Chapter 8 - XYZ Custom Plus

Linear Equationsand Inequalities8IntroductionImage © 2009 DigitalGlobeChapter Outline8.1 Simplifying Expressions8.2 Addition Property ofEquality8.3 Multiplication Property ofEquality8.4 Solving Linear Equations8.5 Formulas8.6 Applications8.7 More Applications8.8 Linear Inequalities8.9 Compound InequalitiesThe Google Earth image is another view of the Leaning Tower of Pisa. As wementioned in the previous chapter, Pisa is the birthplace of the Italian mathematicianFibonacci. It was in Pisa that Fibonacci wrote the first European algebrabook Liber Abaci. It was published in 1202. Below is the cover of an English translationof that book, published in 2003, along with a quote from the MathematicsAssociation of America.Permission granted by Judith Sigler FellMAA Online, March 2003:“The Liber abaci of Leonardo Pisano(today commonly called Fibonacci)is one of the fundamental works ofEuropean mathematics. No other bookdid more to establish the basic frameworkof arithmetic and algebra as theydeveloped in the Western world.”In his book, Fibonacci shows how to solve linear equations in one variable,which is the main topic in this chapter.485

Chapter PretestSimplify each of the following expressions.1. 5y − 3 − 6y + 4−y + 12. 3x − 4 + x + 34x − 13. 4 − 2(y − 3) − 6−2y + 44. 3(3x − 4) − 2(4x + 5)x − 225. Find the value of 3x + 12 + 2x when x = −3. −3 6. Find the value of x 2 − 3xy + y 2 when x = −2 andy = −4. −4Solve the following equations._7. 3x − 2 = 7 3 8. 4y + 15 = y −5 9. ​ 1 4 ​x − ​1 _12 ​= ​1 _3 ​x − ​1 _​ 1 10. −3(3 − 2x) − 7 = 8 4611. 3x − 9 = −6112. 0.05 + 0.07(100 − x) = 3.25513. 4(t − 3) + 2(t + 4) = 2t − 16−315. What number is 40% of 56? 22.4 16. 720 is 24% of what number? 3,00014. 4x − 2(3x − 1) = 2x − 8​ 5_2 ​17. If 3x − 4y = 16, find y when x = 4. −1 18. If 3x − 4y = 16, find x when y = 2. 819. Solve 2x + 6y = 12 for y. y = −​ 1_ 3 ​x + 2 20. Solve x 2 = v 2 + 2ad for a. a = ​_x 2 − v 22d​Solve each inequality.21. ​_1 ​x − 2 > 3 x > 10 22. −6y ≤ 24 y ≥ −4 23. 0.3 − 0.2x < 1.1 x > −4 24. 3 − 2(n − 1) ≥ 9 n ≤ −22Write an inequality whose solution is the given graph.25. x > −3 26. −1 ≤ x ≤ 3–3 0 –1 3Getting Ready for Chapter 8Simplify each of the following.1. −3 + 742. −10 − 4−143. 9 − (−24)334. −6(5) − 5−355. −3(2y + 1)−6y − 36. 8.1 + 2.710.87. −​ 3 _4 ​+ ​ −​ 1 _2 ​ ​_8. −​ 7_−​ 5 4 ​ _−​ 6 5 ​10 ​+ ​ −​ 1 _2 ​ ​9. ​_1 5 ​(5x)x10. 4 ​ ​ 1 _4 ​x ​x11. ​ 3 _2 ​​ ​ 2 _3 ​x ​x12. ​ 5 _2 ​​ ​ 2 _5 ​x ​x13. −1(3x + 4)−3x − 414. −2.4 + ( − 7.3)−9.715. 0.04(x + 7,000)0.04x + 28016. 0.09(x + 2,000)0.09x + 180Apply the distributive property and then simplify if possible.17. 3(x − 5) + 43x − 1118. 5(x − 3) + 25x − 1319. 5(2x − 8) − 310x − 4320. 4(3x − 5) − 212x − 22486Chapter 8 Linear Equations and Inequalities

Simplifying Expressions 8.1If a cellular phone company charges $35 per month plus $0.25 for each minute,or fraction of a minute, that you use one of their cellular phones, then the amountof your monthly bill is given by the expression 35 + 0.25t. To find the amount youwill pay for using that phone 30 minutes in one month, you substitute 30 for t andsimplify the resulting expression. This process is one of the topics we will study inthis section.A Combining Similar TermsAs you will see in the next few sections, the first step in solving an equation is tosimplify both sides as much as possible. In the first part of this section, we will practicesimplifying expressions by combining what are called similar (or like) terms.For our immediate purposes, a term is a number, or a number and one or morevariables multiplied together. For example, the number 5 is a term, as are theexpressions 3x, −7y, and 15xy.ObjectivesA Simplify expressions by combiningsimilar terms.B Simplify expressions by applyingthe distributive property and thencombining similar terms.C Calculate the value of an expressionfor a given value of the variable.DefinitionTwo or more terms with the same variable part are called similar (or like)terms.The terms 3x and 4x are similar because their variable parts are identical. Likewise,the terms 18y, −10y, and 6y are similar terms.To simplify an algebraic expression, we simply reduce the number of terms inthe expression. We accomplish this by applying the distributive property alongwith our knowledge of addition and subtraction of positive and negative realnumbers. The following examples illustrate the procedure.Example 1Simplify 3x + 4x by combining similar terms.Practice ProblemsSimplify by combining similarterms.1. 5x + 2xSolution 3x + 4x = (3 + 4)x Distributive property= 7x Addition of 3 and 4Example 2Simplify 7a − 10a by combining similar terms.Solution 7a − 10a = (7 − 10)a Distributive property= −3a Addition of 7 and −102. 12a − 7aExample 3Simplify 18y − 10y + 6y by combining similar terms.Solution 18y − 10y + 6y = (18 − 10 + 6)y Distributive property= 14y Addition of 18, −10, and 63. 6y − 8y + 5yWhen the expression we intend to simplify is more complicated, we use thecommutative and associative properties first.Answers1. 7x 2. 5a 3. 3y8.1 Simplifying Expressions487

488Chapter 8 Linear Equations and InequalitiesSimplify each expression.4. 2x + 6 + 3x − 5Example 4Simplify the expression 3x + 5 + 2x − 3.Solution 3x + 5 + 2x − 3 = 3x + 2x + 5 − 3 Commutative property= (3x + 2x) + (5 − 3) Associative property= (3 + 2)x + (5 − 3) Distributive property= 5x + 2 Addition5. 8a − 2 − 4a + 5Example 5Simplify the expression 4a − 7 − 2a + 3.Solution 4a − 7 − 2a + 3 = (4a − 2a) + (−7 + 3) Commutative and associativeproperties= (4 − 2)a + (−7 + 3) Distributive property= 2a − 4 Addition6. 9x + 1 − x − 6Example 6Simplify the expression 5x + 8 − x − 6.Solution5x + 8 − x − 6 = (5x − x) + (8 − 6) Commutative and associative properties= (5 − 1)x + (8 − 6) Distributive property= 4x + 2 AdditionBSimplifying Expressions Containing ParenthesesIf an expression contains parentheses, it is often necessary to apply the distributiveproperty to remove the parentheses before combining similar terms.7. Simplify the expression.4(3x − 5) − 2Example 7Simplify the expression 5(2x − 8) − 3.Solution We begin by distributing the 5 across 2x − 8. We then combine similarterms:5(2x − 8) − 3 = 10x − 40 − 3 Distributive property= 10x − 438. Simplify 5 − 2(4y + 1).Example 8Simplify 7 − 3(2y + 1).Solution By the rule for order of operations, we must multiply before weadd or subtract. For that reason, it would be incorrect to subtract 3 from 7 first.Instead, we multiply −3 and 2y + 1 to remove the parentheses and then combinesimilar terms:7 − 3(2y + 1) = 7 − 6y − 3 Distributive property= −6y + 49. Simplify 6(x − 3) − (7x + 2).Example 9Simplify 5(x − 2) − (3x + 4).Solution We begin by applying the distributive property to remove the parentheses.The expression −(3x + 4) can be thought of as −1(3x + 4). Thinking of itin this way allows us to apply the distributive property:−1(3x + 4) = −1(3x) + (−1)(4)= −3x − 4Answers4. 5x + 1 5. 4a + 3 6. 8x − 57. 12x − 22 8. −8y + 39. −x − 20The complete solution looks like this:5(x − 2) − (3x + 4) = 5x − 10 − 3x − 4 Distributive property= 2x − 14 Combine similar terms

8.1 Simplifying Expressions489As you can see from the explanation in Example 9, we use the distributiveproperty to simplify expressions in which parentheses are preceded by a negativesign. In general we can write−(a + b) = −1(a + b)= −a + (−b)= −a − bThe negative sign outside the parentheses ends up changing the sign of eachterm within the parentheses. In words, we say “the opposite of a sum is the sumof the opposites.”C The Value of an ExpressionRecall from Chapter 1, an expression like 3x + 2 has a certain value depending onwhat number we assign to x. For instance, when x is 4, 3x + 2 becomes 3(4) + 2,or 14. When x is −8, 3x + 2 becomes 3(−8) + 2, or −22. The value of an expressionis found by replacing the variable with a given number.ExamplesFind the value of the following expressions by replacingthe variable with the given number.Value ofValue ofExpression the Variable the Expression10. 3x − 1 x = 2 3(2) − 1 = 6 − 1 = 511. 7a + 4 a = −3 7(−3) + 4 = −21 + 4 = −1712. 2x − 3 + 4x x = −1 2(−1) − 3 + 4(−1)= −2 − 3 + (−4) = −913. 2x − 5 − 8x x = 5 2(5) − 5 − 8(5)= 10 − 5 − 40 = −3514. y 2 − 6y + 9 y = 4 4 2 − 6(4) + 9 = 16 − 24 + 9 = 110. Find the value of 4x − 7 whenx = 3.11. Find the value of 2a + 4 whena = −5.12. Find the value of 2x − 5 + 6xwhen x = −2.13. Find the value of 7x − 3 − 4xwhen x = 10.14. Find the value of y 2 − 10y + 25when y = −2.Simplifying an expression should not change its value; that is, if an expressionhas a certain value when x is 5, then it will always have that value no matter howmuch it has been simplified, as long as x is 5. If we were to simplify the expressionin Example 13 first, it would look like2x − 5 − 8x = −6x − 5When x is 5, the simplified expression −6x − 5 is−6(5) − 5 = −30 − 5 = −35It has the same value as the original expression when x is 5.We also can find the value of an expression that contains two variables if weknow the values for both variables.Answers10. 5 11. −6 12. −21 13. 2714. 49

8.1 Problem Set491Problem Set 8.1All answers appear in the Instructors Edition only.A Simplify the following expressions. [Examples 1–6]1. 3x − 6x−3x2. 7x − 5x2x3. −2a + a−a4. 3a − a2a5. 7x + 3x + 2x12x6. 8x − 2x − x5x7. 3a − 2a + 5a6a8. 7a − a + 2a8a9. 4x − 3 + 2x6x − 310. 5x + 6 − 3x2x + 611. 3a + 4a + 57a + 512. 6a + 7a + 813a + 813. 2x − 3 + 3x − 25x − 514. 6x + 5 − 2x + 34x + 815. 3a − 1 + a + 34a + 216. −a + 2 + 8a − 77a − 517. −4x + 8 − 5x − 10−9x − 218. −9x − 1 + x − 4−8x − 519. 7a + 3 + 2a + 3a12a + 320. 8a − 2 + a + 5a14a − 2B Apply distributive property, then simplify. [Examples 7–9]21. 5(2x − 1) + 410x − 122. 2(4x − 3) + 28x − 423. 7(3y + 2) − 821y + 624. 6(4y + 2) − 724y + 525. −3(2x − 1) + 5−6x + 826. −4(3x − 2) − 6−12x + 227. 5 − 2(a + 1)−2a + 328. 7 − 8(2a + 3)−16a − 17

492Chapter 8 Linear Equations and InequalitiesB Simplify the following expressions. [Examples 7–9]29. 6 − 4(x − 5)−4x + 2630. 12 − 3(4x − 2)−12x + 1831. −9 − 4(2 − y) + 14y − 1632. −10 − 3(2 − y) + 33y − 1333. −6 + 2(2 − 3x) + 1−6x − 134. −7 − 4(3 − x) + 14x − 1835. (4x − 7) − (2x + 5)2x − 1236. (7x − 3) − (4x + 2)3x − 537. 8(2a + 4) − (6a − 1)10a + 3338. 9(3a + 5) − (8a − 7)19a + 5239. 3(x − 2) + (x − 3)4x − 940. 2(2x + 1) − (x + 4)3x − 241. 4(2y − 8) − (y + 7)7y − 3942. 5(y − 3) − (y − 4)4y − 1143. −9(2x + 1) − (x + 5)−19x − 1444. −3(3x − 2) − (2x + 3)−11x + 3C Evaluate the following expressions when x is 2. [Examples 10–14]45. 3x − 1546. 4x + 31147. −2x − 5−948. −3x + 6049. x 2 − 8x + 16450. x 2 − 10x + 25951. (x − 4) 2452. (x − 5) 29C Find the value of each expression when x is −5. Then simplify the expression, and check to see that it has the samevalue for x = −5.53. 7x − 4 − x − 3−3754. 3x + 4 + 7x − 6−5255. 5(2x + 1) + 4−4156. 2(3x − 10) + 5−45

8.1 Problem Set493C Find the value of each expression when x is −3 and y is 5. [Examples 15–16]57. x 2 − 2xy + y 26458. x 2 + 2xy + y 2459. (x − y) 26460. (x + y) 2461. x 2 + 6xy + 9y 214462. x 2 + 10xy + 25y 248463. (x + 3y) 214464. (x + 5y) 2484C Find the value of 12x − 3 for each of the following values of x.65. ​ 1 _2 ​366. ​ 1 _3 ​167. ​ 1 _4 ​068. ​ 1 _6 ​−169. ​ 3 _2 ​1570. ​ 2 _3 ​571. ​ 3 _4 ​672. ​ 5 _6 ​773. Fill in the tables below to find the sequences formedby substituting the first four counting numbers into theexpressions 3n and n 3 .74. Fill in the tables below to find the sequences formedby substituting the first four counting numbers into theexpressions 2n − 1 and 2n + 1.a.n 1 2 3 4a.n 1 2 3 43n 3 6 9 122n − 1 1 3 5 7b.n 1 2 3 4b.n 1 2 3 4n 3 1 8 27 642n + 1 3 5 7 9C Find the sequences formed by substituting the counting numbers, in order, into the following expressions. [Examples 17]75. 3n − 21, 4, 7, 10, . . . an arithmetic sequence76. 2n − 3−1, 1, 3, 5, . . . an arithmeticsequence77. n 2 − 2n + 10, 1, 4, 9, . . . a sequence ofsquares78. (n − 1) 20, 1, 4, 9, . . . a sequence ofsquaresHere are some problems you will see later in the book.Simplify.79. 7 − 3(2y + 1)−6y + 480. 4(3x − 2) − (6x − 5)6x −381. 0.08x + 0.09x0.17x82. 0.04x + 0.05x0.09x83. (x + y) + (x − y)2x84. (−12x − 20y) + (25x + 20y)13x85. 3x + 2(x − 2)5x − 486. 2(x − 2) + 3(5x)17x − 487. 4(x + 1) + 3(x − 3)7x − 588. 5(x + 2) + 3(x − 1)8x + 789. x + (x + 3)(−3)−2x − 990. x − 2(x + 2)−x − 4

8.1 Problem Set495113. Cellular Phone Rates A cellular phone companycharges $35 per month plus $0.25 for each minute, orfraction of a minute, that you use one of their cellularphones. The expression 35 + 0.25t gives the amountof money you will pay for using one of their phonesfor t minutes a month. Find the monthly bill for usingone of their phones.a. 10 minutes in a month $37.50b. 20 minutes in a month $40.00c. 30 minutes in a month $42.50114. Cost of Bottled Water A water bottling company charges$7.00 per month for their water dispenser and $1.10 foreach gallon of water delivered. If you have g gallons ofwater delivered ina month, then theWBCWATER BOTTLE CO.expressionMONTHLY BILL7 + 1.1g gives the234 5th StreetDUE 07/23/08amount of yourbill for that month.Find the monthlybill for each of thefollowingdeliveries.Glendora, CA 91740Water dispenserGallons of water18$7.00$2.00$23.00a. 10 gallons $18b. 20 gallons $29c. 30 gallons $40115. Taxes We all have to pay taxes. Suppose that 21%of your monthly pay is withheld for federal incometaxes and another 8% is withheld for Social Security,state income tax, and other miscellaneous items. IfG is your monthly pay before any money is deducted(your gross pay), then the amount of money that youtake home each month is given by the expressionG − 0.21G − 0.08G. Simplify this expression and thenfind your take-home pay if your gross pay is $1,250per month.0.71G; $887.50116. Taxes If you work H hours a day and you pay 21% ofyour income for federal income tax and another 8%for miscellaneous deductions, then the expression0.21H + 0.08H tells you how many hours a day youare working to pay for those taxes and miscellaneousdeductions. Simplify this expression. If you work 8hours a day under these conditions, how many hoursdo you work to pay for your taxes and miscellaneousdeductions?0.29H; 2.32 hours117. Cars The chart shows the fastest cars in America.How fast is a car traveling if its speed is 56 miles perhour less than half the speed of the Saleen S 7 TwinTurbo?74 miles per hour118. Sound The chart shows the decibel level of varioussounds. The loudness of a loud rock concert is 5 decibelsless than twice the loudness of normal conversation.Find the sound level of a rock concert.115 dBReady for the RacesSounds Around UsFord GT 205 mphNormal Conversation60 dBEvans 487 210 mphSaleen S7 Twin Turbo 260 mphFootball Stadium117 dBSSC Ultimate Aero 273 mphBlue Whale188 dBSource: Forbes.comSource: www.4to40.com

