. Next, write and solve anequation that will allow you to find the value of b. Next,write and solve equations that will give you c and d.a = −1, b = 2, c = 5, d = −34 d ba 1 30 c −243. Wages JoAnn works in the publicity office at the stateuniversity. She is paid $14 an hour for the first 35hours she works each week and $21 an hour for everyhour after that. If she makes $574 one week, howmany hours did she work?39 hours44. Ticket Sales Stacey is selling tickets to the school play.The tickets are $6 for adults and $4 for children. Shesells twice as many adult tickets as children’s ticketsand brings in a total of $112. How many of each kind ofticket did she sell?7 children’s tickets, 14 adult tickets45. Cars The chart shows the fastest cars in America. Themaximum speed of an Evans 487 is twice the sum ofthe speed of a trucker and 45 miles per hour. What isthe speed of the trucker?60 miles per hour46. Skyscrapers The chart shows the heights of the three tallestbuildings in the world. The Sears Tower is 80 feetless than 5 times the height of the Statue of Liberty. Whatis the height of the Statue of Liberty?306 feetReady for the RacesFord GT 205 mphEvans 487 210 mphSaleen S7 Twin Turbo 260 mphSSC Ultimate Aero 273 mphSuch Great HeightsTaipei 101Taipei, TaiwanPetronas Tower 1 & 2Kuala Lumpur, Malaysia1,483 ft1,670 ft Sears TowerChicago, USA1,450 ftSource: Forbes.comSource: www.tenmojo.com

458Chapter 7 Solving EquationsGetting Ready for the Next SectionSimplify.47. _5 (95 − 32)93548. _5 (77 − 32)92549. Find the value of 90 − x when x = 25.6550. Find the value of 180 − x when x = 25.15551. Find the value of 2x + 6 when x = −2252. Find the value of 2x + 6 when x = 0.6Solve.53. 40 = 2l + 121454. 80 = 2l + 123455. 6 + 3y = 456. 8 + 3y = 4_− 2 3 _− 4 3 Maintaining Your SkillsThe problems below review some of the work you have done with percents.Change each fraction to a decimal and then to a percent.57. 3 _4 0.75, 75%58. 5 _8 0.625, 62.5%59. 1 1 _5 1.2, 120%60. 7 _10 0.7, 70%Change each percent to a fraction and a decimal.61. 37%62. 18%63. 3.4%64. 125%_37100 , 0.37 _ 950 , 0.18 _ 17500 , 0.034 _ 5 , 1.25465. What number is 15% of 135?20.2566. 19 is what percent of 38?50%67. 12 is 16% of what number?75

Introduction . . .Evaluating FormulasIn mathematics a formula is an equation that contains more than one variable.The equation P = 2w + 2l is an example of a formula. This formula tells us therelationship between the perimeter P of a rectangle, its length l, and its width w.There are many formulas with which you may be familiar already. Perhaps youhave used the formula d = r ⋅ t to find out how far you would go if you traveled at50 miles an hour for 3 hours. If you take a chemistry class while you are in college,you will certainly use the formula that gives the relationship between thetwo temperature scales, Fahrenheit and Celsius:7.6ObjectivesA Solve a formula for a given variable.B Solve problems using the rateequation.F = _9 C + 325Although there are many kinds of problems we can work using formulas, wewill limit ourselves to those that require only substitutions. The examples that followillustrate this type of problem.AFormulasExample 1The perimeter P of a rectangular livestock pen is 40 feet. Ifthe width w is 6 feet, find the length.Practice Problems1. Suppose the livestock pen inExample 1 has a perimeter of 80feet. If the width is still 6 feet,what is the new length?6 feetlSolution First we substitute 40 for P and 6 for w in the formula P = 2l + 2w.Then we solve for l :When P = 40 and w = 6the formula P = 2l + 2wbecomes 40 = 2l + 2(6)or 40 = 2l + 12 Multiply 2 and 628 = 2l Add −12 to each side14 = l Multiply each side by _1 2 To summarize our results, if a rectangular pen has a perimeter of 40 feet and awidth of 6 feet, then the length must be 14 feet.Answer1. 34 feet7.6 Evaluating Formulas459

460Chapter 7 Solving Equations2. Use the formula in Example 2 tofind C when F is 77 degrees.NoteThe formula we areusing here,C = _5 (F − 32),9is an alternative form of the formulawe mentioned in the introductionto this section:F = _9 C + 325Both formulas describe the samerelationship between the twotemperature scales. If you go onto take an algebra class, you willlearn how to convert one formulainto the other.Example 2Use the formula C = 5 _degrees.(F − 32) to find C when F is 959Solution Substituting 95 for F in the formula gives us the following:When F = 95the formula C = _5 (F − 32)9becomesC = _5 (95 − 32)9= 5 _9 (63)= _5 ⋅ 63_9 1 = 315 _9 = 35A temperature of 95 degrees Fahrenheit is the same as a temperature of 35degrees Celsius.3. Use the formula in Example 3 tofind y when x is 0.Example 3Use the formula y = 2x + 6 to find y when x is −2.Solution Proceeding as we have in the previous examples, we have:When x = −2the formula y = 2x + 6becomes y = 2(−2) + 6= −4 + 6= 2In some cases evaluating a formula also involves solving an equation, as the nextexample illustrates.4. Use the formula in Example 4 tofind y when x is −3.Example 4Find y when x is 3 in the formula 2x + 3y = 4.Solution First we substitute 3 for x ; then we solve the resulting equationfor y.When x = 3the equation 2x + 3y = 4becomes 2(3) + 3y = 46 + 3y = 43y = −2Add −6 to each sideAnswers2. 25 degrees Celsius 3. 64. _103 BRate Equation_y = − 2 Divide each side by 33Now we will look at some problems that use what is called the rate equation. Youuse this equation on an intuitive level when you are estimating how long it willtake you to drive long distances. For example, if you drive at 50 miles per hour for2 hours, you will travel 100 miles. Here is the rate equation:Distance = rate ⋅ time, or d = r ⋅ t