496Chapter 8 Linear Equations and InequalitiesGetting Ready for the Next SectionProblems under this heading, Getting Ready for the Next Section, are problems that you must be able to work to understandthe material in the next section. The problems below are exactly the types of problems you will see in the explanationsand examples in the next section.Simplify.119. 17 − 5120. 12 + (−2)121. 2 − 5122. 25 − 201210−35123. −2.4 + (−7.3)−9.7124. 8.1 + 2.710.8125. −​ 1 _2 ​+ ​ −​ 3 _4 ​ ​_126. −​ 1_−​ 5 4 ​ _−​ 5 6 ​6 ​+ ​ −​ 2 _3 ​ ​127. 4(2 ⋅ 9 − 3) − 7 ⋅ 9128. 5(3 ⋅ 45 − 4) − 14 ⋅ 45129. 4(2a − 3) − 7a130. 5(3a − 4) − 14a−325a − 12a − 20131. −3 − ​_1 2 ​132. −5 − ​_1 3 ​133. ​_4 5 ​+ ​1 _10 ​+ ​3 _8 ​134. ​_310 ​+ ​7 _25 ​+ ​3 _4 ​_−​ 7 2 ​ _−​ 163 ​ ​_ 5140 ​ ​_ 133100 ​135. Find the value of 2x − 3 when x is 5.7136. Find the value of 3x + 4 when x is −2.−2Maintaining Your SkillsFrom this point on, each problem set will contain a number of problems under the heading Maintaining Your Skills. Theseproblems cover the most important skills you have learned in previous sections and chapters. By working these problemsregularly, you will keep yourself current on all the topics we have covered and possibly need less time to study for testsand quizzes.137. ​ 1 _8 ​− ​1 _6 ​138. ​_x_−​ 1 24 ​8 ​− ​ x _6 ​−​ x _24 ​139. ​ 5 _9 ​− ​4 _3 ​140. ​_x_−​ 7 9 ​9 ​− ​x _3 ​−​ 2x _9 ​141. −​ 7 _30 ​+ ​5 _28 ​_142. −​ 11105420 ​ ​_ 1142 ​−​ 23 _​+ ​11_30 ​

Addition Property of EqualityWhen light comes into contact with any object, it is reflected, absorbed, andtransmitted, as shown here.Reflected8.2ObjectivesA Check the solution to an equationby substitution.B Use the addition property ofequality to solve an equation.AbsorbedTransmittedFor a certain type of glass, 88% of the light hitting the glass is transmitted throughto the other side, whereas 6% of the light is absorbed into the glass. To find thepercent of light that is reflected by the glass, we can solve the equation88 + R + 6 = 100Solving equations of this type is what we study in this section. To solve an equationwe must find all replacements for the variable that make the equation a truestatement.ASolutions to EquationsDefinitionThe solution set for an equation is the set of all numbers that when used inplace of the variable make the equation a true statement.For example, the equation x + 2 = 5 has solution set {3} because when x is 3the equation becomes the true statement 3 + 2 = 5, or 5 = 5.Example 1Is 5 a solution to 2x − 3 = 7?Solution We substitute 5 for x in the equation, and then simplify to see if atrue statement results. A true statement means we have a solution; a false statementindicates the number we are using is not a solution.When x = 5the equation 2x − 3 = 7becomes 2(5) − 3 ≟ 710 − 3 ≟ 77 = 7 A true statementPractice Problems1. Is 4 a solution to 2x + 3 = 7?NoteWe can use a questionmark over the equalsigns to show that wedon’t know yet whether the twosides of the equation are equal.Because x = 5 turns the equation into the true statement 7 = 7, we know 5 is asolution to the equation.Example 2Is −2 a solution to 8 = 3x + 4?SolutionSubstituting −2 for x in the equation, we have2. Is ​_4 ​a solution to 8 = 3x + 4?38 ≟ 3(−2) + 48 ≟ −6 + 48 = −2 A false statementSubstituting −2 for x in the equation produces a false statement. Therefore,x = −2 is not a solution to the equation.Answers1. No 2. Yes8.2 Addition Property of Equality497

498Chapter 8 Linear Equations and InequalitiesThe important thing about an equation is its solution set. We therefore make thefollowing definition to classify together all equations with the same solution set.DefinitionTwo or more equations with the same solution set are said to be equivalentequations.Equivalent equations may look different but must have the same solution set.3. a. Are the equations x + 3 = 9and x = 6 equivalent?b. Are the equations a − 5 = 3and a = 6 equivalent?c. Are y + 2 = 5, y − 1 = 2, andy = 3 equivalent equations?Example 3a. x + 2 = 5 and x = 3 are equivalent equations because both havesolution set {3}.b. a − 4 = 3, a − 2 = 5, and a = 7 are equivalent equationsbecause they all have solution set {7}.c. y + 3 = 4, y − 8 = −7, and y = 1 are equivalent equationsbecause they all have solution set {1}.BAddition Property of EqualityIf two numbers are equal and we increase (or decrease) both of them by the sameamount, the resulting quantities are also equal. We can apply this concept toequations. Adding the same amount to both sides of an equation always producesan equivalent equation—one with the same solution set. This fact about equationsis called the addition property of equality and can be stated more formally asfollows.NoteWe will use this propertymany times inthe future. Be sure youunderstand it completely by thetime you finish this section.Addition Property of EqualityFor any three algebraic expressions A, B, and C,ifA = Bthen A + C = B + CIn words: Adding the same quantity to both sides of an equation will notchange the solution set.This property is just as simple as it seems. We can add any amount to bothsides of an equation and always be sure we have not changed the solution set.Consider the equation x + 6 = 5. We want to solve this equation for the valueof x that makes it a true statement. We want to end up with x on one side of theequal sign and a number on the other side. Because we want x by itself, we willadd −6 to both sides:x + 6 + (−6) = 5 + (−6)x + 0 = −1Addition property of equalityAdditionx = −1All three equations say the same thing about x. They all say that x is −1. All threeequations are equivalent. The last one is just easier to read.Here are some further examples of how the addition property of equality can beused to solve equations.Answer3. a. Yes b. No c. Yes

8.2 Addition Property of Equality499Example 4Solve the equation x − 5 = 12 for x.4. Solve x − 3 = 10 for x.Solutionboth sides:Because we want x alone on the left side, we choose to add +5 tox − 5 + 5 = 12 + 5 Addition property of equalityx + 0 = 17x = 17To check our solution, we substitute 17 for x in the original equation:When x = 17the equation x − 5 = 12becomes 17 − 5 ≟ 1212 = 12 A true statementAs you can see, our solution checks. The purpose for checking a solution to anequation is to catch any mistakes we may have made in the process of solving theequation.Example 5Solve for a: a + ​ 3 _4 ​= −​1 _2 ​.Solution Because we want a by itself on the left side of the equal sign, we addthe opposite of ​_3 ​to each side of the equation.4a + ​ 3 _4 ​+ ​ −​ 3 _4 ​ ​= −​ 1 _2 ​+ ​ −​ 3 _4 ​ ​ Addition property of equality5. Solve for a: a + ​ 2 _3 ​= −​ 1 _6 ​.a + 0 = −​ 1 _2 ​⋅ ​2 _2 ​+ ​ −​ 3 _4 ​ ​ LCD on the right side is 4a = −​ 2 _4 ​+ ​ −​ 3 _4 ​ ​ ​ 2 _4 ​ is equivalent to ​1 _2 ​_a = −​ 5 ​ Add fractions4_The solution is a = −​ 5 4 ​. To check our result, we replace a with −​ _ 5 ​in the original4_equation. The left side then becomes −​ 5 4 ​+ ​3 _4 ​, which simplifies to −​ _ 1 ​, so our2solution checks.Example 6Solve for x: 7.3 + x = −2.4.Solution Again, we want to isolate x, so we add the opposite of 7.3 to bothsides:6. Solve −2.7 + x = 8.1.7.3 + (−7.3) + x = −2.4 + (−7.3) Addition property of equality0 + x = −9.7x = −9.7The addition property of equality also allows us to add variable expressions toeach side of an equation.Example 7Solve for x: 3x − 5 = 4x.SolutionAdding −3x to each side of the equation gives us our solution.7. Solve for x :−4x + 2 = −3x3x − 5 = 4x3x + (−3x) − 5 = 4x + (−3x)−5 = xDistributive propertyAnswers_4. 13 5. −​ 5 ​ 6. 10.8 7. 26

500Chapter 8 Linear Equations and InequalitiesSometimes it is necessary to simplify each side of an equation before using theaddition property of equality. The reason we simplify both sides first is that wewant as few terms as possible on each side of the equation before we use theaddition property of equality. The following examples illustrate this procedure.8. Solve 5(3a − 4) − 14a = 25.Example 8Solve 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.NoteAgain, we placea question markover the equal signbecause we don’t know yetwhether the expressions on theleft and right side of the equal signwill be equal.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 = 9AdditionTo check our solution, we replace a with 9 in the original equation.4(2 ⋅ 9 − 3) − 7 ⋅ 9 ≟ 2 − 54(15) − 63 ≟ −360 − 63 ≟ −3−3 = −3 A true statement9. Solve 5x − 4 = 4x + 6.NoteIn my experienceteaching algebra,I find that studentsmake fewer mistakes if they thinkin terms of addition rather thansubtraction. So, you are probablybetter off if you continue to usethe addition property just the waywe have used it in the examples inthis section. But, if you are curiousas to whether you can subtract thesame number from both sides ofan equation, the answer is yes.Example 9Solve 3x − 5 = 2x + 7.Solution We can solve this equation in two steps. First, we add −2x to bothsides of the equation. When this has been done, x appears on the left side only.Second, we add 5 to both sides:3x + (−2x) − 5 = 2x + (−2x) + 7 Add −2x to both sidesx − 5 = 7Simplify each sidex − 5 + 5 = 7 + 5Add 5 to both sidesx = 12Simplify each sideA Note on SubtractionAlthough the addition property of equality is stated for addition only, we cansubtract the same number from both sides of an equation as well. Becausesubtraction is defined as addition of the opposite, subtracting the same quantityfrom both sides of an equation does not change the solution.x + 2 = 12 Original equationx + 2 − 2 = 12 − 2 Subtract 2 from each sidex = 10 SubtractionGetting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is a solution to an equation?2. What are equivalent equations?3. Explain in words the addition property of equality.Answers8. 45 9. 104. How do you check a solution to an equation?

8.2 Problem Set501Problem Set 8.2BFind the solution for the following equations. Be sure to show when you have used the addition property of equality.[Examples 4–7]1. x − 3 = 8112. x − 2 = 793. x + 2 = 644. x + 5 = 4−15. a + ​_1 2 ​= −​1 _4 ​6. a + ​_1 3 ​= −​5 _6 ​7. x + 2.3 = −3.5_−​ 3 4 ​ _−​ 7 6 ​ −5.88. x + 7.9 = −3.4−11.39. y + 11 = −6−1710. y − 3 = −1211. x − ​ 5 _8 ​= −​3 _4 ​12. x − ​_2_−​ 1 8 ​ ​_ 310 ​5 ​= −​ 1 _10 ​13. m − 6 = 2m14. 3m − 10 = 4m15. 6.9 + x = 3.316. 7.5 + x = 2.2−6−10−3.6−5.317. 5a = 4a − 7−718. 12a = −3 + 11a−319. −​ 5 _9 ​= x − ​2 _5 ​_20. −​ 7_−​ 7 45 ​ _−​ 3 40 ​8 ​= x − ​4 _5 ​B Simplify both sides of the following equations as much as possible, and then solve. [Example 7]21. 4x + 2 − 3x = 4 + 1322. 5x + 2 − 4x = 7 − 3223. 8a − ​_1 2 ​− 7a = ​3 _4 ​+ ​1 _8 ​​_ 118 ​24. 9a − ​_4 5 ​− 8a = ​3 _10 ​− ​1 _5 ​25. −3 − 4x + 5x = 18​_9 2110 ​26. 10 − 3x + 4x = 2010

502Chapter 8 Linear Equations and InequalitiesB Simplify both sides of the following equations as much as possible, and then solve. [Example 7]27. −11x + 2 + 10x + 2x = 9728. −10x + 5 − 4x + 15x = 0−529. −2.5 + 4.8 = 8x − 1.2 − 7x3.530. −4.8 + 6.3 = 7x − 2.7 − 6x4.231. 2y − 10 + 3y − 4y = 18 − 62232. 15 − 21 = 8x + 3x − 10x−6B The following equations contain parentheses. Apply the distributive property to remove the parentheses, then simplifyeach side before using the addition property of equality. [Example 8]33. 2(x + 3) − x = 4−234. 5(x + 1) − 4x = 2−335. −3(x − 4) + 4x = 3 − 7−1636. −2(x − 5) + 3x = 4 − 9−1537. 5(2a + 1) − 9a = 8 − 6−338. 4(2a − 1) − 7a = 9 − 5839. −(x + 3) + 2x − 1 = 61040. −(x − 7) + 2x − 8 = 4541. 4y − 3(y − 6) + 2 = 8−1242. 7y − 6(y − 1) + 3 = 9043. −3(2m − 9) + 7(m − 4) = 12 − 9444. −5(m − 3) + 2(3m + 1) = 15 − 8−10

8.2 Problem Set503B Solve the following equations by the method used in Example 9 in this section. Check each solution in the originalequation.45. 4x = 3x + 2246. 6x = 5x − 4−447. 8a = 7a − 5−548. 9a = 8a − 3−349. 2x = 3x + 1−150. 4x = 3x + 5551. 2y + 1 = 3y + 4−352. 4y + 2 = 5y + 6−453. 2m − 3 = m + 5854. 8m − 1 = 7m − 3−255. 4x − 7 = 5x + 1−856. 3x − 7 = 4x − 6−157. 4x + ​_4 3 ​= 5x − ​2 _3 ​258. 2x + ​_1 4 ​= 3x − ​5 _4 ​ 59. 8a − 7.1 = 7a + 3.9​_ 3 112 ​60. 10a − 4.3 = 9a + 4.7961. Solve each equation.a. 2x = 3 ​_3 2 ​b. 2 + x = 3 1_c. 2x + 3 = 0 −​ 3 2 ​d. 2x + 3 = −5 −4e. 2x + 3 = 7x − 5 ​_8 5 ​ 62. Solve each equation.a. 5t = 10 2b. 5 + t = 10 5c. 5t + 10 = 0 −2d. 5t + 10 = 12 ​_2 5 ​e. 5t + 10 = 8t + 12_−​ 2 3 ​

8.2 Problem Set50567. Sound The chart shows the decibel level of varioussounds. The sum of the loudness of a motorcycle and17 decibels is the loudness of a football stadium. Whatis the loudness of a motorcycle?100 dB68. Skyscrapers The chart shows the heights of the three tallestbuildings in the world. The height of the Sears Toweris 358 feet taller than the Tokyo Tower in Japan. How tallis the Tokyo Tower?1,092 feetSounds Around UsNormal ConversationFootball Stadium60 dB117 dBSuch Great HeightsTaipei 101Taipei, TaiwanPetronas Tower 1 & 2Kuala Lumpur, Malaysia1,483 ft1,670 ft Sears TowerChicago, USA1,450 ftBlue Whale188 dBSource: www.4to40.comSource: www.tenmojo.comGetting Ready for the Next SectionTo understand all of the explanations and examples in the next section you must be able to work the problems below.Simplify.69. ​ 3 _2 ​​ ​ 2 _3 ​y ​y70. ​ 5 _2 ​​ ​ 2 _5 ​y ​y71. ​ 1 _5 ​(5x)x72. −​ 1 _4 ​(−4a)a73. ​ 1 _5 ​(30)674. −​ 1 _4 ​(24)−675. ​ 3 _2 ​(4)676. ​ 1 _26 ​(13)​ 1 _2 ​77. 12 ​ −​ 3 _4 ​ ​−978. 12 ​ ​ 1 _2 ​ ​679. ​ 3 _2 ​​ −​ 5 _4 ​ ​−​ 15 _8 ​ 80. ​ 5 _3 ​​ −​ 6 _5 ​ ​−281. −13 + (−5)−1882. −14 + (−3)−1783. −​ 3 _4 ​+ ​ −​ 1 _2 ​ ​_84. −​ 7_−​ 5 4 ​ _−​ 6 5 ​10 ​+ ​ −​ 1 _2 ​ ​85. 7x + (−4x)3x86. 5x + (−2x)3x

506Chapter 8 Linear Equations and InequalitiesMaintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.87. 3(6x)18x88. 5(4x)20x89. ​_1 5 ​(5x)x90. ​_1 3 ​(3x)x91. 8 ​ ​ 1 _8 ​y ​y92. 6 ​ ​ 1 _6 ​y ​y93. −2 ​ −​ 1 _2 ​x ​x94. −4 ​ −​ 1 _4 ​x ​x95. −​ 4 _3 ​​ −​ 3 _4 ​a ​a96. −​ 5 _2 ​​ −​ 2 _5 ​a ​a