7.6 Evaluating Formulas461The rate equation has two equivalent forms, one of which is obtained by solvingfor r, while the other is obtained by solving for t. Here they are:r = d _t and t = d _r The rate in this equation is also referred to as average speed.Example 5At 1 p.m., Jordan leaves her house and drives at an averagespeed of 50 miles per hour to her sister’s house. She arrives at 4 p.m.Solutiona. How many hours was the drive to her sister’s house?b. How many miles from her sister does Jordan live?a. If she left at 1:00 p.m. and arrived at 4:00 p.m., we simply subtract1 from 4 for an answer of 3 hours.b. We are asked to find a distance in miles given a rate of 50 milesper hour and a time of 3 hours. We will use the rate equation,d = r ⋅ t, to solve this. We have:5. At 9 a.m. Maggie leaves herhouse and drives at an averagespeed of 60 miles per hour toher sister’s house. She arrives at11 a.m.a. How many hours was thedrive to her sister’s house?b. How many miles from hersister does Maggie live?d = 50 miles per hour ⋅ 3 hoursd = 50(3)d = 150 milesNotice that we were asked to find a distance in miles, so our answer has a unit ofmiles. When we are asked to find a time, our answer will include a unit of time,like days, hours, minutes, or seconds.When we are asked to find a rate, our answer will include units of rate, likemiles per hour, feet per second, problems per minute, and so on.facts from geometryEarlier we defined complementary angles as angles that add to 90°. That is, ifx and y are complementary angles, thenx + y = 90°If we solve this formula for y, we obtain a formula equivalent to our originalformula:y = 90° − x90˚−xxComplementary anglesBecause y is the complement of x, we can generalize by saying that the complementof angle x is the angle 90° − x. By a similar reasoning process, wecan say that the supplement of angle x is the angle 180° − x. To summarize,if x is an angle, thenthe complement of x is 90° − x, andthe supplement of x is 180° − x180˚−xxIf you go on to take a trigonometry class, you will see these formulas again.Supplementary anglesAnswer5. a. 2 hours b. 120 miles

462Chapter 7 Solving Equations6. Find the complement and thesupplement of 35°.Example 6Find the complement and the supplement of 25°.Solution We can use the formulas above with x = 25°.The complement of 25° is 90° − 25° = 65°.The supplement of 25° is 180° − 25° = 155°.Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is a formula?2. How do you solve a formula for one of its variables?3. What are complementary angles?4. What is the formula that converts temperature on the Celsius scale totemperature on the Fahrenheit scale?Answer6. Complement = 55°;Supplement = 145°

7.6 Problem Set463Problem Set 7.6A The formula for the area A of a rectangle with length l and width w is A = l ⋅ w. Find A if: [Examples 1–4]1. l = 32 feet and w = 22 feet2. l = 22 feet and w = 12 feet704 ft 2 264 ft 23. l = 3 _2 inch and w = 3 _4 inch4. l = _3_9 8 in2 _ 950 in25 inch and w = 3 _10 inchThe formula G = H ⋅ R tells us how much gross pay G a person receives for working H hours at an hourly rate of pay R. InProblems 5-8, find G.5. H = 40 hours and R = $6$2406. H = 36 hours and R = $8$2887. H = 30 hours and R = $9_1 2 $2858. H = 20 hours and R = $6_3 4 $135Because there are 3 feet in every yard, the formula F = 3 ⋅ Y will convert Y yards into F feet. In Problems 9-12, find F.9. Y = 4 yards12 ft10. Y = 8 yards24 ft11. Y = 2_2 3 yards8 ft12. Y = 6_1 3 yards19 ftIf you invest P dollars (P is for principal ) at simple interest rate R for T years, the amount of interest you will earn is givenby the formula I = P ⋅ R ⋅ T. In Problems 13 and 14, find I.13. P = $1,000, R = _7 , and T = 2 years100$14014. P = $2,000, R = _6100 , and T = 21 _2 years$300In Problems 15-18, use the formula P = 2w + 2l to find P.15. w = 10 inches and l = 19 inches58 in.16. w = 12 inches and l = 22 inches68 in.17. w = 3 _4 foot and l = 7 _8 foot3 1 _4= 13_18. w = _14 ft4 ft2 foot and l = 3 _2 feet

464Chapter 7 Solving EquationsWe have mentioned the two temperature scales, Fahrenheit and Celsius. Table 1 is intended to give you a more intuitiveidea of the relationship between the two temperatures scales.Table 1Comparing Two Temperature ScalesTemperature TemperatureSituation (Fahrenheit) (Celsius)Water freezes 32°F 0°CRoom temperature 68°F 20°CNormal body temperature98_3 °F537°CWater boils 212°F 100°CBake cookies 365°F 185°CTable 2 gives the formulas, in both symbols and words, that are used to convert between the two scales.Table 2Formulas for Converting Between Temperature ScalesTo Convert From formula In Symbols Formula In WordsFahrenheit to CelsiusC = 5 _9 (F − 32) Subtract 32, then multiply by 5 _9 .Celsius to FahrenheitF = _9 5 C + 32 Multiply by 9 _, then add 32.519. Let F = 212 in the formula C = _5 (F − 32), and solve for9C. Does the value of C agree with the information inTable 1?C = 100°C; yes20. Let C = 100 in the formula F = _9 C + 32, and solve for F.5Does the value of F agree with the information in Table 1?F = 212°F; yes21. Let F = 68 in the formula C = _5 (F − 32), and solve for9C. Does the value of C agree with the information inTable 1?C = 20°C; yes22. Let C = 37 in the formula F = _9 C + 32, and solve for F.5Does the value of F agree with the information in Table 1?F = 98_ 3 °F; yes523. Find C when F is 32°.0°C24. Find C when F is −4°.−20°C25. Find F when C is −15°.5°F26. Find F when C is 35°.95°F