Multiplication Property of Equality 8.3As we have mentioned before, we all have to pay taxes. According to Figure 1,people have been paying taxes for quite a long time.ObjectivesA Use the multiplication property ofequality to solve an equation.B Use the addition and multiplicationproperties of equality together tosolve an equation.Figure 1 Collection of taxes, ca. 3000 b.c. Clerks and scribes appear at the right,with pen and papyrus, and officials and taxpayers appear at the left.If 21% of your monthly pay is withheld for federal income taxes and another 8%is withheld for Social Security, state income tax, and other miscellaneous items,leaving you with $987.50 a month in take-home pay, then the amount you earnedbefore the deductions were removed from your check is given by the equationG − 0.21G − 0.08G = 987.5In this section we will learn how to solve equations of this type.A Multiplication Property of EqualityIn the previous section, we found that adding the same number to both sides ofan equation never changed the solution set. The same idea holds for multiplicationby numbers other than zero. We can multiply both sides of an equation bythe same nonzero number and always be sure we have not changed the solutionset. (The reason we cannot multiply both sides by zero will become apparentlater.) This fact about equations is called the multiplication property of equality,which can be stated formally as follows.Multiplication Property of EqualityFor any three algebraic expressions A, B, and C, where C ≠ 0,if A = Bthen AC = BCIn words: Multiplying both sides of an equation by the same nonzero numberwill not change the solution set.NoteThis property is alsoused many timesthroughout the book.Make every effort to understand itcompletely.Suppose we want to solve the equation 5x = 30. We have 5x on the left sidebut would like to have just x. We choose to multiply both sides by ​_1 5 ​because ​ _ ​ 1 5 ​ ​(5) = 1. Here is the solution:5x = 30​_1 5 ​(5x) = ​1 _​(30) Multiplication property of equality5​ ​ 1 _5 ​⋅ 5 ​x = ​_1 ​(30) Associative property of multiplication51x = 6x = 68.3 Multiplication Property of Equality507

508Chapter 8 Linear Equations and InequalitiesPractice Problems1. Solve for a: 3a = −27.We chose to multiply by ​_1 ​ because it is the reciprocal of 5. We can see that5multiplication by any number except zero will not change the solution set. If,however, we were to multiply both sides by zero, the result would always be0 = 0 because multiplication by zero always results in zero. Although the statement0 = 0 is true, we have lost our variable and cannot solve the equation. Thisis the only restriction of the multiplication property of equality. We are free tomultiply both sides of an equation by any number except zero.Here are some more examples that use the multiplication property of equality.Example 1Solve for a: −4a = 24.Solution Because we want a alone on the left side, we choose to multiply_both sides by −​ 1 4 ​:_−​ 1 4 ​(−4a) = −​1 _​(24) Multiplication property of equality4​ −​ 1 _4 ​(−4) _ ​a = −​ 1 ​(24) Associative property4a = −62. Solve for t : ​_t ​= 6.4Example 2Solve for t: −​ t _​= 5.3Solution Because division by 3 is the same as multiplication by ​_1 ​, we can3_write −​ t 3 ​as −​1 _​t. To solve the equation, we multiply each side by the reciprocal of3_−​ 1 3 ​, which is −3. _−​ t ​= 5 Original equation3−​ 1 _3 ​t = 5 Dividing by 3 is equivalent to multiplying by ​1 _3 ​−3 ​ −​ 1 _3 ​t ​= −3(5) Multiply each side by −3t = −15Multiplication3. Solve ​_2 ​y = 8.5NoteNotice in Examples1 through 3 that ifthe variable is beingmultiplied by a number like −4or ​ 2_ 3​, we always multiply by thenumber’s reciprocal, −​ 1_ 4 ​or ​3_ 2​, toend up with just the variable onone side of the equation.4. Solve.3 + 7 = 6x + 8x − 9xAnswers1. −9 2. 24 3. 20 4. 2Example 3Solve ​ 2 _Solution​y = 4.3We can multiply both sides by ​_3 ​and have 1y on the left side:2​_3 2 ​​ _ ​ 2 3 ​y ​= ​_3 ​(4)2Multiplication property of equality_​ ​ 3 2 ​⋅ ​2 _3 ​y ​= ​_3 ​(4)2Associative propertyy = 6 Simplify ​_3 2 ​ (4) = ​3 _2 ​ _​ 4 1 ​ ​ = ​_122 ​ = 6Example 4Solve 5 + 8 = 10x + 20x − 4x.SolutionOur first step will be to simplify each side of the equation:13 = 26x Simplify both sides first​_126 ​(13) = ​ _ 1 ​(26x) Multiplication property of equality26_ 13​= x Multiplication261_​= x Reduce to lowest terms2

510Chapter 8 Linear Equations and InequalitiesNoteOur original equationhas denominators of3, 2, and 4. The LCDfor these three denominators is12, and it has the property thatall three denominators will divideit evenly. Therefore, if we multiplyboth sides of our equation by 12,each denominator will divide into12 and we will be left with anequation that does not contain anydenominators other than 1.Method 2 Eliminating the fractions in the beginning.12 ​ ​ 2 _3 ​x + ​1 _2 ​ ​= 12 ​ −​ 3 _4 ​ ​ Multiply each side by the LCD 1212 ​ ​ 2 _3 ​x ​+ 12 ​ ​ 1 _2 ​ ​= 12 ​ −​ 3 _4 ​ ​ Distributive property on the left side8x + 6 = −98x = −15MultiplyAdd −6 to each sidex = −​ 15 _8 ​ Multiply each side by ​1 _8 ​As the third line in Method 2 indicates, multiplying each side of the equation bythe LCD eliminates all the fractions from the equation.As you can see, both methods yield the same solution.PropertyBecause division is defined as multiplication by the reciprocal, multiplyingboth sides of an equation by the same number is equivalent to dividing bothsides of the equation by the reciprocal of that number; that is, multiplyingeach side of an equation by ​_1 ​and dividing each side of the equation by 3 are3equivalent operations. If we were to solve the equation 3x = 18 using divisioninstead of multiplication, the steps would look like this:3x = 18 Original equation_ 3x​= ​18_​ Divide each side by 33 3x = 6 DivisionUsing division instead of multiplication on a problem like this may save yousome writing. However, with multiplication, it is easier to explain “why” weend up with just one x on the left side of the equation. (The “why” has todo with the associative property of multiplication.) My suggestion is that youcontinue to use multiplication to solve equations like this one until you understandthe process completely. Then, if you find it more convenient, you canuse division instead of multiplication.Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. Explain in words the multiplication property of equality.2. If an equation contains fractions, how do you use the multiplicationproperty of equality to clear the equation of fractions?3. Why is it okay to divide both sides of an equation by the same nonzeronumber?4. Explain in words how you would solve the equation 3x = 7 using themultiplication property of equality.Answer8. −2

8.3 Problem Set511Problem Set 8.3A Solve the following equations. Be sure to show your work. [Examples 1–3]1. 5x = 1022. 6x = 1223. 7a = 2844. 4a = 3695. −8x = 46. −6x = 27. 8m = −16_−​ 1 2 ​ _−​ 1 3 ​ −28. 5m = −25−59. −3x = −9310. −9x = −36411. −7y = −28412. −15y = −30213. 2x = 0014. 7x = 0015. −5x = 0016. −3x = 0017. ​_x 3 ​= 2618. ​_x 4 ​= 312_19. −​ m ​= 105−50_20. −​ m 7 ​= 1−7_21. −​ x 2 ​= −​ _ 3 4 ​_22. −​ x 3 ​= ​5 _6 ​23. ​_2 3 ​a = 8​_3 2 ​ _−​ 5 2 ​ 1224. ​_3 4 ​a = 68_25. −​ 3 5 ​x = ​9 _5 ​−3_26. −​ 2 5 ​x = ​6 _15 ​−1_27. −​ 5 ​y = −20832_28. −​ 7 ​y = −1424A Simplify both sides as much as possible, and then solve. [Example 4]29. −4x − 2x + 3x = 24−830. 7x − 5x + 8x = 20231. 4x + 8x − 2x = 15 − 1032. 5x + 4x + 3x = 4 + 8​_1 2 ​ 1

512Chapter 8 Linear Equations and Inequalities33. −3 − 5 = 3x + 5x − 10x434. 10 − 16 = 12x − 6x − 3x−235. 18 − 13 = ​ 1 _2 ​a + ​3 _4 ​a − ​5 _8 ​a836. 20 − 14 = ​ 1 _3 ​a + ​5 _6 ​a − ​2 _3 ​a12Solve the following equations by multiplying both sides by −1.37. −x = 438. −x = −339. −x = −440. −x = 3−434−341. 15 = −a−1542. −15 = −a1543. −y = ​_1 2 ​_44. −y = −​ 3 4 ​_−​ 1 2 ​ ​_ 3 4 ​B Solve each of the following equations. [Examples 5–8]45. 3x − 2 = 746. 2x − 3 = 947. 2a + 1 = 348. 5a − 3 = 7361249. ​_1 8 ​+ ​1 _2 ​x = ​1 _4 ​50. ​_1 3 ​+ ​1 _7 ​x = −​ _ 821 ​​_1 4 ​ −551. 6x = 2x − 12−352. 8x = 3x − 10−253. 2y = −4y + 1854. 3y = −2y − 1555. −7x = −3x − 856. −5x = −2x − 123−324

8.3 Problem Set51357. 2x − 5 = 8x + 458. 3x − 6 = 5x + 6_−​ 3 −62 ​59. x + ​_1 2 ​= ​1 _4 ​x − ​5 _8 ​ 60. ​_1 3 ​x + ​2 _5 ​= ​1 _5 ​x − ​2 _5 ​_−​ 3 2 ​ −661. m + 2 = 6m − 3162. m + 5 = 6m − 5263. ​_1 2 ​m − ​1 _4 ​= ​1 _12 ​m + ​1 _6 ​164. ​_1 2 ​m − ​5 _12 ​= ​1 _12 ​m + ​5 _12 ​265. 6y − 4 = 9y + 2−266. 2y − 2 = 6y + 14−467. ​_3 2 ​y + ​1 _3 ​= y − ​2 _3 ​−268. ​_3 2 ​y + ​7 _2 ​= ​1 _2 ​y − ​1 _2 ​−469. 5x + 6 = 270. 2x + 15 = 3_−​ 4 −65 ​71. ​ x _2 ​= ​6 _12 ​172. ​ x _4 ​= ​6 _8 ​373. ​ 3 _x ​= ​6 _7 ​74. ​_2​_7 2 ​ 369 ​= ​8 _x ​75. ​ a _3 ​= ​5 _12 ​76. ​_a​_ 5 4 ​ ​_ 710 ​2 ​= ​7 _20 ​77. ​_10 ​= ​20_20 x ​4078. ​_15 ​= ​60_60 x ​24079. ​ 2 _x ​= ​6 _7 ​80. ​_4​_ 7 3 ​x ​= ​6 _7 ​​ 14 _3 ​

514Chapter 8 Linear Equations and InequalitiesApplying the Concepts81. Break-Even Point Movie theaters pay a certain price forthe movies that you and I see. Suppose a theater pays$1,500 for each showing of a popular movie. If theycharge $7.50 for each ticket they sell, then the equation7.5x = 1,500 gives the number of tickets they mustsell to equal the $1,500 cost of showing the movie.This number is called the break-even point. Solve theequation for x to find the break-even point.200 tickets82. Basketball Laura plays basketball for her communitycollege. In one game she scored 13 points total, witha combination of free throws, field goals, and threepointers.Each free throw is worth 1 point, each field goalis 2 points, and each three-pointer is worth 3 points. Ifshe made 1 free throw and 3 field goals, then solving theequation1 + 3(2) + 3x = 13will give us the number of three-pointers she made.Solve the equation to find the number of three-pointshots Laura made. 2 three-pointers83. Taxes If 21% of your monthly pay is withheld for federalincome taxes and another 8% is withheld for SocialSecurity, state income tax, and other miscellaneousitems, leaving you with $987.61 a month in take-homepay, then the amount you earned before the deductionswere removed from your check is given by the equationG − 0.21G − 0.08G = 987.61Solve this equation to find your gross income.$1,391 per month84. Rhind Papyrus The Rhind Papyrus is an ancient documentthat contains mathematical riddles. One problem asksthe reader to find a quantity such that when it is addedto one-fourth of itself the sum is 15. The equation thatdescribes this situation isx + ​_1 ​x = 154Solve this equation. 12British Museum/Bridgeman Art Library85. Skyscrapers The chart shows the heights of the threetallest buildings in the world. Two times the height ofthe Hoover Dam is the height of the Sears Tower. Howtall is the Hoover Dam?725 feet86. Cars The chart shows the fastest cars in America. Thespeed of the Ford GT in miles per hour is ​_15 ​of its speed22in feet per second. What is its speed in feet per second?Round to the nearest foot per second.301 feet per secondSuch Great HeightsTaipei 101Taipei, TaiwanPetronas Tower 1 & 2Kuala Lumpur, Malaysia1,483 ft1,670 ft Sears TowerChicago, USA1,450 ftReady for the RacesFord GT 205 mphEvans 487 210 mphSaleen S7 Twin Turbo 260 mphSSC Ultimate Aero 273 mphSource: www.tenmojo.comSource: Forbes.com

8.3 Problem Set515Getting Ready for the Next SectionTo understand all of the explanations and examples in the next section you must be able to work the problems below.Solve each equation.87. 2x = 4288. 3x = 24889. 30 = 5x690. 0 = 5x091. 0.17x = 5103,00092. 0.1x = 4004,000Apply the distributive property and then simplify if possible.93. 3(x − 5) + 494. 5(x − 3) + 295. 0.09(x + 2,000)96. 0.04(x + 7,000)3x − 115x − 130.09x + 1800.04x + 28097. 7 − 3(2y + 1)98. 4 − 2(3y + 1)99. 3(2x − 5) − (2x − 4)100. 4(3x − 2) − (6x − 5)−6y + 4−6y + 24x − 116x − 3Simplify.101. 10x + (−5x)102. 12x + (−7x)103. 0.08x + 0.09x104. 0.06x + 0.04x5x5x0.17x0.10xMaintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.Apply the distributive property, and then simplify each expression as much as possible.105. 2(3x − 5)6x − 10106. 4(2x − 6)8x − 24107. ​_1 ​(3x + 6)2​_ 3 2 ​x + 3108. ​_1 ​(2x + 8)4109. ​_1 ​(−3x + 6)3​_1 2 ​x + 2 −x + 2110. ​_1 ​(−2x + 6)2−x + 3Simplify each expression.111. 5(2x − 8) − 310x − 43112. 4(3x − 1) + 712x + 3113. −2(3x + 5) + 3(x − 1)−3x − 13114. 6(x + 3) − 2(2x + 4)2x + 10115. 7 − 3(2y + 1)−6y + 4116. 8 − 5(3y − 4)−15y + 28

Solving Linear Equations 8.4A Solving Linear EquationsWe will now use the material we have developed in the first three sections of thischapter to build a method for solving any linear equation.ObjectivesA Solve a linear equation in onevariable.DefinitionA linear equation in one variable is any equation that can be put in theform ax + b = 0, where a and b are real numbers and a is not zero.Each of the equations we will solve in this section is a linear equation in onevariable. The steps we use to solve a linear equation in one variable are listed here.Strategy Solving Linear Equations in One VariableStep 1a: Use the distributive property to separate terms, if necessary.1b: If fractions are present, consider multiplying both sides by theLCD to eliminate the fractions. If decimals are present, considermultiplying both sides by a power of 10 to clear the equation ofdecimals.1c: Combine similar terms on each side of the equation.Step 2: Use the addition property of equality to get all variable terms onone side of the equation and all constant terms on the other side. Avariable term is a term that contains the variable (for example, 5x).A constant term is a term that does not contain the variable (thenumber 3, for example).NoteYou may have someprevious experiencesolving equations.Even so, you should solve theequations in this section usingthe method developed here.Your work should look like theexamples in the text. If you havelearned shortcuts or a differentmethod of solving equationssomewhere else, you always cango back to them later. What isimportant now is that you are ableto solve equations by the methodsshown here.Step 3: Use the multiplication property of equality to get x (that is, 1x) byitself on one side of the equation.Step 4: Check your solution in the original equation to be sure that you havenot made a mistake in the solution process.As you will see as you work through the examples in this section, it is notalways necessary to use all four steps when solving equations. The number ofsteps used depends on the equation. In Example 1 there are no fractions or decimalsin the original equation, so step 1b will not be used. Likewise, after applyingthe distributive property to the left side of the equation in Example 1, there areno similar terms to combine on either side of the equation, making step 1c alsounnecessary.Example 1Solve 2(x + 3) = 10.Practice Problems1. Solve 3(x + 4) = 6Solution To begin, we apply the distributive property to the left side of theequation to separate terms:Step 1a: 2x + 6 = 10 Distributive property2x + 6 + (−6) = 10 + (−6) Addition property of equalityStep 2: {2x = 4Step 3:{ ​1 _2 ​(2x) = ​1 _2 ​(4) Multiply each side by ​1 _2 ​x = 2 The solution is 28.4 Solving Linear Equations517

518Chapter 8 Linear Equations and InequalitiesThe solution to our equation is 2. We check our work (to be sure we have notmade either a mistake in applying the properties or an arithmetic mistake) bysubstituting 2 into our original equation and simplifying each side of the resultseparately.Check: When x = 2{the equation 2(x + 3) = 10Step 4: becomes 2(2 + 3) ≟ 102(5) ≟ 1010 = 10 A true statementOur solution checks.The general method of solving linear equations is actually very simple. It isbased on the properties we developed in Chapter 1 and on two very simple newproperties. We can add any number to both sides of the equation and multiplyboth sides by any nonzero number. The equation may change in form, but thesolution set will not. If we look back to Example 1, each equation looks a little differentfrom each preceding equation. What is interesting and useful is that eachequation says the same thing about x. They all say x is 2. The last equation, ofcourse, is the easiest to read, and that is why our goal is to end up with x by itself.2. Solve for x : 2(x − 3) + 3 = 9.Example 2Solve for x : 3(x − 5) + 4 = 13.Solution Our first step will be to apply the distributive property to the left sideof the equation:Step 1a: 3x − 15 + 4 = 13 Distributive propertyStep 1c: 3x − 11 = 13 Simplify the left side3x − 11 + 11 = 13 + 11 Add 11 to both sidesStep 2:{ 3x = 24Step 3:{ ​1 _3 ​(3x) = ​1 _3 ​(24) Multiply both sides by ​1 _3 ​x = 8 The solution is 8Check: When x = 8{the equation 3(x − 5) + 4 = 13becomes 3(8 − 5) + 4 ≟ 13Step 4:3(3) + 4 ≟ 139 + 4 ≟ 1313 = 13 A true statement3. Solve.7(x − 3) + 5 = 4(3x − 2) − 8Example 3Solve 5(x − 3) + 2 = 5(2x − 8) − 3.Solution In this case we apply the distributive property on each side of theequation:Step 1a: 5x − 15 + 2 = 10x − 40 − 3 Distributive propertyStep 1c: 5x − 13 = 10x − 43 Simplify each sideStep 2:5x + (−5x) − 13 = 10x + (−5x) − 43 Add −5x to both sides{−13 = 5x − 43−13 + 43 = 5x − 43 + 43 Add 43 to both sides30 = 5xAnswers1. −2 2. 6