7.6 Problem Set465B Maximum Heart Rate In exercise physiology, a person’s maximum heart rate, in beats per minute, is found by subtractinghis age in years from 220. So, if A represents your age in years, then your maximum heart rate isM = 220 − AUse this formula to complete the following tables.27.Age Maximum Heart Rate28.(years) (beats per minute)18 20219 20120 20021 19922 19823 197Age Maximum Heart Rate(years) (beats per minute)15 20520 20025 19530 19035 18540 180Training Heart Rate A person’s training heart rate, in beats per minute, is the person’s resting heart rate plus 60% of the differencebetween maximum heart rate and his resting heart rate. If resting heart rate is R and maximum heart rate is M, thenthe formula that gives training heart rate isT = R + 3 _5 (M − R )Use this formula along with the results of Problems 27 and 28 to fill in the following two tables.29. For a 20-year-old person 30. For a 40-year-old personResting Heart Rate(beats per minute)Training Heart Rate(beats per minute)Resting Heart Rate(beats per minute)Training Heart Rate(beats per minute)60 14465 14670 14875 15080 15285 15460 13265 13470 13675 13880 14085 142B Use the rate equation d = r ⋅ t to solve Problems 31 and 32. [Example 5]31. At 2:30 p.m. Shelly leaves her house and drives at anaverage speed of 55 miles per hour to her sister’shouse. She arrives at 6:30 p.m.a. How many hours was the drive to her sister’shouse?4 hrb. How many miles from her sister does Shelly live?220 mi32. At 1:30 p.m. Cary leaves his house and drives at anaverage speed of 65 miles per hour to his brother’shouse. He arrives at 5:30 p.m.a. How many hours was the drive to his brother’shouse?4 hrb. How many miles from his brother’s house does Carylive?260 mi

466Chapter 7 Solving EquationsUse the rate equation r = _d to solve Problems 33 and 34.t33. At 2:30 p.m. Brittney leaves her house and drives 260miles to her sister’s house. She arrives at 6:30 p.m.a. How many hours was the drive to her sister’shouse?4 hrb. What was Brittney’s average speed?65 mi/hr34. At 8:30 a.m. Ethan leaves his house and drives 220 milesto his brother’s house. He arrives at 12:30 p.m.a. How many hours was the drive to his brother’shouse?4 hrb. What was Ethan’s average speed?55 mi/hrAs you know, the volume V enclosed by a rectangular solid with length l, width w, and height h is V = l ⋅ w ⋅ h. In Problems35-38, find V if:35. l = 6 inches, w = 12 inches, and h = 5 inches36. l = 16 inches, w = 22 inches, and h = 15 inches360 in 3 5,280 in 337. l = 6 yards, w = _1 2 yard, and h = 1 _3 yard38. l = 30 yards, w = _5 2 yards, and h = 5 _3 yards1 yd 3 125 yd 3Suppose y = 3x − 2. In Problems 39–44, find y if:39. x = 3y = 740. x = −5y = −17_41. x = − 1 3 y = −342. x = _2 3 y = 043. x = 0y = −244. x = 5y = 13Suppose x + y = 5. In Problems 45–50, find x if:45. y = 246. y = −247. y = 048. y = 549. y = −350. y = 3x = 3x = 7x = 5x = 0x = 8x = 2Suppose x + y = 3. In Problems 51–56, find y if:51. x = 2y = 152. x = −2y = 553. x = 0y = 354. x = 3y = 055. x = _1 2 _56. x = − 1 2 y = _ 5 2 y = _ 7 2 Suppose 4x + 3y = 12. In Problems 57–62, find y if:57. x = 3y = 0y = _ 323 y = _ 133 y = 258. x = −5_59. x = − 1 4 60. x = _3 2 61. x = 0y = 462. x = −3y = 8Suppose 4x + 3y = 12. In Problems 63-68, find x if:63. y = 4x = 064. y = −4x = 6_65. y = − 1 3 66. y = _5 3 67. y = 0x = 3x = _ 134 x = _ 7 4 68. y = −3x = 21 _4 Find the complement and supplement of each angle. [Example 6]69. 45°Complement is 45°,supplement is 135°70. 75°Complement is 15°,supplement is 105°71. 31°Complement is 59°,supplement is 149°72. 59°Complement is 31°,supplement is 121°

468Chapter 7 Solving EquationsMaintaining Your SkillsSimplify._3 5 _577. _7 _4 78. _5 6 79. _7 _3 4 _ 5 36 1 + 1 __2 1 − 1 _2 1 + _180. _ 3 1 − _1 3 2_1 2 + 1 _81. _4 _1 4 + 1 _8 2_1 2 − 1 _82. _4 _1 4 − 1 _8 2_3 5 + 3 _83. _ 7 _3 5 − 3 _7 6_5 7 + 5 _84. _ 8 _5 7 − 5 _8 15

Introduction . . .Paired Data and the RectangularCoordinate SystemThe table and line graph below are similar to those that we have used previouslyto discuss temperature. Note that we have extended both the table and the linegraph to show some temperatures below zero on both scales.7.7ObjectivesA Plot ordered pairs on a coordinatesystem.B Name the coordinates of a point onthe rectangular coordinate system.C Graph a line given two points.Table 1Comparing Temperatureson Two ScalesTemperatureIn DegreesCelsiusTemperatureIn Degreesfahrenheit250200−100°C−75°C−50°C−25°C−148°F−103°F−58°F−13°F0°C 32°F25°C 77°F50°C 122°FFahrenheit temperature (°F)150100500501001502001007550 25 0 25 50Celsius temperature (°C)7510075°C 167°FFigure 1 A line graph of the data in Table 1100°C 212°FThe data in Table 1 are called paired data because it is organized so that eachnumber in the first column is paired with a specific number in the second column:a Celsius temperature of 100°C (first column) corresponds to a Fahrenheittemperature of 212°F (second column). The information in Figure 1 is also paireddata because each dot on the line graph comes from one of the pairs of numbersin the table: The upper rightmost dot on the line graph corresponds to 100°C and212°F. In order to standardize the way in which we present paired data visually,we use the rectangular coordinate system.A The Rectangular Coordinate SystemThe rectangular coordinate system can be used to plot (or graph) pairs of numbers(see Figure 2 on the following page). It consists of two number lines, calledaxes, which intersect at right angles. (A right angle is a 90° angle.) The point atwhich the axes intersect is called the origin.7.7 Paired Data and the Rectangular Coordinate System469

470Chapter 7 Solving Equationsy-axisQuadrantII54321−5−4 −3 −2 −1−1−2−3Quadrant−4III−5QuadrantI1 2 3 4 5OriginQuadrantIVx-axisFigure 2The horizontal number line is exactly the same as the real number line and iscalled the x-axis. The vertical number line is also the same as the real numberline with the positive direction up and the negative direction down. It is called they-axis. As you can see, the axes divide the plane into four regions, called quadrants,which are numbered 1 through IV in a counterclockwise direction.Because the rectangular coordinate system consists of two number lines, onecalled the x-axis and the other called the y-axis, we can plot pairs of numberssuch as x = 2 and y = 3. As a matter of fact, each point in the rectangular coordinatesystem is named by exactly one pair of numbers. We call the pair of numbersthat name a point the coordinates of that point. To find the point that is associatedwith the pair of numbers x = 2 and y = 3, we start at the origin and move 2 unitshorizontally to the right and then 3 units vertically up (see Figure 3). The placewhere we end up is the point named by the pair of numbers x = 2, y = 3, whichwe write in shorthand form as the ordered pair (2, 3).y5432154 3 2 112345(2, 3)1 2 3 4 5xFigure 3In general, to graph an ordered pair (a, b ) on the rectangular coordinate system,we start at the origin and move a units right or left (right if a is positive, left if a isnegative). From there we move b units up or down (up if b is positive, down if b isnegative). The point where we end up is the graph of the ordered pair (a, b ).Practice Problems1. Plot the ordered pairs (3, 2),(3, −2), (−3, −2), and (−3, 2) onthe coordinate system used inExample 1.Example 1Plot (graph) the ordered pairs (2, 3), (−2, 3), (−2, −3),and (2, −3).Solution To graph the ordered pair (2, 3), we start at the origin and move 2units to the right, then 3 units up. We are now at the point whose coordinates are

7.7 Paired Data and the Rectangular Coordinate System471(2, 3). We plot the other three ordered pairs in the same manner (Figure 4).y(2, 3)54321(2, 3)54 3 2111 2 3 4 5(2, 3)23(2, 3)45xFigure 4Note Looking at Example 1, we see that any point in quadrant I must have positivex- and y-coordinates, (+, +). In quadrant II, x-coordinates are negative andy-coordinates are positive, (−, +). In quadrant III, both coordinates are negative,(−, −). Finally, in quadrant IV, all ordered pairs must have the form (+, −).Example 2Plot the ordered pairs (1, −4), 1 _2 , 3 , (2, 0), (0, −2), and(−3, 0).Solutiony2. Plot the ordered pairs (−1, −4), − 1 _2 , 3 , (0, 2), (5, 0), (0, −5),and (−1, 0) on the coordinatesystem used in Example 2.(3, 0)54 32 154321123451( 2,3)(2, 0)1 2 3 4 5(0, 2)(1, 4)xB Points on a Rectangular Coordinate SystemExample 3Give the coordinates of each point in Figure 5.D5432154 3 2 112B345yAFigure 51 2 3 4 5Cx3. Give the coordinates of eachpoint in the figure below.y5D4321−5 −4−3 −2−1−11 2 3A4 5−2−3−4−5B CAnswers1. See solutions section.2. See solutions section.x