8.4 Solving Linear Equations519Step 3:Check:Step 4:{5(3){ ​1 _5 ​(30) = ​1 _5 ​(5x) Multiply both sides by ​1 _5 ​6 = x The solution is 6Replacing x with 6 in the original equation, we have5(6 − 3) + 2 ≟ 5(2 ⋅ 6 − 8) − 35(3) + 2 ≟ 5(12 − 8) − 3+ 2 ≟ 5(4) − 315 + 2 ≟ 20 − 317 = 17 A true statementNoteIt makes no differenceon which side of theequal sign x ends up.Most people prefer to have x onthe left side because we read fromleft to right, and it seems to soundbetter to say x is 6 rather than 6 isx. Both expressions, however, haveexactly the same meaning.Example 4Solve the equation 0.08x + 0.09(x + 2,000) = 690.Solution We can solve the equation in its original form by working with thedecimals, or we can eliminate the decimals first by using the multiplication propertyof equality and solving the resulting equation. Both methods follow.4. Solve the equation.0.06x + 0.04(x + 7,000) = 680Method 1Working with the decimals0.08x + 0.09(x + 2,000) = 690 Original equationStep 1a: 0.08x + 0.09x + 0.09(2,000) = 690 Distributive propertyStep 1c: 0.17x + 180 = 690 Simplify the left side0.17x + 180 + (−180) = 690 + (−180) Add −180 to each sideStep 2:{0.17x = 510Step 3:{ ​0.17x _​= ​_510​ Divide each side by 0.170.17 0.17x = 3,000Note that we divided each side of the equation by 0.17 to obtain the solution. Thisis still an application of the multiplication property of equality because dividingby 0.17 is equivalent to multiplying by ​_10.17 ​.Method 2 Eliminating the decimals in the beginning0.08x + 0.09(x + 2,000) = 690 Original equationStep 1a: 0.08x + 0.09x + 180 = 690 Distributive propertyStep 1b:100(0.08x + 0.09x + 180) = 100(690) Multiply both sides by 100{ 8x + 9x + 18,000 = 69,000Step 1c: 17x + 18,000 = 69,000 Simplify the left sideStep 2: 17x = 51,000 Add −18,000 to each sideStep 3:Check:{ ​17x _​= ​51,000_​ Divide each side by 1717 17x = 3,000Substituting 3,000 for x in the original equation, we have0.08(3,000) + 0.09(3,000 + 2,000) ≟ 690{0.08(3,000) + 0.09(5,000) ≟ 690Step 4:240 + 450 ≟ 690690 = 690 A true statementAnswers3. 0 4. 4,000

520Chapter 8 Linear Equations and Inequalities5. Solve 4 − 2(3y + 1) = −16.Example 5Solve 7 − 3(2y + 1) = 16.Solution We begin by multiplying −3 times the sum of 2y and 1:Step 1a: 7 − 6y − 3 = 16 Distributive propertyStep 1c: −6y + 4 = 16 Simplify the left sideStep 2:−6y + 4 + (−4) = 16 + (−4) Add −4 to both sides{ −6y = 12Step 3:{ −​1 _6 ​(−6y) = −​1 _6 ​(12) Multiply both sides by −​1 _6 ​y = −2Step 4: Replacing y with −2 in the original equation yields a true statement.There are two things to notice about the example that follows: first, the distributiveproperty is used to remove parentheses that are preceded by a negativesign, and, second, the addition property and the multiplication property are notshown in as much detail as in the previous examples.6. Solve.4(3x − 2) − (6x − 5) = 6 − (3x + 1)Example 6Solve 3(2x − 5) − (2x − 4) = 6 − (4x + 5).Solution When we apply the distributive property to remove the groupingsymbols and separate terms, we have to be careful with the signs. Remember, wecan think of −(2x − 4) as −1(2x − 4), so that−(2x − 4) = −1(2x − 4) = −2x + 4It is not uncommon for students to make a mistake with this type of simplificationand write the result as −2x − 4, which is incorrect. Here is the complete solutionto our equation:3(2x − 5) − (2x − 4) = 6 − (4x + 5) Original equation6x − 15 − 2x + 4 = 6 − 4x − 5 Distributive property4x − 11 = −4x + 1 Simplify each side8x − 11 = 1Add 4x to each side8x = 12Add 11 to each sidex = ​_128 ​ Multiply each side by ​1 _8 ​x = ​_3 ​ Reduce to lowest terms2The solution, ​_3 ​, checks when replacing x in the original equation.2AnswersGetting 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 linear equation containingparentheses?2. What is the last step in solving a linear equation?3. Explain in words how you would solve the equation 2x − 3 = 8.4. If an equation contains decimals, what can you do to eliminate thedecimals?5. 3 6. ​ 8 _9 ​

8.4 Problem Set521Problem Set 8.4A Solve each of the following equations using the four steps shown in this section. [Examples 1–2]1. 2(x + 3) = 1232. 3(x − 2) = 643. 6(x − 1) = −18−24. 4(x + 5) = 16−15. 2(4a + 1) = −6−16. 3(2a − 4) = 1247. 14 = 2(5x − 3)28. −25 = 5(3x + 4)−39. −2(3y + 5) = 14−410. −3(2y − 4) = −6311. −5(2a + 4) = 0−212. −3(3a − 6) = 0213. 1 = ​_1 ​(4x + 2)2014. 1 = ​_1 ​(6x + 3)3015. 3(t − 4) + 5 = −4116. 5(t − 1) + 6 = −9−2A Solve each equation. [Examples 2–6]17. 4(2y + 1) − 7 = 118. 6(3y + 2) − 8 = −2​_1 2 ​ _−​ 1 3 ​19. ​_1 2 ​(x − 3) = ​1 _​(x + 1)4720. ​_1 3 ​(x − 4) = ​1 _​(x − 6)21021. −0.7(2x − 7) = 0.3(11 − 4x)822. −0.3(2x − 5) = 0.7(3 − x)623. −2(3y + 1) = 3(1 − 6y) − 924. −5(4y − 3) = 2(1 − 8y) + 11_−​ 1 3 ​ ​_ 1 2 ​25. ​_3 4 ​(8x − 4) + 3 = ​2 _​(5x + 10) − 1526. ​_5 6 ​(6x + 12) + 1 = ​2 _​(9x − 3) + 53​_3 4 ​ 827. 0.06x + 0.08(100 − x) = 6.57528. 0.05x + 0.07(100 − x) = 6.240

522Chapter 8 Linear Equations and Inequalities29. 6 − 5(2a − 3) = 1230. −8 − 2(3 − a) = 0731. 0.2x − 0.5 = 0.5 − 0.2(2x − 13)632. 0.4x − 0.1 = 0.7 − 0.3(6 − 2x)533. 2(t − 3) + 3(t − 2) = 28834. −3(t − 5) − 2(2t + 1) = −8335. 5(x − 2) − (3x + 4) = 3(6x − 8) + 10036. 3(x − 1) − (4x − 5) = 2(5x − 1) − 7137. 2(5x − 3) − (2x − 4) = 5 − (6x + 1)38. 3(4x − 2) − (5x − 8) = 8 − (2x + 3)​_3 7 ​ ​_ 1 3 ​39. −(3x + 1) − (4x − 7) = 4 − (3x + 2)140. −(6x + 2) − (8x − 3) = 8 − (5x + 1)−​ 2 _3 ​41. x + (2x − 1) = 2142. x + (5x + 2) = 20343. x − (3x + 5) = −3−144. x − (4x − 1) = 7−245. 15 = 3(x − 1)646. 12 = 4(x − 5)847. 4x − (−4x + 1) = 548. −2x − (4x − 8) = −149. 5x − 8(2x − 5) = 7​_ 3 4 ​ ​_ 3 2 ​ 350. 3x + 4(8x − 15) = 10251. 7(2y − 1) − 6y = −152. 4(4y − 3) + 2y = 3​_ 3 4 ​ ​_ 5 6 ​53. 0.2x + 0.5(12 − x) = 3.6854. 0.3x + 0.6(25 − x) = 121055. 0.5x + 0.2(18 − x) = 5.4656. 0.1x + 0.5(40 − x) = 32−3057. x + (x + 3)(−3) = x − 3−258. x − 2(x + 2) = x − 2−159. 5(x + 2) + 3(x − 1) = −9−260. 4(x + 1) + 3(x − 3) = 2161. 3(x − 3) + 2(2x) = 52

8.4 Problem Set52362. 2(x − 2) + 3(5x) = 30263. 5(y + 2) = 4(y + 1)−664. 3(y − 3) = 2(y − 2)565. 3x + 2(x − 2) = 6266. 5x −(x − 5) = 25567. 50(x − 5) = 30(x + 5)2068. 34(x − 2) = 26(x + 2)1569. 0.08x + 0.09(x + 2,000) = 8604,00070. 0.11x + 0.12(x + 4,000) = 9402,00071. 0.10x + 0.12(x + 500) = 21470072. 0.08x + 0.06(x + 800) = 10440073. 5x + 10(x + 8) = 2451174. 5x + 10(x + 7) = 175775. 5x + 10(x + 3) + 25(x + 5) = 435776. 5(x + 3) + 10x + 25(x + 7) = 3905The next two problems are intended to give you practice reading, and paying attention to, the instructions that accompanythe problems you are working. Working these problems is a excellent way to get ready for a test or a quiz.78. Work each problem according to the instructions.77. Work each problem according to the instructions.f. Solve: 4(x − 5) = 2x + 25 ​_452 ​ = 22.5 f. Solve: 3(x + 6) = 7x + 4 ​_7 2 ​a. Solve: 4x − 5 = 0 ​_5 4 ​ = 1.25a. Solve: 3x + 6 = 0 −2b. Solve: 4x − 5 = 25 ​_152 ​ = 7.5b. Solve: 3x + 6 = 4_−​ 2 3 ​c. Add: (4x − 5) + (2x + 25) 6x + 20c. Add: (3x + 6) + (7x + 4) 10x + 10d. Solve: 4x − 5 = 2x + 25 15d. Solve: 3x + 6 = 7x + 4 ​_1 2 ​e. Multiply: 4(x − 5) 4x − 20e. Multiply: 3(x + 6) 3x + 18Applying the Concepts79. Wind Energy The graph shows the world’s capacity forgenerating power from wind energy. The year thatproduced 48,000 megawatts is given by the equation48 = 7x − 13980. What year is this?200480. Camera Phones The chart shows the estimated number ofcamera phones and non-camera phones sold from 2004to 2010. The year in which 730 million phones were soldcan be given by the equation 730 = 10(5x − 9,952). Whatyear is this?2005World Wind Electricity-Generating CapacityNumber of Camera Phones in the WorldMegawatts70,00060,00050,00040,00030,00020,00010,000Millions of Phones10008006004002000 2004CameraPhonesNon-CameraPhones1985 1990 1995 2000 2005 2010Source: GWEC, WorldWatch 2006Source: http://www.InfoTrends.com Estimates result of interviews of 4,782 people in US, UK, France,Germany, Spain, Japan, and China

524Chapter 8 Linear Equations and InequalitiesGetting Ready for the Next SectionTo understand all of the explanations and examples in the next section you must be able to work the problems below.Solve each equation.81. 40 = 2x + 121482. 80 = 2x + 123483. 12 + 2y = 6−384. 3x + 18 = 6−485. 24x = 686. 45 = 0.75x​_1 4 ​ 6087. 70 = x ⋅ 21088. 15 = x ⋅ 80​_ 1 3 ​ ​_ 316 ​Apply the distributive property.89. ​_1 _​(−3x + 6)90. −​ 1 2_−​ 3 2 ​x + 3​(−5x + 20)4​_ 5 4 ​x − 5Maintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.Multiply.91. ​_1 2 ​(3)92. ​_1 3 ​(2)93. ​_2 3 ​(6)​_3 2 ​ ​_ 2 3 ​ 494. ​ 3 _2 ​(4)695. ​ 5 _9 ​⋅ ​9 _5 ​196. ​ 3 _7 ​⋅ ​7 _3 ​1Fill in the tables by finding the value of each expression for the given values of the variables.97.x 3(x + 2) 3x + 2 3x + 698.0 6 2 61 9 5 92 12 8 123 15 11 1599.a (2a + 1) 2 4a 2 + 4a + 1100.1 9 92 25 253 49 49x 7(x − 5) 7x − 5 7x − 35−3 −56 −26 −56−2 −49 −19 −49−1 −42 −12 −420 −35 −5 −35a (a + 1) 3 a 3 + 3a 2 + 3a + 11 8 82 27 273 64 64

Formulas 8.5In this section we continue solving equations by working with formulas. To begin,here is the definition of a formula.DefinitionIn mathematics, a formula is an equation that contains more than onevariable.The equation P = 2l + 2w, which tells us how to find the perimeter of a rectangle,is an example of a formula.ObjectivesA Find the value of a variable in aformula given replacements for theother variables.B Solve a formula for one of itsvariables.C Find the complement andsupplement of an angle.D Solve simple percent problems.lwwA Solving FormulaslTo begin our work with formulas, we will consider some examples in which weare given numerical replacements for all but one of the variables.Example 1The perimeter P of a rectangular livestock pen is 40 feet.If the width w is 6 feet, find the length.P = 40Practice Problems1. Suppose the livestock pen inExample 1 has a perimeter of 80feet. If the width is still 6 feet,what is the new length?w = 6 ftlSolution 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 formulaP = 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 feet8.5 Formulas525

526Chapter 8 Linear Equations and Inequalities2. Find x when y is 9 in the formula3x + 2y = 6.Example 2Find y when x = 4 in the formula 3x + 2y = 6.Solution We substitute 4 for x in the formula and then solve for y :BWhen x = 4the formula 3x + 2y = 6becomes 3(4) + 2y = 6or 12 + 2y = 6 Multiply 3 and 4Solving for a Variable2y = −6y = −3Add −12 to each sideMultiply each side by ​ 1 _2 ​In the next examples we will solve a formula for one of its variables withoutbeing given numerical replacements for the other variables.Consider the formula for the area of a triangle:A =12bhhbwhere A = area, b = length of the base, and h = height of the triangle.Suppose we want to solve this formula for h. What we must do is isolate thevariable h on one side of the equal sign. We begin by multiplying both sides by 2:2 ⋅ A = 2 ⋅ ​ 1 _2 ​bh2A = bhThen we divide both sides by b:2A_b ​= ​bh _b ​h = ​_2Ab ​The original formula A = ​_1 2 ​bh and the final formula h = ​2A _​both give the samebrelationship among A, b, and h. The first one has been solved for A and the secondone has been solved for h.RuleTo solve a formula for one of its variables, we must isolate that variable oneither side of the equal sign. All other variables and constants will appear onthe other side.Answer2. −4

8.5 Formulas527Example 3Solve 3x + 2y = 6 for y.3. Solve 5x − 4y = 20 for y.Solution To solve for y, we must isolate y on the left side of the equation. Tobegin, we use the addition property of equality to add −3x to each side:3x + 2y = 6Original formula3x + (−3x) + 2y = (−3x) + 6 Add −3x to each side2y = −3x + 6 Simplify the left side​_1 2 ​(2y) = ​1 _2 ​(−3x + 6) Multiply each side by ​1 _2 ​_y = −​ 3 ​x + 3 Multiplication2Example 4Solve h = vt − 16t 2 for v.4. Solve P = 2w + 2l for l.Solution Let’s begin by interchanging the left and right sides of the equation.That way, the variable we are solving for, v, will be on the left side.vt − 16t 2 = hExchange sidesvt − 16t 2 + 16t 2 = h + 16t 2 Add 16t 2 to each sidevt = h + 16t 2​_vt2+ 16t​= ​h_​ Divide each side by tt tv = ​_h + 16t 2​tWe know we are finished because we have isolated the variable we are solvingfor on the left side of the equation and it does not appear on the other side.Example 5Solve for y: ​_y − 1 ​= ​_3x 2 ​.5. Solve for y : ​_y − 2 ​= ​_4x 3 ​.Solution Although we will do more extensive work with formulas of this formlater in the book, we need to know how to solve this particular formula for y inorder to understand some things in the next chapter. We begin by multiplyingeach side of the formula by x. Doing so will simplify the left side of the equationand make the rest of the solution process simple.​_y − 1 ​= ​_3 ​ Original formulax 2x ⋅ ​_y − 1 ​= ​_3 ​⋅ x Multiply each side by xx 2y − 1 = ​_3 ​x Simplify each side2y = ​_3 ​x + 1 Add 1 to each side2This is our solution. If we look back to the first step, we can justify our result onthe left side of the equation this way: Dividing by x is equivalent to multiplying byits reciprocal ​_1 ​. Here is what it looks like when written out completely:xx ⋅ ​_y − 1 ​= x ⋅ ​_1x x ​⋅ ( y − 1) = 1( y − 1) = y − 1 Answers3. y = ​_5 4 ​x − 5 4. l = ​P _ − 2w ​25. y = ​ 4 _3 ​x + 2