472Chapter 7 Solving EquationsSolution A is named by the ordered pair (2, 3). B is named by the ordered pair(−3, −2). C is named by the ordered pair (4, −3). D is named by the ordered pair(−2, 5).4. The points (2, 0), (−3, 0), 5 _2 , 0 ,and (−8, 0) all lie on which axis?Example 4Where are all the points that have coordinates of theform (x, 0)?Solution Because the y-coordinate is 0, these points must lie on the x-axis.Remember, the y-coordinate tells us how far up or down we move to find thepoint in question. If the y-coordinate is 0, then we don’t move up or down at all.Therefore, we must stay on the x-axis.CGraphing Lines5. a. Does the point (−3, −2) lie onthe line shown in Example 5?b. Does the point (3, 2) lie on theline shown in Example 5?Example 5Graph the points (1, 2) and (3, 4), and draw a linethrough them. Then use your result to answer these questions.a. Does the graph of (2, 3) lie on this line?b. Does the graph of (−3, −5) lie on this line?Solution Figure 6 shows the graphs of (1, 2) and (3, 4) and the line that connectsthem. The line does not pass through the point (−3, −5) but does passthrough (2, 3).y543215 4 3 2 1123451 2 3 4 5xFigure 6Getting Ready for ClassAfter reading the preceding section and working the practice problemsin the margin, answer the following questions.1. What is an ordered pair of numbers?2. Explain in words how you would graph the ordered pair (2, 3).3. How do you construct a rectangular coordinate system?4. Where is the origin on a rectangular coordinate system?Answers3. A is (3, 0); B is (0, −3); C is(3, −3); D is (−3, 3).4. The x-axis 5. a. Yes b. No

7.7 Problem Set473Problem Set 7.7A Graph each of the following ordered pairs. [Examples 1, 2]1. (4, 2) 2. (4, −2) 3. (−4, 2) 4. (−4, −2) 5. (3, 4) 6. (−3, 4)7. (−3, −4) 8. (3, −4) 9. (4, 3) 10. (−4, 3) 11. 5, 1 _2 12. −5, − 1 _2 13. 1, − 3 _2 14. − 3 _2 , 1 15. (2, 0) 16. (0, −5) 17. (−2, 0) 18. (0, 5)yy54(–4, 2) 321(–2, 0)543 2 11234(–3, –4)5(3, 4)(4, 3)(4, 2)1(2, 0) (5,2)1 2 3 4 5(1, – 32 )x(–5, – 12 )(–3, 4)(–4, 3)(– 3 , 12 )54321543 2 112(–4, –2) 345(0, 5)1 2 3 4 5(4, –2)(0, –5)(3, –4)xUSE FOR ODD-NUMBERED PROBLEMSUSE FOR EVEN-NUMBERED PROBLEMSB 19–25. Give the coordinates of each point in the figure below. [Examples 3, 4]y5 20(4, 4)21(–3, 2)25(–4, 0)432119(2, 2)24 (4, 0)543 2 111 2 3 4 52 23(3, –3)22(–3, –4) 345x19. (2, 2)20. (4, 4)21. (−3, 2)22. (−3, −4)23. (3, −3)24. (4, 0)25. (−4, 0)26. Where will you find all the ordered pairs of the form (0, y )?On the y-axis

474Chapter 7 Solving EquationsC Graph the points (1, 3) and (2, 4), and draw a line through them. Use that graph to answer Problems 27–30. [Example 5]27. Does the ordered pair (−1, 1) lie on this line?yYes54321543 2 1123451 2 3 4 5x28. Does the ordered pair (3, 4) lie on this line?No29. Does the ordered pair (−3, −5) lie on this line?No30. Does the ordered pair (3, 5) lie on this line?YesGraph the points (−3, 2) and (1, 3), and draw a line through them. Use that graph to answer Problems 31–34.31. Does the ordered pair (5, 4) lie on this line?yYes54321543 2 1123451 2 3 4 5x32. Does the ordered pair (3, −1) lie on this line?No33. Does the ordered pair (4, 5) lie on this line?No34. Does the ordered pair (−7, 1) lie on this line?YesApplying the Concepts35. Horse Racing The graph shows the total amount ofmoney wagered on the Kentucky Derby over the years.List three ordered pairs that lie on the graph.(1985, 20), (1990, 34), (1995, 45)36. Solar Energy The graph shows the annual number ofsolar thermal collector shipments in the United States.Use the graph to answer the following questions.Betting the PoniesSolar Thermal CollectorsMillions of dollars100755025020198534199045199565200010420052500020000150001000050000‘97 ‘98 ‘99 ‘00 ‘01 ‘02 ‘03 ‘04 ‘05 ‘06Source: http://www.kentuckyderby.comSource: Energy Information Association 2006a. Does the graph contain the point (2000, 7,500)? Yesb. Does the graph contain the point (2004, 15,000)? Noc. Does the graph contain the point (2005, 15,000)? Yes

7.7 Problem Set47537. Hourly Wages Jane takes a job at the local Marcy’sdepartment store. Her job pays $8.00 per hour. Thegraph shows how much Jane earns for working from 0to 40 hours in a week.38. Hourly Wages Judy takes a job at Gigi’s boutique. Her jobpays $6.00 per hour plus $50 per week in commission.The graph shows how much Judy earns for working from0 to 40 hours in a week.Amount earned ($)350300250200150100504080120160200240280320Amount earned ($)400300200100508011014017020023026029005 10 15 20 25 30 35 40Hours worked05 10 15 20Hours worked25 30 35 40a. List three ordered pairs that lie on the line graph.(5, 40), (10, 80), (20, 160)b. How much will she earn for working 40 hours?$320c. If her check for one week is $240, how many hoursdid she work?30 hoursd. She works 35 hours one week, but her paycheckbefore deductions are subtracted out is for $260. Isthis correct? ExplainNo, if she works 35 hours, she should be paid $280.a. List three ordered pairs that lie on the line graph.(5, 80), (10, 110), (20, 170)b. How much will she earn for working 40 hours?$290c. If her check for one week is $230, how many hoursdid she work?30 hoursd. She works 35 hours one week, but her paycheckbefore deductions are subtracted out is for $260. Isthis correct? Explain.Yes, this is the correct amount.Maintaining Your SkillsMultiply or divide as indicated.39. _x 2 3⋅ y_y x 40. _a 3⋅ 12_15 a 41. x 2 3_÷ x_2 3y y 2xy 2 _ 4a5 _ 1 xy 42. _a 2 bc ÷ ab _23 c 3_ a b Add or subtract as indicated.43. x _5 + 3 _4 44. 5 + 2 _x 45. 3 − 1 _x 46. x _8 − 5 _6 4x + 15_20_ 5x + 2 x_ 3x − 1 x3x − 20_24