528Chapter 8 Linear Equations and InequalitiesCComplementary and Supplementary Angles90° – xxComplementary angles180° – xxSupplementary anglesFACTS FROM GEOMETRY: More on Complementary and Supplementary AnglesIn Chapter 1 we defined complementary angles as angles that add to 90°; thatis, if x 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° − xBecause 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, we cansay that the supplement of angle x is the angle 180° − x. To summarize, if x isan angle, thenThe complement of x is 90° − x, andThe supplement of x is 180° − xIf you go on to take a trigonometry class, you will see this formula again.6. 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°.DBasic Percent ProblemsThe last examples in this section show how basic percent problems can be translateddirectly into equations. To understand these examples, you must recall thatpercent means “per hundred.” That is, 75% is the same as ​_75 ​, 0.75, and, in reduced100fraction form, ​_3 ​. Likewise, the decimal 0.25 is equivalent to 25%. To change a decimalto a percent, we move the decimal point two places to the right and write the %4symbol. To change from a percent to a decimal, we drop the % symbol and movethe decimal point two places to the left. The table that follows gives some of themost commonly used fractions and decimals and their equivalent percents.Fraction Decimal Percent​_1 ​ 0.5 50%2​_1 ​ 0.25 25%4​_3 ​ 0.75 75%4​ 1 _3 ​ 0.​_ 3​ 33​ 1 _3 ​%​_2 _3 ​ 0.​ 6​66​ 2 _3 ​%​_1 ​ 0.2 20%5​_2 ​ 0.4 40%5Answer6. 55°; 145°​_3 ​ 0.6 60%5​_4 ​ 0.8 80%5

8.5 Formulas529Example 7What number is 25% of 60?Solution To solve a problem like this, we let x = the number in question (thatis, the number we are looking for). Then, we translate the sentence directly intoan equation by using an equal sign for the word “is” and multiplication for theword “of.” Here is how it is done:7. What number is 25% of 74?What number is 25% of 60?x = 0.25 ⋅ 60x = 15Notice that we must write 25% as a decimal to do the arithmetic in the problem.The number 15 is 25% of 60.Example 8What percent of 24 is 6?8. What percent of 84 is 21?Solution Translating this sentence into an equation, as we did in Example 7,we have:What percent of 24 is 6?x ⋅ 24 = 6or 24x = 6Next, we multiply each side by ​_1 ​. (This is the same as dividing each side by 24.)24​ 1 _24 ​(24x) = ​ 1 _24 ​(6)x = ​ 6 _24 ​= ​ 1 _4 ​The number 6 is 25% of 24.= 0.25, or 25%Example 945 is 75% of what number?SolutionAgain, we translate the sentence directly:9. 35 is 40% of what number?45 is 75% of what number?45 = 0.75 ⋅ xNext, we multiply each side by ​_1 ​(which is the same as dividing each side by0.750.75):1_​0.75 ​(45) = ​ 1_0.75 ​(0.75x)​_450.75 ​= x60 = xThe number 45 is 75% of 60.We can solve application problems involving percent by translating each probleminto one of the three basic percent problems shown in Examples 7, 8, and 9.Answers7. 18.5 8. 25% 9. 87.5

530Chapter 8 Linear Equations and Inequalities10. In a third granola bar, 28% ofthe calories are from fat. If oneserving of this bar contains 150calories, how many of thosecalories are from fat?Example 10The American Dietetic Association (ADA) recommendseating foods in which the calories from fat are less than 30% of the total calories.The nutrition labels from two kinds of granola bars are shown in Figure 1. Foreach bar, what percent of the total calories come from fat?Solution The information needed to solve this problem is located toward thetop of each label. Each serving of Bar I contains 210 calories, of which 70 caloriescome from fat. To find the percent of total calories that come from fat, we mustanswer this question:70 is what percent of 210?Nutrition FactsServing Size 2 bars (47g)Servings Per Container 6Amount Per ServingCalories 210Calories from Fat 70% Daily Value*Total Fat 8g 12%Saturated Fat 1g 5%Cholesterol 0mg 0%Sodium 150mg 6%Total Carbohydrate 32g 11%Dietary Fiber 2g 10%Sugars 12gProtein 4g* Percent Daily Values are based on a 2,000 caloriediet. Your daily values may be higher or lower dependingon your calorie needs.Nutrition FactsServing Size 1 bar (21g)Servings Per Container 8Amount Per ServingCalories 80Calories from Fat 15% Daily Value*Total Fat 1.5g 2%Saturated Fat 0g 0%Cholesterol 0mg 0%Sodium 60mg 3%Total Carbohydrate 16g 5%Dietary Fiber 1g 4%Sugars 5gProtein 2g* Percent Daily Values are based on a 2,000 caloriediet. Your daily values may be higher or lower dependingon your calorie needs.Figure 1For Bar II, one serving contains 80 calories, of which 15 calories come fromfat. To find the percent of total calories that come from fat, we must answer thisquestion:15 is what percent of 80?Translating each equation into symbols, we have70 is what percent of 210 15 is what percent of 8070 = x ⋅ 210 15 = x ⋅ 80x = ​_70210 ​ x = ​15 _80 ​x = 0.33 to the nearest hundredth x = 0.19 to the nearest hundredthx = 33% x = 19%Comparing the two bars, 33% of the calories in Bar I are fat calories, whereas19% of the calories in Bar II are fat calories. According to the ADA, Bar II is thehealthier choice.Answer10. 42 caloriesGetting 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 does percent mean?

8.5 Problem Set531Problem Set 8.5A Use the formula P = 2l + 2w to find the length l of a rectangular lot if [Examples 1–2]1. The width w is 50 feet and the perimeter P is 300 feet100 feet2. The width w is 75 feet and the perimeter P is 300 feet75 feet3. For the equation 2x + 3y = 6a. Find y when x is 0. 2b. Find x when y is 1. ​_3 2 ​c. Find y when x is 3. 04. For the equation 2x − 5y = 20a. Find y when x is 0. −4b. Find x when y is 0. 10c. Find x when y is 2. 15_5. For the equation y = −​ 1 3 ​x + 2a. Find y when x is 0. 2b. Find x when y is 3. −3c. Find y when x is 3. 1_6. For the equation y = −​ 2 3 ​x + 1a. Find y when x is 0. 1b. Find x when y is −1. 3c. Find y when x is −3. 3AUse the formula 2x + 3y = 6 to find y if7. x is 308. x is −29. x is 0​_ 103 ​ 210. x is −34AUse the formula 2x − 5y = 20 to find x if11. y is 212. y is −413. y is 014. y is −615010−5AUse the equation y = (x + 1) 2 − 3 to find the value of y when15. x = −216. x = −117. x = 118. x = 2−2−316

532Chapter 8 Linear Equations and Inequalities19. Use the formula y = ​_20 ​to find y whenxa. x = 10 2b. x = 5 420. Use the formula y = 2x 2 to find y whena. x = 5 50b. x = −6 7221. Use the formula y = Kx to find K whena. y = 15 and x = 3 5b. y = 72 and x = 4 1822. Use the formula y = Kx 2 to find K whena. y = 32 and x = 4 2b. y = 45 and x = 3 523. If y = ​_K ​, find K ifxa. x is 5 and y is 4. 20b. x is 5 and y is 15. 7524. If I = ​_Kd 2​,find K ifa. I = 200 and d = 10. 20,000b. I = 200 and d = 5. 5,000B Solve each of the following for the indicated variable. [Examples 3–4]25. A = lw for l26. d = rt for r27. V = lwh for h28. PV = nRT for Pl = ​ A _w ​r = ​ d _t ​h = ​ V _lw ​P = ​ nRT _V ​29. P = a + b + c for aa = P − b − c30. P = a + b + c for bb = P − a − c31. x − 3y = −1 for xx = 3y − 132. x + 3y = 2 for xx = −3y + 233. −3x + y = 6 for yy = 3x + 634. 2x + y = −17 for yy = −2x − 1735. 2x + 3y = 6 for y 36. 4x + 5y = 20 for y_y = −​ 2 3 ​x + 2 _y = −​ 4 5 ​x + 437. y − 3 = −2(x + 4) for yy = −2x − 538. y + 5 = 2(x + 2) for yy = 2x − 1_39. y − 3 = −​ 2 ​(x + 3) for y3_y = −​ 2y = −​ 1 _2 ​x − 13 ​x + 1 40. y − 1 = −​ 1 _​(x + 4) for y2

8.5 Problem Set533B Solve for the indicated variable. [Examples 3–5]41. P = 2l + 2w for ww = ​_P − 2l ​242. P = 2l + 2w for ll = ​_ P − 2w ​243. h = vt + 16t 2 for vv = ​_ h − 16t 2​t44. h = vt − 16t 2 for vv = ​_h + 16t 2​t45. A = πr 2 + 2πrh for hh = ​_ A − πr 2​2πr46. A = 2πr 2 + 2πrh for hh = ​_ A − 2πr 2​2πr47. Solve for y.a. y − 3 = −2(x + 4)y = −2x − 5b. y − 5 = 4(x − 3)y = 4x − 748. Solve for y._a. y + 1 = −​ 2 ​(x − 3)3_y = −​ 2 3 ​x + 149. Solve for y.y = ​_3 4 ​x + ​1 _4 ​3 ​x + 1_b. y − 3 = −​ 2 ​(x + 3)3_y = −​ 2a. y − 1 = ​_3 ​(x − 1)4b. y + 2 = ​_3 ​(x − 4)4y = ​_3a. y + 3 = ​_3 ​(x − 2)2y = ​_3 2 ​x − 6b. y + 4 = ​_4 ​(x − 3)3y = ​_4 3 ​x − 84 ​x − 5 50. Solve for y.51. Solve for y.52. Solve for y.a. ​_y − 1 ​= ​_3x 5 ​ b. ​y _ − 2 ​= ​_1x 2 ​a. ​ _ y + 1xy = ​_3 5 ​x + 1 y = ​ _ 1 2 ​x + 2​= −​ 3 _5​ b. ​y _ + 2x​= −​ 1 _2 ​y = −​ 3 _5 ​x − 1 y = −​1 _2 ​x − 2B Solve each formula for y.53. ​ x _7 ​− ​y _3 ​= 1y = ​_3 7 ​x − 3 y = ​_ 9 4 ​x − 9 y = 2x + 854. ​_x 4 ​− ​y _9 ​= 1_55. −​ 14 ​x + ​1 _8 ​y = 156. −​ 1 _9 ​x + ​1 _3 ​y = 1y = ​ 1 _3 ​x + 3The next two problems are intended to give you practice reading, and paying attention to, the instructions that accompanythe problems you are working. As we have mentioned previously, working these problems is an excellent way to getready for a test or a quiz.57. Work each problem according to the instructions. 58. Work each problem according to the instructions.a. Solve: 4x + 5 = 20 ​_154 ​ = 3.75_a. Solve: −2x + 1 = 4 −​ 3 2 ​ = −1.5b. Find the value of 4x + 5 when x is 3. 17b. Find the value of −2x + 1 when x is 8. −15_c. Solve for y: 4x + 5y = 20 y = −​ 4 5 ​x + 4c. Solve for y: −2x + y = 20 y = 2x + 20_d. Solve for x: 4x + 5y = 20 x = −​ 5 4 ​y + 5 d. Solve for x: −2x + y = 20 x = ​_1 ​y − 102C Find the complement and the supplement of each angle. [Example 6]59. 30°60°; 150°60. 60°30°; 120°61. 45°45°; 135°62. 15°75°; 165°

534Chapter 8 Linear Equations and InequalitiesD Translate each of the following into an equation, and then solve that equation. [Examples 7–9]63. What number is 25% of 40?1064. What number is 75% of 40?3065. What number is 12% of 2,000?24066. What number is 9% of 3,000?27067. What percent of 28 is 7?25%68. What percent of 28 is 21?75%69. What percent of 40 is 14?35%70. What percent of 20 is 14?70%71. 32 is 50% of what number?6472. 16 is 50% of what number?3273. 240 is 12% of what number?2,00074. 360 is 12% of what number?3,000Applying the ConceptsMore About Temperatures As we mentioned in Chapter 1, in the U.S. system, temperature is measured on the Fahrenheitscale. In the metric system, temperature is measured on the Celsius scale. On the Celsius scale, water boils at 100 degreesand freezes at 0 degrees. To denote a temperature of 100 degrees on the Celsius scale, we write100°C, which is read “100 degrees Celsius”Table 1 is intended to give you an intuitive idea of the relationship between the two temperature scales. Table 2 gives theformulas, in both symbols and words, that are used to convert between the two scales.Table 1temperatureSituation Fahrenheit CelsiusWater freezes 32°F 0°CRoom temperature 68°F 20°CNormal body temperature 98.6°F 37°CWater boils 212°F 100°CBake cookies 365°F 185°CTable 2To Convert from Formula in Symbols Formula in WordsFahrenheit to CelsiusCelsius to FahrenheitC = ​_5 ​(F − 32) Subtract 32, multiply9by 5, then divide by 9.F = ​_9 5 ​C + 32 Multiply by ​9 _​, then5add 32.75. Let F = 212 in the formula C = ​_5 ​(F − 32), and solve for9C. Does the value of C agree with the information inTable 1?100°C; yes77. Let F = 68 in the formula C = ​_5 ​(F − 32), and solve for9C. Does the value of C agree with the information inTable 1?20°C; yes76. 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?212°F; yes78. 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?98.6°F; yes79. Solve the formula F = ​_9 ​C + 32 for C.80. Solve the formula C = ​_5 ​(F − 32) for F.5C = ​_5 99 ​(F − 32) F = ​_ 9 ​C + 325

8.5 Problem Set53581. Population The map shows the most populated citiesin the United States. If the population of California in2006 was about 36.5 million, what percentage of thepopulation are found in Los Angeles and San Diego?Round to the nearest tenth of a percent.13.9%82. Winter Activities About 22.6 million Americans participatein winter activities each year. The pie graph showswhich winter activities Americans enjoy. Use the graphto find how many more people enjoy snowboarding thanice-hockey. Round to the nearest tenth.2.4 million peopleWhere Is Everyone?Let It Snow!Los Angeles, CA3.80San Diego, CA 1.26Phoeniz, AZ 1.37Dallas, TX 1.21Houston, TX 2.01Chicago, IL2.89Philadelphia, PA 1.49New York City, NY* Polpulation in MillionsSource: U.S. Census Bureau8.08Alpine-ski32.74%Ice-skate29.65%Snowboard19.03%Ice-Hockey8.40%Crosscountry-ski10.18%Source: U.S. Census BureauNutrition Labels The nutrition label in Figure 2 is from a quart of vanilla ice cream. The label in Figure 3 is from a pint ofvanilla frozen yogurt. Use the information on these labels for problems 83–86. Round your answers to the nearest tenth ofa percent. [Example 10]Nutrition FactsServing Size 1/2 cup (65g)Servings 8Amount/ServingCalories 150 Calories from Fat 90% Daily Value*Total Fat 10g 16%Saturated Fat 6g 32%Cholesterol 35mg 12%Sodium 30mg 1%Total Carbohydrate 14g 5%Dietary Fiber 0g 0%Sugars 11gProtein 2gVitamin A 6% • Vitamin C 0%Calcium 6% • Iron 0%* Percent Daily Values are based on a 2,000 calorie diet.Nutrition FactsServing Size 1/2 cup (98g)Servings Per Container 4Amount Per ServingCalories 160 Calories from Fat 25% Daily Value*Total Fat 2.5g 4%Saturated Fat 1.5g 7%Cholesterol 45mg 15%Sodium 55mg 2%Total Carbohydrate 26g 9%Dietary Fiber 0g 0%Sugars 19gProtein 8gVitamin A 0% • Vitamin C 0%Calcium 25% • Iron 0%* Percent Daily Values are based on a 2,000 calorie diet.Figure 2Vanilla ice creamFigure 3Vanilla frozen yogurt83. What percent of the calories in one serving of thevanilla ice cream are fat calories?60%85. One serving of frozen yogurt is 98 grams, of which 26grams are carbohydrates. What percent of one servingare carbohydrates?26.5%84. What percent of the calories in one serving of the frozenyogurt are fat calories?15.6%86. One serving of vanilla ice cream is 65 grams. Whatpercent of one serving is sugar?16.9%Circumference The circumference of a circle is given by the formula C = 2πr. Find r if87. The circumference C is 44 meters and π is ​ 22 _7 ​7 meters88. The circumference C is 176 meters and π is ​ 22 _7 ​28 meters89. The circumference is 9.42 inches and π is 3.1490. The circumference is 12.56 inches and π is 3.14​_3 2 ​or 1.5 inches 2 inches

536Chapter 8 Linear Equations and InequalitiesVolume The volume of a cylinder is given by the formula V = πr 2 h. Find the height h if91. The volume V is 42 cubic feet, the radius is ​_7 ​feet, and22π is ​_227 ​132 feet92. The volume V is 84 cubic inches, the radius is ​_711 ​inches,and π is ​_227 ​66 inches94. The volume is 12.56 cubic centimeters, the radius is93. The volume is 6.28 cubic centimeters, the radius is​_2 9 ​centimeters 1 centimeter3 centimeters, and π is 3.142 centimeters, and π is 3.14Getting Ready for the Next SectionTo understand all of the explanations and examples in the next section you must be able to work the problems below.Write an equivalent expression in English. Include the words sum and difference when possible.95. 4 + 1 = 5The sum of 4 and 1 is 5.96. 7 + 3 = 10The sum of 7 and 3 is 10.97. 6 − 2 = 4The difference of 6 and 2 is 4.98. 8 − 1 = 7The difference of 8 and 1 is 7.99. x − 5 = −12The difference of a number and5 is −12.100. 2x + 3 = 7The sum of twice a numberand 3 is 7.101. x + 3 = 4(x − 3)The sum of a number and 3is four times the difference ofthat number and 3102. 2(2x − 5) = 2x − 34Twice the difference of twice anumber and 5 is the difference oftwice that number and 34.For each of the following expressions, write an equivalent equation.103. Twice the sum of 6 and 3 is 18.2(6 + 3) = 18104. Four added to the product of 3 and −1 is 1.3(−1) + 4 = 1105. The sum of twice 5 and 3 is 13.2(5) + 3 = 13106. Twice the difference of 8 and 2 is 12.2(8 − 2) = 12107. The sum of a number and five is thirteen.x + 5 = 13108. The difference of ten and a number is negative eight.10 − x = −8109. Five times the sum of a number and seven is thirty.5(x + 7) = 30110. Five times the difference of twice a number and six isnegative twenty.5(2x − 6) = −20Maintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.111. a. 27 − (−68) 95b. 27 + (−68) −41c. −27 − 68 −95d. −27 + 68 41112. a. 55 − (−29) 84b. 55 + (−29) 26c. −55 − 29 −84d. −55 + 29 −26113. a. −32 − (−41) 9b. −32 + (−41) −73c. −32 + 41 9d. −32 − 41 −73114. a. −56 − (−35) −21b. −56 + (−35) −91c. −56 + 35 −21d. −56 − 35 −91