476Chapter 7 Solving EquationsExtending the Concepts47. Right triangle ABC has legs of length 5. Point C is theordered pair (6, 2). Find the coordinates of A and B.A = (1, 2), B = (6, 7)yB48. Right triangle ABC has legs of length 7. Point C is theordered pair (−8, −3). Find the coordinates of A and B.A = (−1, −3), B = (−8, −10)xCAACxBy49. Rectangle ABCD has a length of 5 and a width of 3.Point D is the ordered pair (7, 2). Find points A, B,and C.A = (2, 2), B = (2, 5), C = (7, 5)y50. Rectangle ABCD has a length of 5 and a width of 3. PointD is the ordered pair (−1, 1). Find points A, B, and C.A = (−6, 1), B = (−6, 4), C = (−1, 4)yBCBCADxxAD

Chapter 7 SummaryCombining Similar Terms [7.1]Two terms are similar terms if they have the same variable part. The expressions7x and 2x are similar because the variable part in each is the same. Similar termsare combined by using the distributive property.EXAMPLEs1. 7x + 2x = (7 + 2)x= 9xFinding the Value of an Algebraic Expression [7.1]An algebraic expression is a mathematical expression that contains numbersand variables. Expressions that contain a variable will take on different valuesdepending on the value of the variable.2. When x = 5, the expression2x + 7 becomes2(5) + 7 = 10 + 7 = 17The Solution to an Equation [7.2]A solution to an equation is a number that, when used in place of the variable,makes the equation a true statement.The Addition Property of Equality [7.2]Let A, B, and C represent algebraic expressions.IfA = Bthen A + C = B + CIn words: Adding the same quantity to both sides of an equation will not changethe solution.3. We solve x − 4 = 9 by adding 4to each side.x − 4 = 9x − 4 + 4 = 9 + 4x + 0 = 13x = 13The Multiplication Property of Equality [7.3]Let A, B, and C represent algebraic expressions, with C not equal to 0.IfA = Bthen AC = BCIn words: Multiplying both sides of an equation by the same nonzero number willnot change the solution to the equation. This property holds for division as well.4. Solve _1 x = 5.3 1 _3 x = 53 ⋅ _1 3 x = 3 ⋅ 5x = 15Steps Used to Solve a Linear Equation in One Variable [7.4]Step 1 Simplify each side of the equation.Step 2 Use the addition property of equality to get all variable terms on one sideand all constant terms on the other side.5. 2(x − 4) + 5 = −112x − 8 + 5 = −112x − 3 = −112x − 3 + 3 = −11 + 32x = −8_2x = −8_2 2 x = −4Chapter 7Summary477

478Chapter 7 Solving EquationsStep 3 Use the multiplication property of equality to get just one x isolated oneither side of the equation.Step 4 Check the solution in the original equation if necessary.If the original equation contains fractions, you can begin by multiplying each sideby the LCD for all fractions in the equation.Evaluating Formulas [7.6]6. When w = 8 and l = 13the formula P = 2w + 2lbecomes P = 2 ⋅ 8 + 2 ⋅13= 16 + 26= 42In mathematics, a formula is an equation that contains more than one variable.For example, the formula for the perimeter of a rectangle is P = 2l + 2w. Weevaluate a formula by substituting values for all but one of the variables and thensolving the resulting equation for that variable.The Rectangular Coordinate System [7.7]The rectangular coordinate system consists of two number lines, called axes,which intersect at right angles. The point at which the axes intersect is called theorigin.y-axisQuadrantII−5 −4 −3 −2QuadrantIII54321−1−1−2−3−4−5QuadrantI1 2 3 4 5OriginQuadrantIVx-axisGraphing Ordered Pairs [7.7]To graph an ordered pair (a, b) on the rectangular coordinate system, we start atthe origin and move a units right or left (right if a is positive, left if a is negative).From there we move b units up or down (up if b is positive, down if b is negative).The point where we end up is the graph of the ordered pair (a, b).

Chapter 7 ReviewSimplify the expressions by combining similar terms. [7.1]1. 10x + 7x2. 8x − 12x3. 2a + 9a + 3 − 64. 4y − 7y + 8 − 1017x−4x11a − 3−3y − 25. 6x − x + 46. −5a + a + 4 − 37. 2a − 6 + 8a + 28. 12y − 4 + 3y − 95x + 4−4a + 110a − 415y − 13Find the value of each expression when x is 4. [7.1]9. 10x + 210. 5x − 1211. −2x + 912. −x + 842814Find the value of each expression when x is −5.13. 3y + 614. 9 − 2y15. 12 + y16. −6y − 20−91971017. Is x = −3 a solution to 5x − 2 = −17? [7.2]Yes18. Is x = 4 a solution to 3x − 2 = 2x + 1? [7.2]NoSolve the equations. [7.2, 7.3, 7.4]19. x − 5 = 420. −x + 3 + 2x = 6 − 721. 2x + 1 = 722. 3x − 5 = 19−43223. 2x + 4 = 3x − 524. 4x + 8 = 2x − 1025. 3(x − 2) = 926. 4(x − 3) = −209−95−227. 3(2x + 1) − 4 = −7−128. 4(3x + 1) = −2(5x − 2)029. 5x + 3 _8 = −1 _4 30. _7_− 1 8 5x − 2 _5 = 131. 3(2x − 5) = 4x + 3932. _2 (6x − 9) = 6x − 23−233. 3x − 11 = 2(x − 2)734. 4(3x − 9) = 2(4x + 6)1235. 5x + 9 = 4x − 7−1636. 4(x − 6) = 8837. 5x − 7 = 3238. 3 _x − 1 _4 = 2 4 _3 Chapter 7Review479