ApplicationsThe title of the section is “Applications,” but many of the problems here don’tseem to have much to do with “real life.” Example 3 is what we refer to as an“age problem.” But imagine a conversation in which you ask someone how oldher children are and she replies, “Bill is 6 years older than Tom. Three years agothe sum of their ages was 21. You figure it out.” Although many of the “application”problems in this section are contrived, they are also good for practicing thestrategy we will use to solve all application problems.8.6ObjectivesA Apply the Blueprint for ProblemSolving to a variety of applicationproblems.A Problem SolvingTo 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.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 cases thiswill amount to letting x = the item that is asked for in the problem.)Then translate the other information in the problem to expressionsinvolving 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 visualize the relationship between the items in the problem. In othercases, a table helps organize the information. As you gain more experience usingthe blueprint to solve application problems, you will find additional techniquesthat expand the blueprint.To help with problems of the type shown next in Example 1, here are somecommon English words and phrases and their mathematical translations.EnglishThe sum of a and bThe difference of a and bThe product of a and bAlgebraa + ba − 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 − 48.6 Applications537

538Chapter 8 Linear Equations and InequalitiesNumber ProblemsPractice Problems1. The sum of twice a number andfive is eleven. Find the number.Example 1The sum of twice a number and three is seven. Find thenumber.Solution Using the 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.Known items: The numbers 3 and 7Unknown items:The number in questionStep 2: Assign a variable to one of the unknown items. Then translate theother information in the problem to expressions involving the variable.Let x = the number asked for in the problem,then “The sum of twice a number andthree” translates to 2x + 3.Step 3: Reread the problem, and then write an equation, using the items andvariables listed in steps 1 and 2, that describes the situation.With all word problems, the word is translates to =.The sum of twice x and 3 is 72x + 3 = 7Step 4: Solve the equation found in step 3.2x + 3 = 72x + 3 + (−3) = 7 + (−3)2x = 4​ 1 _2 ​(2x) = ​1 _2 ​(4)x = 2Step 5: Write your answer using a complete sentence.The number is 2.Step 6: Reread the problem, and check your solution with the original wordsin the problem.The sum of twice 2 and 3 is 7. A true statement.You may find some examples and problems in this section that you can solvewithout using algebra or our blueprint. It is very important that you solve theseproblems using the methods we are showing here. The purpose behind theseproblems is to give you experience using the blueprint as a guide to solving problemswritten in words. Your answers are much less important than the work thatyou show to obtain your answer. You will be able to condense the steps in theblueprint later in the course. For now, though, you need to show your work in thesame detail that we are showing in the examples in this section.Answer1. 3

8.6 Applications539Example 2One number is three more than twice another; their sumis eighteen. Find the numbers.SolutionStep 1: Read and list.2. If three times the differenceof a number and two weredecreased by six, the resultwould be three. Find thenumber.Known items: Two numbers that add to 18. One is 3 more than twicethe other.Unknown items: The numbers in question.Step 2: Assign a variable, and translate information.Let x = the first number. The other is 2x + 3.Step 3: Reread, and write an equation.Their sum is 18x + (2x + 3) = 18Step 4: Solve the equation.x + (2x + 3) = 183x + 3 = 183x + 3 + (−3) = 18 + (−3)3x = 15x = 5Step 5: Write the answer.The first number is 5. The other is 2 ⋅ 5 + 3 = 13.Step 6: Reread, and check.The sum of 5 and 13 is 18, and 13 is 3 more than twice 5.Age Problem}Remember as you read through the steps in the solutions to the examples in thissection that step 1 is done mentally. Read the problem, and then mentally list theitems that you know and the items that you don’t know. The purpose of step 1 isto give you direction as you begin to work application problems. Finding the solutionto an application problem is a process; it doesn’t happen all at once. The firststep is to read the problem with a purpose in mind. That purpose is to mentallynote the items that are known and the items that are unknown.Example 3Bill is 6 years older than Tom. Three years ago Bill’s agewas four times Tom’s age. Find the age of each boy now.Solution We apply the Blueprint for Problem Solving.3. Pete is five years older thanMary. Five years ago Pete wastwice as old as Mary. Find theirages now.Step 1: Read and list.Known items: Bill is 6 years older than Tom. Three years ago Bill’s agewas four times Tom’s age.Unknown items: Bill’s age and Tom’s ageStep 2: Assign a variable, and translate information.Let x = Tom’s age now. That makes Bill x + 6 years old now. A table likethe one shown here can help organize the information in an age prob-Answer2. 5

540Chapter 8 Linear Equations and Inequalitieslem. Notice how we placed the x in the box that corresponds to Tom’sage now.Three YearsAgoNowBill x + 6TomxIf Tom is x years old now, 3 years ago he was x − 3 years old. If Bill isx + 6 years old now, 3 years ago he was x + 6 − 3 = x + 3 years old. Weuse this information to fill in the remaining entries in the table.Three YearsAgoNowBill x + 3 x + 6Tom x − 3 xStep 3: Reread, and write an equation.Reading the problem again, we see that 3 years ago Bill’s age was fourtimes Tom’s age. Writing this as an equation, we have Bill’s age 3 yearsago = 4 ⋅ (Tom’s age 3 years ago):Step 4: Solve the equation.x + 3 = 4(x − 3)x + 3 = 4(x − 3)x + 3 = 4x − 12x + (−x) + 3 = 4x + (−x) − 123 = 3x − 123 + 12 = 3x − 12 + 1215 = 3xx = 5Step 5: Write the answer.Tom is 5 years old. Bill is 11 years old.Step 6: Reread, and check.If Tom is 5 and Bill is 11, then Bill is 6 years older than Tom. Three yearsago Tom was 2 and Bill was 8. At that time, Bill’s age was four timesTom’s age. As you can see, the answers check with the original problem.Geometry ProblemTo understand Example 4 completely, you need to recall from Chapter 1 that theperimeter of a rectangle is the sum of the lengths of the sides. The formula for theperimeter is P = 2l + 2w.4. The length of a rectangle isthree more than twice thewidth. The perimeter is 36inches. Find the length and thewidth.Answer3. Mary is 10; Pete is 15Example 4The length of a rectangle is 5 inches more than twice thewidth. The perimeter is 34 inches. Find the length and width.Solution When working problems that involve geometric figures, a sketch ofthe figure helps organize and visualize the problem.

8.6 Applications541Step 1: Read and list.Known items: The figure is a rectangle. The length is 5 inches morethan twice the width. The perimeter is 34 inches.Unknown items: The length and the widthStep 2: Assign a variable, and translate information.Because the length is given in terms of the width (the length is 5 morethan twice the width), we let x = the width of the rectangle. The lengthis 5 more than twice the width, so it must be 2x + 5. The diagram belowis a visual description of the relationships we have listed so far.x2x + 5Step 3: Reread, and write an equation.The equation that describes the situation isStep 4: Solve the equation.Twice the length + twice the width is the perimeter2(2x + 5) + 2x = 342(2x + 5) + 2x = 34 Original equation4x + 10 + 2x = 34Distributive property6x + 10 = 34Add 4x and 2x6x = 24Add −10 to each sidex = 4 Divide each side by 6Step 5: Write the answer.The width x is 4 inches. The length is 2x + 5 = 2(4) + 5 = 13 inches.Step 6: Reread, and check.If the length is 13 and the width is 4, then the perimeter must be2(13) + 2(4) = 26 + 8 = 34, which checks with the original problem.Coin ProblemExample 5Jennifer has $2.45 in dimes and nickels. If she has 8 moredimes than nickels, how many of each coin does she have?Solution5. Amy has $1.75 in dimes andquarters. If she has 7 moredimes than quarters, how manyof each coin does she have?Step 1:Read and list.Step 2:Known items: The type of coins, the total value of the coins, and thatthere are 8 more dimes than nickelsUnknown items: The number of nickels and the number of dimesAssign a variable, and translate information.If we let x = the number of nickels, then x + 8 = the number of dimes.Because the value of each nickel is 5 cents, the amount of money inAnswer4. Width is 5; length is 13

542Chapter 8 Linear Equations and Inequalitiesnickels is 5x. Similarly, because each dime is worth 10 cents, theamount of money in dimes is 10(x + 8). Here is a table that summarizesthe information we have so far:nickelsDimesNumber x x + 8Value 5x 10(x + 8)(in cents)Step 3:Reread, and write an equation.Because the total value of all the coins is 245 cents, the equation thatdescribes this situation isAmount of money Amount of money Total amount+=in nickelsin dimesof money5x + 10(x + 8) = 245Step 4:Solve the equation.To solve the equation, we apply the distributive property first.5x + 10x + 80 = 245 Distributive property15x + 80 = 245 Add 5x and 10x15x = 165 Add −80 to each sidex = 11 Divide each side by 15Step 5:Step 6:Write the answer.The number of nickels is x = 11.The number of dimes is x + 8 = 11 + 8 = 19.Reread, and check.To check our results11 nickels are worth 5(11) = 55 cents19 dimes are worth 10(19) = 190 centsThe total value is 245 cents = $2.45When you begin working the problems in the problem set that follows, thereare a couple of things to remember. The first is that you may have to read theproblems a number of times before you begin to see how to solve them. The secondthing to remember is that word problems are not always solved correctly thefirst time you try them. Sometimes it takes a couple of attempts and some wronganswers before you can set up and solve these problems correctly.Answer5. 3 quarters, 10 dimesGetting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is the first step in the Blueprint for Problem Solving?2. What is the last thing you do when solving an application problem?3. What good does it do you to solve application problems even when theydon’t have much to do with real life?4. Write an application problem whose solution depends on solving theequation 2x + 3 = 7.

544Chapter 8 Linear Equations and InequalitiesAge Problems [Example 3]11. Shelly is 3 years older than Michele. Four years agothe sum of their ages was 67. Find the age of eachperson now.Shelly is 39; Michele is 36four years ago NowShelly x − 1 x + 3Michele x − 4 x12. Cary is 9 years older than Dan. In 7 years the sum oftheir ages will be 93. Find the age of each man now.Dan is 35; Cary is 44now in seven yearsCary x + 9 x + 16Dan x x + 713. Cody is twice as old as Evan. Three years ago the sumof their ages was 27. Find the age of each boy now.Evan is 11; Cody is 2214. Justin is 2 years older than Ethan. In 9 years the sum oftheir ages will be 30. Find the age of each boy now.Ethan is 5; Justin is 7three years agoNownowin nine yearsCody 2x − 3 2xEvan x − 3 xJustin x + 2 x + 11Ethan x x + 915. Fred is 4 years older than Barney. Five years ago thesum of their ages was 48. How old are they now?(Begin by filling in the table.)Barney is 27; Fred is 31five years ago NowFred x − 1 x + 4Barney x − 5 x16. Tim is 5 years older than JoAnn. Six years from now thesum of their ages will be 79. How old are they now?JoAnn is 31; Tim is 36now six years from nowTim x + 5 x + 11JoAnn x x + 617. Jack is twice as old as Lacy. In 3 years the sum of theirages will be 54. How old are they now?Lacy is 16; Jack is 3218. John is 4 times as old as Martha. Five years ago the sumof their ages was 50. How old are they now?Martha is 12; John is 4819. Pat is 20 years older than his son Patrick. In 2 yearsPat will be twice as old as Patrick. How old are theynow?Patrick is 18; Pat is 3820. Diane is 23 years older than her daughter Amy. In 6years Diane will be twice as old as Amy. How old arethey now?Amy is 17; Diane is 40

8.6 Problem Set545Geometry Problems [Example 4]21. The perimeter of a square is 36 inches. Find the lengthof one side.s = 9 inches22. The perimeter of a square is 44 centimeters. Find thelength of one side.s = 11 centimeters23. The perimeter of a square is 60 feet. Find the length ofone side.s = 15 feet24. The perimeter of a square is 84 meters. Find the lengthof one side.s = 21 meters25. One side of a triangle is three times the shortest side.The third side is 7 feet more than the shortest side.The perimeter is 62 feet. Find all three sides.11 feet, 18 feet, 33 feet26. One side of a triangle is half the longest side. The thirdside is 10 meters less than the longest side. The perimeteris 45 meters. Find all three sides.22 meters, 11 meters, 12 meters27. One side of a triangle is half the longest side. The thirdside is 12 feet less than the longest side. The perimeteris 53 feet. Find all three sides.26 feet, 13 feet, 14 feet28. One side of a triangle is 6 meters more than twice theshortest side. The third side is 9 meters more than theshortest side. The perimeter is 75 meters. Find all threesides.15 meters, 24 meters, 36 meters

546Chapter 8 Linear Equations and Inequalities29. The length of a rectangle is 5 inches more than thewidth. The perimeter is 34 inches. Find the length andwidth.l = 11 inches; w = 6 inches30. The width of a rectangle is 3 feet less than the length.The perimeter is 10 feet. Find the length and width.l = 4 feet; w = 1 footxx + 531. The length of a rectangle is 7 inches more than twicethe width. The perimeter is 68 inches. Find the lengthand width.l = 25 inches; w = 9 inches32. The length of a rectangle is 4 inches more than threetimes the width. The perimeter is 72 inches. Find thelength and width.l = 28 inches; w = 8 inches33. The length of a rectangle is 6 feet more than threetimes the width. The perimeter is 36 feet. Find thelength and width.l = 15 feet; w = 3 feet34. The length of a rectangle is 3 feet less than twice thewidth. The perimeter is 54 feet. Find the length andwidth.l = 17 feet; w = 10 feetCoin Problems [Example 5]35. Marissa has $4.40 in quarters and dimes. If she has5 more quarters than dimes, how many of each coindoes she have?9 dimes; 14 quarters36. Kendra has $2.75 in dimes and nickels. If she has twiceas many dimes as nickels, how many of each coin doesshe have?11 nickels; 22 dimesdimesquartersnickelsdimesNumber x x + 5Value (cents) 10(x) 25(x + 5)Number x 2xValue (cents) 5(x) 10(2x)

8.6 Problem Set54737. Tanner has $4.35 in nickels and quarters. If he has 15more nickels than quarters, how many of each coindoes he have?12 quarters; 27 nickels38. Connor has $9.00 in dimes and quarters. If he has twiceas many quarters as dimes, how many of each coin doeshe have?15 dimes; 30 quartersnickelsquartersdimesquartersNumber x + 15 xValue (cents) 5(x + 15) 25(x)Number x 2xValue (cents) 10(x) 25(2x)39. Sue has $2.10 in dimes and nickels. If she has 9 moredimes than nickels, how many of each coin does shehave?8 nickels; 17 dimes40. Mike has $1.55 in dimes and nickels. If he has 7 morenickels than dimes, how many of each coin does hehave?8 dimes; 15 nickels41. Katie has a collection of nickels, dimes, and quarterswith a total value of $4.35. There are 3 more dimesthan nickels and 5 more quarters than nickels. Howmany of each coin is in her collection?7 nickels; 10 dimes; 12 quarters42. Mary Jo has $3.90 worth of nickels, dimes, and quarters.The number of nickels is 3 more than the numberof dimes. The number of quarters is 7 more than thenumber of dimes. How many of each coin is in hercollection?8 nickels; 5 dimes; 12 quartersnickels dimes quartersNumber x x + 3 x + 5Value 5x 10(x + 3) 25(x + 5)nickels dimes quartersNumber x + 3 x x + 7Value 5(x + 3) 10(x ) 25(x + 7)43. Cory has a collection of nickels, dimes, and quarterswith a total value of $2.55. There are 6 more dimesthan nickels and twice as many quarters as nickels.How many of each coin is in her collection?3 nickels; 9 dimes; 6 quarters44. Kelly has a collection of nickels, dimes, and quarterswith a total value of $7.40. There are four more nickelsthan dimes and twice as many quarters as nickels. Howmany of each coin is in her collection?12 nickels; 8 dimes; 24 quartersnickels dimes quartersNumber x x + 6 2xValue 5x 10(x + 6) 25(2x)nickels dimes quartersNumber x + 4 x 2(x + 4)Value 5(x + 4) 10(x) 25[2(x + 4)]

548Chapter 8 Linear Equations and InequalitiesGetting Ready for the Next SectionTo understand all of the explanations and examples in the next section you must be able to work the problems below.Simplify the following expressions.45. x + 2x + 2x46. x + 2x + 3x47. x + 0.075x48. x + 0.065x5x6x1.075x1.065x49. 0.09(x + 2,000)50. 0.06(x + 1,500)0.09x + 1800.06x + 90Solve each of the following equations.51. 0.05x + 0.06(x − 1,500) = 5706,00052. 0.08x + 0.09(x + 2,000) = 6903,00053. x + 2x + 3x = 1803054. 2x + 3x + 5x = 18018Maintaining Your SkillsWrite an equivalent statement in English.55. 4 < 1056. 4 ≤ 1057. 9 ≥ −558. x − 2 > 44 is less than 10.4 is less than or equal to 10.9 is greater than or equal to−5.The difference of x and 2 isgreater than 4.Place the symbol < or the symbol > between the quantities in each expression.59. 12 2060. −12 2061. −8 −662. −10 −20

More Applications 8.7Now that you have worked through a number of application problems using ourblueprint, you probably have noticed that step 3, in which we write an equationthat describes the situation, is the key step. Anyone with experience solvingapplication problems will tell you that there will be times when your firstattempt at step 3 results in the wrong equation. Remember, mistakes are part ofthe process of learning to do things correctly. Many times the correct equationwill become obvious after you have written an equation that is partially wrong.In any case it is better to write an equation that is partially wrong and be activelyinvolved with the problem than to write nothing at all. Application problems, likeother problems in algebra, are not always solved correctly the first time.ObjectivesA Apply the Blueprint for ProblemSolving to a variety of applicationproblems.A Consecutive IntegersOur next example involves consecutive integers. When we ask for consecutiveintegers, we mean integers that are next to each other on the number line, like5 and 6, or 13 and 14, or −4 and −3. In the dictionary, consecutive is defined asfollowing one another in uninterrupted order. If we ask for consecutive odd integers,then we mean odd integers that follow one another on the number line. Forexample, 3 and 5, 11 and 13, and −9 and −7 are consecutive odd integers. As youcan see, to get from one odd integer to the next consecutive odd integer we add 2.If we are asked to find two consecutive integers and we let x equal the first integer,the next one must be x + 1, because consecutive integers always differ by 1.Likewise, if we are asked to find two consecutive odd or even integers, and we let xequal the first integer, then the next one will be x + 2, because consecutive even orodd integers always differ by 2. Here is a table that summarizes this information.In Words using Algebra ExampleTwo consecutive x, x + 1 The sum of two consecutive integers is 15.integersx + (x + 1) = 15 or7 + 8 = 15Three consecutive x, x + 1, x + 2 The sum of three consecutive integers is 24.integersx + (x + 1) + (x + 2) = 24 or7 + 8 + 9 = 24Two consecutive x, x + 2 The sum of two consecutive odd integers is 16.odd integersx + (x + 2) = 16 or7 + 9 = 16Two consecutive x, x + 2 The sum of two consecutive even integers is 18.even integersx + (x + 2) = 18 or8 + 10 = 18Example 1The sum of two consecutive odd integers is 28. Find thetwo integers.SolutionStep 1: Read and list.Known items: Two consecutive odd integers. Their sum is equal to 28.Practice Problems1. The sum of two consecutiveintegers is 27. What are the twointegers?Unknown items: The numbers in question8.7 More Applications549

550Chapter 8 Linear Equations and InequalitiesStep 2: Assign a variable, and translate information.If we let x = the first of the two consecutive odd integers, then x + 2 isthe next consecutive one.Step 3: Reread, and write an equation.Their sum is 28.Step 4: Solve the equation.Step 5: Write the answer.2x + 2 = 282x = 26x = 13x + (x + 2) = 28Simplify the left sideAdd −2 to each sideMultiply each side by ​ 1 _2 ​The first of the two integers is 13. The second of the two integers will betwo more than the first, which is 15.Step 6: Reread, and check.Suppose the first integer is 13. The next consecutive odd integer is 15.The sum of 15 and 13 is 28.Interest2. On July 1, 1992, Alfredo openeda savings account that earned4% in annual interest. At thesame time, he put $3,000 morethan he had in his savings intoa mutual fund that paid 5%annual interest. One year later,the total interest from bothaccounts was $510. How muchmoney did Alfredo put in hissavings account?Example 2Suppose you invest a certain amount of money in anaccount that earns 8% in annual interest. At the same time, you invest $2,000more than that in an account that pays 9% in annual interest. If the total interestfrom both accounts at the end of the year is $690, how much is invested in eachaccount?SolutionStep 1: Read and list.Known items: The interest rates, the total interest earned, and howmuch more is invested at 9%Unknown items: The amounts invested in each accountStep 2: Assign a variable, and translate information.Let x = the amount of money invested at 8%. From this, x + 2,000 = theamount of money invested at 9%. The interest earned on x dollarsinvested at 8% is 0.08x. The interest earned on x + 2,000 dollarsinvested at 9% is 0.09(x + 2,000).Here is a table that summarizes this information:Dollars Invested Dollars Investedat 8% at 9%Number of x x + 2,000Interest on 0.08x 0.09(x + 2,000)Step 3: Reread, and write an equation.Because the total amount of interest earned from both accounts is $690,the equation that describes the situation isAnswer1. 13 and 14Interest earned Interest earned Total interest+=at 8%at 9%earned0.08x + 0.09(x + 2,000) = 690

8.7 More Applications551Step 4: Solve the equation.0.08x + 0.09(x + 2,000) = 6900.08x + 0.09x + 180 = 690 Distributive property0.17x + 180 = 690 Add 0.08x and 0.09x0.17x = 510 Add −180 to each sidex = 3,000 Divide each side by 0.17Step 5: Write the answer :The amount of money invested at 8% is $3,000, whereas the amount ofmoney invested at 9% is x + 2,000 = 3,000 + 2,000 = $5,000.Step 6: Reread, and check.The interest at 8% is 8% of 3,000 = 0.08(3,000) = $240The interest at 9% is 9% of 5,000 = 0.09(5,000) = $450The total interest is $690FACTS FROM GEOMETRY: Labeling Triangles and the Sum of the Angles in a TriangleOne way to label the important parts of a triangle is to label the verticeswith capital letters and the sides with small letters, as shown in Figure 1.BcaAbFigure 1CIn Figure 1, notice that side a is opposite vertex A, side b is opposite vertexB, and side c is opposite vertex C. Also, because each vertex is the vertex ofone of 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 1, the relationship is writtenA + B + C = 180°TrianglesExample 3The angles in a triangle are such that one angle is twicethe smallest angle, whereas the third angle is three times as large as the smallestangle. Find the measure of all three angles.SolutionStep 1: Read and list.Known items:The sum of all three angles is 180°; one angle is twice thesmallest angle; the largest angle is three times the smallest angle.3. The angles in a triangle aresuch that one angle is threetimes the smallest angle,whereas the largest angle is fivetimes the smallest angle. Findthe measure of all three angles.Answer2. $4,000

552Chapter 8 Linear Equations and InequalitiesUnknown items: 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 anotherangle and 3x will be the measure of the largest angle.Step 3: Reread, and write an equation.When working with geometric objects, drawing a generic diagramsometimes will help us visualize what it is that we are asked to find. InFigure 2, we draw a triangle with angles A, B, and C.CbaAcFigure 2BWe can let the value of A = x, the value of B = 2x, and the value ofC = 3x. We know that the sum of angles A, B, and C will be 180°, so ourequation 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 checkGetting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. How do we label triangles?2. What rule is always true about the three angles in a triangle?3. Write an application problem whose solution depends on solving theequation x + 0.075x = 500.Answer3. 20°, 60°, and 100°4. Write an application problem whose solution depends on solving theequation 0.05x + 0.06(x + 200) = 67.

8.7 Problem Set553Problem Set 8.7A Consecutive Integer Problems [Example 1]1. The sum of two consecutive integers is 11. Find thenumbers.5 and 62. The sum of two consecutive integers is 15. Find thenumbers.7 and 83. The sum of two consecutive integers is −9. Find thenumbers.−4 and −54. The sum of two consecutive integers is −21. Find thenumbers.−10 and −115. The sum of two consecutive odd integers is 28. Findthe numbers.13 and 156. The sum of two consecutive odd integers is 44. Find thenumbers.21 and 237. The sum of two consecutive even integers is 106. Findthe numbers.52 and 548. The sum of two consecutive even integers is 66. Find thenumbers.32 and 349. The sum of two consecutive even integers is −30. Findthe numbers.−14 and −1610. The sum of two consecutive odd integers is −76. Findthe numbers.−37 and −3911. The sum of three consecutive odd integers is 57. Findthe numbers.17, 19, and 2112. The sum of three consecutive odd integers is −51. Findthe numbers.−15, −17, and −1913. The sum of three consecutive even integers is 132.Find the numbers.42, 44, and 4614. The sum of three consecutive even integers is −108.Find the numbers.−34, −36, and −38

554Chapter 8 Linear Equations and InequalitiesA Interest Problems [Example 2]15. Suppose you invest money in two accounts. One ofthe accounts pays 8% annual interest, whereas theother pays 9% annual interest. If you have $2,000 moreinvested at 9% than you have invested at 8%, howmuch do you have invested in each account if the totalamount of interest you earn in a year is $860? (Beginby completing the following table.)$4,000 invested at 8%, $6,000 invested at 9%DollarsDollarsinvestedinvestedat 8% at 9%Number of x x + 2,00016. Suppose you invest a certain amount of money in anaccount that pays 11% interest annually, and $4,000more than that in an account that pays 12% annually.How much money do you have in each account if thetotal interest for a year is $940?$2,000 invested at 11%, $6,000 invested at 12%DollarsDollarsinvestedinvestedat 11% at 12%Number of x x + 4,000Interest on .11x .12 (x + 4,000)Interest on .08x .09 (x + 2,000)17. Tyler has two savings accounts that his grandparentsopened for him. The two accounts pay 10% and 12% inannual interest; there is $500 more in the account thatpays 12% than there is in the other account. If the totalinterest for a year is $214, how much money does hehave in each account?$700 invested at 10%, $1,200 invested at 12%18. Travis has a savings account that his parents opened forhim. It pays 6% annual interest. His uncle also opened anaccount for him, but it pays 8% annual interest. If thereis $800 more in the account that pays 6%, and the totalinterest from both accounts is $104, how much money isin each of the accounts?$1,200 at 6%, $400 at 8%19. A stockbroker has money in three accounts. The interestrates on the three accounts are 8%, 9%, and 10%.If she has twice as much money invested at 9% as shehas invested at 8%, three times as much at 10% as shehas at 8%, and the total interest for the year is $280,how much is invested at each rate? (Hint: Let x = theamount invested at 8%.)$500 at 8%, $1,000 at 9%, $1,500 at 10%20. An accountant has money in three accounts that pay9%, 10%, and 11% in annual interest. He has twice asmuch invested at 9% as he does at 10% and three timesas much invested at 11% as he does at 10%. If the totalinterest from the three accounts is $610 for the year,how much is invested at each rate? (Hint: Let x = theamount invested at 10%.)$2,000 at 9%, $1,000 at 10%, $3,000 at 11%

8.7 Problem Set555A Triangle Problems [Example 3]21. 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?45°, 45°, 90°22. One angle in a triangle measures twice the smallestangle, whereas the largest angle is six times the smallestangle. Find the measures of all three angles.20°, 40°, 120°23. The smallest angle in a triangle is ​_1 ​as large as the5largest angle. The third angle is twice the smallestangle. Find the three angles.22.5°, 45°, 112.5°24. One angle in a triangle is half the largest angle but threetimes the smallest. Find all three angles.18°, 54°, 108°25. A right triangle has one 37° angle. Find the other twoangles.53°, 90°26. In a right triangle, one of the acute angles is twice aslarge as the other acute angle. Find the measure of thetwo acute angles.30°, 60°27. One angle of a triangle measures 20° more than thesmallest, while a third angle is twice the smallest. Findthe measure of each angle.80°, 60°, 40°28. One angle of a triangle measures 50° more than thesmallest, while a third angle is three times the smallest.Find the measure of each angle.78°, 76°, 26°Miscellaneous Problems29. Ticket Prices Miguel is selling tickets to a barbecue.Adult tickets cost $6.00 and children’s tickets cost$4.00. He sells six more children’s tickets than adulttickets. The total amount of money he collects is $184.How many adult tickets and how many children’stickets did he sell?16 adult and 22 children’s ticketsAdult ChildNumber x x + 6Income 6(x) 4(x + 6)30. Working Two Jobs Maggie has a job working in an officefor $10 an hour and another job driving a tractor for $12an hour. One week she works in the office twice as longas she drives the tractor. Her total income for that weekis $416. How many hours did she spend at each job?13 hours driving the tractor and 26 hours working at the officeJob Office TractorHours Worked 2x xWages Earned 10(2x) 12x

556Chapter 8 Linear Equations and Inequalities31. Phone Bill The cost of a long-distance phone call is$0.41 for the first minute and $0.32 for each additionalminute. If the total charge for a long-distance call is$5.21, how many minutes was the call?16 minutes32. Phone Bill Danny, who is 1 year old, is playing with thetelephone when he accidentally presses one of the buttonshis mother has programmed to dial her friend Sue’snumber. Sue answers the phone and realizes Danny ison the other end. She talks to Danny, trying to get him tohang up. The cost for a call is $0.23 for the first minuteand $0.14 for every minute after that. If the total chargefor the call is $3.73, how long did it take Sue to convinceDanny to hang up the phone?26 minutes33. Hourly Wages JoAnn works in the publicity office at thestate university. She is paid $12 an hour for the first 35hours she works each week and $18 an hour for everyhour after that. If she makes $492 one week, howmany hours did she work?39 hours34. Hourly Wages Diane has a part-time job that pays her$6.50 an hour. During one week she works 26 hours andis paid $178.10. She realizes when she sees her checkthat she has been given a raise. How much per hour isthat raise?$0.35 per hour raise35. Office Numbers Professors Wong and Gil have officesin the mathematics building at Miami Dade College.Their office numbers are consecutive odd integers witha sum of 14,660. What are the office numbers of thesetwo professors?They are in offices 7329 and 7331.36. Cell Phone Numbers Diana and Tom buy two cell phones.The phone numbers assigned to each are consecutiveintegers with a sum of 11,109,295. If the smaller numberis Diana’s, what are their phone numbers?Diana’s phone number is 555-4647 and Tom’s is 555-4648.37. Age Marissa and Kendra are 2 years apart in age.Their ages are two consecutive even integers. Kendrais the younger of the two. If Marissa’s age is added totwice Kendra’s age, the result is 26. How old is eachgirl?Kendra is 8 years old and Marissa is 10 years old.38. Age Justin’s and Ethan’s ages form two consecutive oddintegers. What is the difference of their ages?The difference is 2 or −2, depending on the order of subtraction.39. Arrival Time Jeff and Carla Cole are driving separatelyfrom San Luis Obispo, California, to the north shoreof Lake Tahoe, a distance of 425 miles. Jeff leaves SanLuis Obispo at 11:00 am and averages 55 miles perhour on the drive, Carla leaves later, at 1:00 pm butaverages 65 miles per hour. Which person arrives inLake Tahoe first?Jeff40. Piano Lessons Tyler is taking piano lessons. Because hedoesn’t practice as often as his parents would like himto, he has to pay for part of the lessons himself. His parentspay him $0.50 to do the laundry and $1.25 to mowthe lawn. In one month, he does the laundry 6 moretimes than he mows the lawn. If his parents pay him$13.50 that month, how many times did he mow thelawn?6 times

8.7 Problem Set557At one time, the Texas Junior College Teachers Association annual conference was held in Austin. At that time a taxi ridein Austin was $1.25 for the first ​_1 5 ​of a mile and $0.25 for each additional ​1 _​of a mile. Use this information for Problems 415and 42.41. Cost of a Taxi Ride If the distance from one of the conventionhotels to the airport is 7.5 miles, how muchwill it cost to take at taxi from that hotel to the airport?$10.3842. Cost of a Taxi Ride Suppose the distance from one of thehotels to one of the western dance clubs in Austin is 12.4miles. If the fare meter in the taxi gives the charge forthat trip as $16.50, is the meter working correctly?yes43. Geometry The length and width of a rectangle are consecutiveeven integers. The perimeter is 44 meters.Find the length and width.l = 12 meters; w = 10 meters44. Geometry The length and width of a rectangle are consecutiveodd integers. The perimeter is 128 meters. Findthe length and width.l = 33 meters; w = 31 meters45. Geometry The angles of a triangle are three consecutiveintegers. Find the measure of each angle.59°, 60°, 61°46. Geometry The angles of a triangle are three consecutiveeven integers. Find the measure of each angle.58°, 60°, 62°Ike and Nancy Lara give western dance lessons at the Elks Lodge on Sunday nights. The lessons cost $3.00 for membersof the lodge and $5.00 for nonmembers. Half of the money collected for the lesson is paid to Ike and Nancy. The ElksLodge keeps the other half. One Sunday night Ike counts 36 people in the dance lesson. Use this information to workProblems 47 through 50.47. Dance Lessons What is the least amount of money Ikeand Nancy will make?$54.0048. Dance Lessons What is the largest amount of money Ikeand Nancy will make?$90.0049. Dance Lessons At the end of the evening, the ElksLodge gives Ike and Nancy a check for $80 to coverhalf of the receipts. Can this amount be correct?Yes50. Dance Lessons Besides the number of people in the dancelesson, what additional information does Ike need toknow to always be sure he is being paid the correctamount?How many of the people taking lessons are members, or how many arenot members

558Chapter 8 Linear Equations and InequalitiesGetting Ready for the Next SectionTo understand all of the explanations and examples in the next section you must be able to work the problems below.Solve the following equations.51. a. x − 3 = 6 9b. x + 3 = 6 3c. −x − 3 = 6 −9d. −x + 3 = 6 −352. a. x − 7 = 16 23b. x + 7 = 16 9c. −x − 7 = 16 −23d. −x + 7 = 16 −953. a. ​_x ​= −2 −854. a. 3a = 15 54_b. −​ x b. 3a = −15 −5​= −2 8c. −3a = 15 −54c. ​_x d. −3a = −15 54 ​= 2 8_d. −​ x 4 ​= 2 −855. 2.5x − 3.48 = 4.9x + 2.07−2.312557. 3(x − 4) = −2​_1056. 2(1 − 3x) + 4 = 4x − 14258. Solve 2x − 3y = 6 for y.3 ​ y = ​_ 2 3 ​x − 2Maintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.Simplify the expression 36x − 12 for each of the following values of x.59. ​ 1 _4 ​−360. ​ 1 _6 ​−661. ​ 1 _9 ​−862. ​ 3 _2 ​4263. ​ 1 _3 ​064. ​ 5 _12 ​365. ​ 5 _9 ​866. ​ 2 _3 ​12Find the value of each expression when x = −4.67. 3(x − 4)−2468. −3(x − 4)2469. −5x + 82870. 5x + 8−1273. ​_16 ​+ 3xx−1671. ​_x − 1436 ​72. ​_x − 1236 ​_−​ 1 2 ​ _−​ 4 9 ​74. ​_16 ​− 3x75. 7x − ​_12x x ​8−2576. 7x + ​_12x ​−3177. 8 ​ ​ x _2 ​+ 5 ​2478. −8 ​ ​ x _2 ​+ 5 ​−24