480Chapter 7 Solving Equations39. Number Problem The sum of a number and −3 is −5.Find the number. [7.5]The number is −2.40. Number Problem If twice a number is added to 3, theresult is 7. Find the number. [7.5]The number is 2.41. Number Problem Three times the sum of a number and2 is −6. Find the number. [7.5]The number is −4.42. Number Problem If 7 is subtracted from twice a number,the result is 5. Find the number. [7.5]The number is 6.43. Geometry The length of a rectangle is twice its width.If the perimeter is 42 meters, find the length and thewidth. [7.5]The length is 14 m, and the width is 7 m.44. Age Problem Patrick is 3 years older than Amy. In 5 yearsthe sum of their ages will be 31. How old are they now?[7.5]Amy is 9 years old; and Patrick is 12 years old.45. Geometry Two angles are complementary angles. Ifone angle is 5 times larger than the other angle, findthe two angles. [7.5]15° and 75°46. Geometry Two angles are supplementary. If one angle is85˚, what is the other angle? [7.5]95°47. Geometry The biggest angle in a triangle is 6 times biggerthan the smallest angle. The third angle is half thelargest angle. Find the three angles. [7.5]18°, 54°, 108°48. Number Problem If two times the sum of a number and 6 isincreased by 12, the result is 16. Find the number. [7.5]−4In Problems 49-52, use the equation 3x + 2y = 6 to find y. [7.6]49. x = −2y = 650. x = 6y = −651. x = 0y = 352. x = 1 _3 y = 5 _2 In Problems 53–55, use the equation 3x + 2y = 6 to find x when y has the given value. [7.6]53. y = 3x = 054. y = −3x = 455. y = 0x = 2In Problems 56–59, use the equation y = _2 x − 4 to find x when y has the given value. [7.6]356. y = 0x = 657. y = −6x = −358. y = 4x = 1259. y = −10x = −9In Problems 9–16, plot the points. [7.7]60. (4, 2) 61. (4, −2)y62. (−4, 2) 63. (−4, −2)64. (4, 0) 65. (0, 4)66. (0, −4) 67. (−4, 0)565.4362.60.2167.64.5 43 21 1 2 3 4 5163.61.2366.45x

Chapter 7 Cumulative ReviewSimplify.1. 5,309 + 687 5,996 2. 7 _11 + 4 _5 _ 79553. 11.09 − 6.531 4.559 4. 4 1 _8 − 13 _4 2 3 _8 = 124 _55 5. 2305(407) 938,135 6. 0.002(230) 0.46_______7. 314) 13,188 42 8. _632 ÷ 9 _48 19. Round the number 435,906 to the nearest tenthousand. 440,00010. Write 0.48 as a fraction in lowest terms._1225 11. Change _76 to a mixed number in lowest terms.126_1 3 12. Find the difference of 0.45 and _2 5 .0.0513. Write the decimal 0.8 as a percent.80%Use the table given in Chapter 7 to make the followingconversion.14. 7 kilograms to pounds15.4 lb15. Write 124% as a fraction or mixed number in lowestterms. 1_625 16. What percent of 60 is 21?35%Simplify.17. _1 33 + _1 29 _ 4 18. 8x + 9 − 9x − 14 −x − 581−3(−8) + 4(−2)19. −| −7 | −7 20. __ 811 − 921. 19 − 5(7 − 4) 4 22. √ — 25 + √ — 16 9Solve.23. 3 _8 y = 21 56 24. −3(2x − 1) = 3(x + 5) −4 _3 25. _3.6 = 4.5_4 x 526. Write the following ratio as a fraction in lowest terms:0.04 to 0.32 _1 8 27. Subtract −3 from 5.828. Surface Area Find the surface area of a rectangularsolid with length 7 inches, width 3 inches, and height 2inches. 82 in 229. Age Ben is 8 years older than Ryan. In 6 years the sumof their ages will be 38. How old are they now?Ben is 17, Ryan is 930. Gas Mileage A truck travels 432 miles on 27 gallons ofgas. What is the rate of gas mileage in miles per gallon.16 mpg31. Discount A surfboard that usually sells for $400 ismarked down to $320. What is the discount? What isthe discount rate?$80, 20%32. Geometry Find the length of the hypotenuse of a righttriangle with sides of 5 and 12 meters.13 m33. Cost of Coffee If coffee costs $6.40 per pound, howmuch will 2 lb 4 oz, cost?$14.4034. Interest If $1,400 is invested at 6% simple interest for90 days, how much interest is earned?$2135. Wildflower Seeds C.J. works in a nursery, and one ofhis tasks is filling packets of wildflower seeds. If eachpacket is to contain _1 pound of seeds, how many4packets can be filled from 16 pounds of seeds?6436. Commission A car stereo salesperson receives a commissionof 8% on all units he sells. If his total sales forMarch are $9,800, how much money in commissionwill he make?$78437. Volume How many 8 fluid ounce glasses of water will ittake to fill a 15-gallon aquarium?240Internet Access Speed The graph below gives the speed ofthe most common modems used for Internet access.Modem Speeds900,000800,000786,000700,000600,000512,000500,000400,000300,000200,000128,000100,00028,000 56,000028K 56K ISDN Cable DSLModem Type38. How much faster is cable than a 56K modem?456,000 bps39. What is the ratio of the speed of cable to the speed ofan ISDN modem?_ 4 1 Speed (bps)Chapter 7Cumulative Review481

Chapter 7 TestSimplify each expression by combining similar terms.1. 9x − 3x + 7 − 126x − 53. 3(2x − 6) − 4x2x − 182. 4b − 1 − b − 33b − 44. 4(5x − 6) − 6x + 1214x − 12Find the value of each expression when x = 3.5. 3x − 12−36. −x + 1297. Is x = −1 a solution to 4x − 3 = −7?Yes8. Is x = 3 a solution to 3x − 7 = −2?No9. Use the equation 4x + 3y = 12 to find y when x = −3.810. Use the equation y = _4 x + 7 to find x when y = −1.3−6Solve each equation.11. x − 7 = −3 4 12. a − 2.9 = −7.8 −4.913. 2 _3 y = 18 27 14. 7 _x − 1 _6 = 1 615. 3x − 7 = 5x + 1 −4 16. 2(x − 5) = −8 124. Age problem Karen is 5 years younger than Susan.Three years ago, the sum of their ages was 11. How oldare they now?Susan is 11 years old, and Karen is 6 years old.25. Geometry The largest angle in a triangle is 3 times biggerthan another angle, which is twice as big as thesmallest angle. Find the three angles.20°, 40°, 120°26. Coin Problem A coin collection has seven more dimesthan quarters. If the collection has a value of $2.10,how many quarters and dimes are in the collection?4 quarters, 11 dimes27. Geometry Two angles are complementary. If one angleis 3 times bigger than the other angle, find the twoangles.22.5°, and 67.5°28. Perimeter of a Rectangle If a rectangle has a length of13 meters and a perimeter of 35 meters, find the widthof the rectangle.4.5 m29. Google Earth The length of the base of the Sphinx is 12times longer than the width. If the perimeter of thebase is 520 feet. Find the length and the width.17. 3(2x + 3) = −3(x − 5) _2 18. 6(3x − 2) − 8 = 4x − 6 1319. _4 (6x − 9) = 2(4x − 6) 20. 3(x − 7) = − 6 53all real numbers21. Number Problem Twice the sum of a number and 3 is−10. Find the number.The number is −8.20 ft, 240 ftImage © 2009 GeoEyeImage © 2009 DigitalGlobe© 2009 Cnes/Spot Image22. Hot Air Balloon The first successful crossing of theAtlantic in a hot air balloon was made in August 1978by Maxie Anderson, Ben Abruzzo, and Larry Newmanof the United States. The 3,100 mile trip took approximately140 hours. Use the formula r = _d to find theirtaverage speed to the nearest whole number.22 mi/hr23. Geometry The length of a rectangle is 4 centimeterslonger than its width. If the perimeter is 28 centimeters,find the length and the width.The length is 9 cm, and the width is 5 cm.30. Plot the following points: (3, 2), (−3, 2), (3, 0), (0, −2).543215432112345y1 2 3 4 5482Chapter 7 Solving Equations

Chapter 7 ProjectsSolving Equationsgroup PROJECTThe Equation GameNumber of PeopleTime NeededEquipmentBackgroundProcedure2–530 minutesPer group: deck of cards, timer or clock, penciland paper, copy of rules.The Equation Game is a fun way to practiceworking with equations.Remove all the face cards from the deck. Aceswill be 1’s. The dealer deals four cards face up,a fifth card face down. Each player writes downthe four numbers that are face up. Set the timerfor 5 minutes, then flip the fifth card. Each playerwrites down equations that use the numbers onthe first four cards to equal the number on thefifth card. When the five minutes are up, figureout the scores. An equation that usesExample1 of the four cards scores 0 points2 of the four cards scores 4 points3 of the four cards scores 9 points4 of the four cards scores 16 pointsCheck the other players’ equations. If you findan error, you get 7 points. The person with themistake gets no points for that equation.The first four cards are a four, a nine, an ace,and a two. The fifth card is a seven. Here aresome equations you could make:9 − 2 = 74 + 2 + 1 = 79 − (4 − 2) = 71One solution (9 points)*This project was adapted from www.exploratorium.edu/math_explorer/fantasticFour.html.Chapter 7Projects483

RESEARCH PROJECTAlgebraic SymbolismAlgebra made a beginning as early as 1850 b.c.in Egypt. However, the symbols we use in algebratoday took some time to develop. For example,the algebraic use of letters for numbersbegan much later with Diophantus, a mathematicianfamous for studying Diophantine equations.In the early centuries, the full words plus,minus, multiplied by, divided by, and equals werewritten out. Imagine how much more difficultyour homework would be if you had to writeout all these words instead of using symbols.Algebraists began to come up with a system ofsymbols to make writing algebra easier. At first,not everyone agreed on the symbols to be used.For example, the present division sign ÷ wasoften used for subtraction. People in differentcountries used different symbols: the Italianspreferred to use p and m for plus and minus,while the less traditional Germans were startingto use + and −.Research the history of algebraic symbolism.Find out when the algebraic symbols we usetoday (such as letters to represent variables, +,−, ÷, ⋅, a/b and a _b came into common use.Summarize your results in an essay.484Chapter 7 Solving Equations

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