Linear Inequalities 8.8Linear inequalities are solved by a method similar to the one used in solving linearequations. The only real differences between the methods are in the multiplicationproperty for inequalities and in graphing the solution set.A Addition Property for InequalitiesAn inequality differs from an equation only with respect to the comparison symbolbetween the two quantities being compared. In place of the equal sign, weuse < (less than), ≤ (less than or equal to), > (greater than), or ≥ (greater than orequal to). The addition property for inequalities is almost identical to the additionproperty for equality.ObjectivesA Use the addition property forinequalities to solve an inequality.B Use the multiplication property forinequalities to solve an inequality.C Use both the addition andmultiplication properties to solve aninequality.D Graph the solution set for aninequality.ETranslate and solve applicationproblems involving inequalities.Addition Property for InequalitiesFor any three algebraic expressions A, B, and C,ifA < Bthen A + C < B + CIn words: Adding the same quantity to both sides of an inequality will notchange the solution set.It makes no difference which inequality symbol we use to state the property.Adding the same amount to both sides always produces an inequality equivalentto the original inequality. Also, because subtraction can be thought of as additionof the opposite, this property holds for subtraction as well as addition.Example 1Solve the inequality x + 5 < 7.Solution To isolate x, we add −5 to both sides of the inequality:x + 5 < 7x + 5 + (−5) < 7 + (−5) Addition property for inequalitiesx < 2We can go one step further here and graph the solution set. The solution set isall real numbers less than 2. To graph this set, we simply draw a straight line andlabel the center 0 (zero) for reference. Then we label the 2 on the right side ofzero and extend an arrow beginning at 2 and pointing to the left. We use an opencircle at 2 because it is not included in the solution set. Here is the graph.Practice Problems1. Solve the inequality x + 3 < 5,and then graph the solution.0 2Example 2Solve x − 6 ≤ −3.Solution Adding 6 to each side will isolate x on the left side:x − 6 ≤ −3x − 6 + 6 ≤ −3 + 6 Add 6 to both sidesx ≤ 3The graph of the solution set is2. Solve x − 4 ≤ 1, and then graphthe solution.0 3Notice that the dot at the 3 is darkened because 3 is included in the solutionset. We always will use open circles on the graphs of solution sets with < or >and closed (darkened) circles on the graphs of solution sets with ≤ or ≥.8.8 Linear InequalitiesAnswers1. x < 2 2. x ≤ 5559

560NoteThis discussion isintended to showwhy the multiplicationproperty for inequalities is writtenthe way it is. You may want to lookahead to the property itself andthen come back to this discussionif you are having trouble makingsense out of it.Chapter 8 Linear Equations and InequalitiesB Multiplication Property for InequalitiesTo see the idea behind the multiplication property for inequalities, we will considerthree true inequality statements and explore what happens when we multiply bothsides by a positive number and then what happens when we multiply by a negativenumber.Consider the following three true statements:3 < 5 −3 < 5 −5 < −3Now multiply both sides by the positive number 4:4(3) < 4(5) 4(−3) < 4(5) 4(−5) < 4(−3)12 < 20 −12 < 20 −20 < −12In each case, the inequality symbol in the result points in the same direction itdid in the original inequality. We say the “sense” of the inequality doesn’t changewhen we multiply both sides by a positive quantity.Notice what happens when we go through the same process but multiply bothsides by −4 instead of 4:3 < 5 −3 < 5 −5 < −3−4(3) > −4(5) −4(−3) > −4(5) −4(−5) > −4(−3)−12 > −20 12 > −20 20 > 12In each case, we have to change the direction in which the inequality symbolpoints to keep each statement true. Multiplying both sides of an inequality by anegative quantity always reverses the sense of the inequality. Our results are summarizedin the multiplication property for inequalities.NoteBecause division isdefined in terms ofmultiplication, thisproperty is also true for division.We can divide both sides of aninequality by any nonzero numberwe choose. If that number happensto be negative, we mustalso reverse the direction of theinequality symbol.Multiplication Property for InequalitiesFor any three algebraic expressions A, B, and C,if A < Bthen AC < BC when C is positiveand AC > BC when C is negativeIn words: Multiplying both sides of an inequality by a positive number doesnot change the solution set. When multiplying both sides of aninequality by a negative number, it is necessary to reverse theinequality symbol to produce an equivalent inequality.We can multiply both sides of an inequality by any nonzero number we choose.If that number happens to be negative, we must also reverse the sense of theinequality.3. Solve 4a < 12, and graph thesolution.Example 3Solve 3a < 15 and graph the solution.Solution We begin by multiplying each side by ​_1 3 ​. Because ​1 _​is a positive3number, we do not reverse the direction of the inequality symbol:3a < 15​ 1 _3 ​(3a) < ​1 _3 ​(15) Multiply each side by ​1 _3 ​a < 5Answer3. a < 30 5

8.8 Linear Inequalities561Example 4Solve −3a ≤ 18, and graph the solution._Solution We begin by multiplying both sides by −​ 1 3 ​. Because −​1 _​is a negative3number, we must reverse the direction of the inequality symbol at the same time_that we multiply by −​ 1 3 ​. −3a ≤ 184. Solve −2x ≤ 10, and graph thesolution._−​ 1 3 ​(−3a) ≥ −​1 _3 ​(18) Multiply both sides by −​ _ 1 ​ and reverse3the direction of the inequality symbola ≥ −6−6 0Example 5Solve −​ x _Solution​> 2 and graph the solution.4To isolate x, we multiply each side by −4. Because −4 is a negativenumber, we also must reverse the direction of the inequality symbol:_5. Solve −​ x ​> 2, and graph the3solution.−​ x _4 ​> 2−4 ​ −​ x _4 ​ ​< −4(2) Multiply each side by −4, and reverse the direction ofthe inequality symbolx < −8−8 0C Solving Linear Inequalities in One VariableTo solve more complicated inequalities, we use the following steps.StrategySolving Linear Inequalities in One VariableStep 1a: Use the distributive property to separate terms, if necessary.1b: If fractions are present, consider multiplying both sides by theLCD to eliminate the fractions. If decimals are present, considermultiplying both sides by a power of 10 to clear the inequality ofdecimals.1c: Combine similar terms on each side of the inequality.Step 2: Use the addition property for inequalities to get all variable terms onone side of the inequality and all constant terms on the other side.Step 3: Use the multiplication property for inequalities to get x by itself onone side of the inequality.Step 4: Graph the solution set.Answers4. x ≥ −5 5. x < −6

562Chapter 8 Linear Equations and Inequalities6. Solve and graph.4.5x + 2.31 > 6.3x − 4.89Example 6Solve 2.5x − 3.48 < −4.9x + 2.07.Solution We have two methods we can use to solve this inequality. We cansimply apply our properties to the inequality the way it is currently written andwork with the decimal numbers, or we can eliminate the decimals to begin withand solve the resulting inequality.Method 1 Working with the decimals.2.5x − 3.48 < −4.9x + 2.07 Original inequality2.5x + 4.9x − 3.48 < −4.9x + 4.9x + 2.07 Add 4.9x to each side7.4x − 3.48 < 2.077.4x − 3.48 + 3.48 < 2.07 + 3.48 Add 3.48 to each side7.4x < 5.55​_7.4x ​< ​5.55_​7.4 7.4Divide each side by 7.4x < 0.75Method 2 Eliminating the decimals in the beginning.Because the greatest number of places to the right of the decimal point in anyof the numbers is 2, we can multiply each side of the inequality by 100 and wewill be left with an equivalent inequality that contains only whole numbers.2.5x − 3.48 < −4.9x + 2.07 Original inequality100(2.5x − 3.48) < 100(−4.9x + 2.07) Multiply each side by 100100(2.5x) − 100(3.48) < 100(−4.9x) + 100(2.07) Distributive property250x − 348 < −490x + 207Multiplication740x − 348 < 207Add 490x to each side740x < 555Add 348 to each side​_740x ​< ​555_​740 740Divide each side by 740x < 0.75The solution by either method is x < 0.75. Here is the graph:0 0.757. Solve and graph.2(x − 5) ≥ −6Example 7Solve 3(x − 4) ≥ −2.Solution3x − 12 ≥ −2 Distributive property3x − 12 + 12 ≥ −2 + 12 Add 12 to both sides3x ≥ 10​ 1 _3 ​(3x) ≥ ​1 _3 ​(10) Multiply both sides by ​1 _3 ​x ≥ ​ 10 _3 ​0 103Answers6. x < 4 7. x ≥ 2

8.8 Linear Inequalities563DGraphing InequalitiesExample 8Solve and graph 2(1 − 3x) + 4 < 4x − 14.Solution8. Solve and graph.3(1 − 2x) + 5 < 3x − 12 − 6x + 4 < 4x − 14 Distributive property−6x + 6 < 4x − 14−6x + 6 + (−6) < 4x − 14 + (−6)SimplifyAdd −6 to both sides−6x < 4x − 20−6x + (−4x) < 4x + (−4x) − 20Add −4x to both sides−10x < −20​ −​ 1 _10 ​ ​(−10x) > ​ −​ 1 _x > 210 ​ _ ​(−20) Multiply by −​ 1 ​, reverse10the sense of the inequality0 2Example 9Solve 2x − 3y < 6 for y.Solution We can solve this formula for y by first adding −2x to each side and_then multiplying each side by −​ 1 3 ​. When we multiply by −​ _ 1 ​we must reverse the3direction of the inequality symbol. Because this is an inequality in two variables,we will not graph the solution.9. Solve 2x − 3y < 6 for x.2x − 3y < 62x + (−2x) − 3y < (−2x) + 6Original inequalityAdd −2x to each side−3y < −2x + 6EApplications−​ 1 _3 ​(−3y) > −​1 _3 ​(−2x + 6) Multiply each side by −​1 _3 ​y > ​_2 ​x − 2 Distributive property3When working application problems that involve inequalities, the phrases “atleast” and “at most” translate as follows:In WordsIn Symbolsx is at least 30 x ≥ 30x is at most 20 x ≤ 20Our next example is similar to an example done earlier in this chapter. This timeit involves an inequality instead of an equation.We can modify our Blueprint for Problem Solving to solve application problemswhose solutions depend on writing and then solving inequalities.Answers8. x > 1 9. x < ​ 3 _2 ​y + 3

564Chapter 8 Linear Equations and Inequalities10. The sum of two consecutiveintegers is at least 27. What arethe possibilities for the first ofthe two integers?Example 10The sum of two consecutive odd integers is at most 28.What are the possibilities for the first of the two integers?Solution When we use the phrase “their sum is at most 28,” we mean thattheir sum is less than or equal to 28.Step 1: Read and list.Known items: Two consecutive odd integers. Their sum is less than orequal to 28.Unknown items: The numbers in questionStep 2: Assign a variable, and translate information.If we let x = the first of the two consecutive odd integers, then x + 2 isthe next consecutive one.Step 3: Reread, and write an inequality.Their sum is at most 28.x + (x + 2) ≤ 28Step 4: Solve the inequality.2x + 2 ≤ 282x ≤ 26x ≤ 13Simplify the left sideAdd −2 to each sideMultiply each side by ​_1 2 ​Step 5: Write the answer.The first of the two integers must be an odd integer that is less thanor equal to 13. The second of the two integers will be two more thanwhatever the first one is.Step 6: Reread, and check.Suppose the first integer is 13. The next consecutive odd integer is 15.The sum of 15 and 13 is 28. If the first odd integer is less than 13, thesum of it and the next consecutive odd integer will be less than 28.Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. State the addition property for inequalities.2. How is the multiplication property for inequalities different from themultiplication property of equality?3. When do we reverse the direction of an inequality symbol?4. Under what conditions do we not change the direction of the inequalitysymbol when we multiply both sides of an inequality by a number?Answer10. The first integer is at least 13;x ≥ 13

8.8 Problem Set565Problem Set 8.8A D Solve the following inequalities using the addition property of inequalities. Graph each solution set. [Examples 1–2]1. x − 5 < 7x < 122. x + 3 < −5x < −83. a − 4 ≤ 8a ≤ 120 12–8 0 0 124. a + 3 ≤ 10a ≤ 75. x − 4.3 > 8.7x > 136. x − 2.6 > 10.4x > 130 70 13 0 137. y + 6 ≥ 10y ≥ 48. y + 3 ≥ 12y ≥ 99. 2 < x − 7x > 90 4 0 9 0 910. 3 < x + 8x > −5−5 0B D Solve the following inequalities using the multiplication property of inequalities. If you multiply both sides by anegative number, be sure to reverse the direction of the inequality symbol. Graph the solution set. [Examples 3–4]11. 3x < 6x < 212. 2x < 14x < 713. 5a ≤ 25a ≤ 50 20 70 514. 4a ≤ 16a ≤ 415. ​_x 3 ​> 5x > 1516. ​_x 7 ​> 1x > 7040 15 0 717. −2x > 6x < −318. −3x ≥ 9x ≤ −319. −3x ≥ −18x ≤ 6−3 0 −3 0 0 6

566Chapter 8 Linear Equations and Inequalities20. −8x ≥ −24x ≤ 3_21. −​ x ​≤ 105x ≥ −50_22. −​ x ​≥ −19x ≤ 90 3−50 0 0 9_23. −​ 2 3 ​y > 4y < −6_24. −​ 3 4 ​y > 6y < −8−6 0 −8 0C D Solve the following inequalities. Graph the solution set in each case. [Examples 6–9]25. 2x − 3 < 9x < 626. 3x − 4 < 17x < 7_27. −​ 1 5 ​y − ​1 _3 ​≤ ​2 _3 ​y ≥ −50 60 7−5 0_28. −​ 1 6 ​y − ​1 _2 ​≤ ​2 _3 ​y ≥ −729. −7.2x + 1.8 > −19.8x < 330. −7.8x − 1.3 > 22.1x < −3−7 00 3−3 031. ​_2 3 ​x − 5 ≤ 7x ≤ 1832. ​_3 4 ​x − 8 ≤ 1x ≤ 12_33. −​ 2 5 ​a − 3 > 5a < −200 180 12−20 0

8.8 Problem Set567_34. −​ 4 ​a − 2 > 105a < −1535. 5 − ​_3 ​y > −105y < 2536. 4 − ​_5 ​y > −116y < 18−15 0 0 250 1837. 0.3(a + 1) ≤ 1.2a ≤ 338. 0.4(a − 2) ≤ 0.4a ≤ 339. 2(5 − 2x) ≤ −20x ≥ ​_ 152 ​0 30 30 15240. 7(8 − 2x) > 28x < 241. 3x − 5 > 8xx < −142. 8x − 4 > 6xx > 20 2−1 0 0 243. ​_1 3 ​y − ​1 _2 ​≤ ​5 _6 ​y + ​1 _2 ​y ≥ −244. ​_7 6 ​y + ​4 _​≤ ​11_3 6 ​y − ​7 _6 ​45. −2.8x + 8.4 < −14x − 2.8y ≥ ​_ 154 ​ x < −1−2 0 0 154−1 046. −7.2x − 2.4 < −2.4x + 12x > −347. 3(m − 2) − 4 ≥ 7m + 14m ≤ −648. 2(3m − 1) + 5 ≥ 8m − 7m ≤ 5−3 0 −6 0 0 5

568Chapter 8 Linear Equations and Inequalities49. 3 − 4(x − 2) ≤ −5x + 6x ≤ −550. 8 − 6(x − 3) ≤ −4x + 12x ≥ 7−5 0 0 7Solve each of the following inequalities for y.51. 3x + 2y < 652. −3x + 2y < 653. 2x − 5y > 1054. −2x − 5y > 5_y < −​ 3 2 ​x + 3 y < ​_ 3 2 ​x + 3 y < ​_ 2 5 ​x − 2 _y < −​ 2 5 ​x − 155. −3x + 7y ≤ 2156. −7x + 3y ≤ 2157. 2x − 4y ≥ −458. 4x − 2y ≥ −8y ≤ ​_3 7 ​x + 3 y ≤ ​_ 7 3 ​x + 7 y ≤ ​_ 1 2 ​x + 1 y ≤ 2x + 4The next two problems are intended to give you practice reading, and paying attention to, the instructions that accompanythe problems you are working.59. Work each problem according to the instructionsgiven.a. Evaluate when x = 0: −5x + 3 3b. Solve: −5x + 3 = −7 2c. Is 0 a solution to −5x + 3 < −7 Nod. Solve: −5x + 3 < −7 x > 260. Work each problem according to the instructions given.a. Evaluate when x = 0: −2x − 5 −5b. Solve: −2x − 5 = 1 −3c. Is 0 a solution to −2x − 5 > 1 Nod. Solve: −2x − 5 > 1 x < −3For each graph below, write an inequality whose solution is the graph.61.62.0 3 0 3x < 3x ≤ 363.64.0 3 0 3x ≥ 3x > 3

8.8 Problem Set569E Applying the Concepts [Example 10]65. Organic Groceries The map shows a range of howmany organic stores are found in each state. Write aninequality that describes the number of organic storesfound in Florida.41 < x < 5066. Population The map shows the most populated cities inthe United States. Write an inequality that describes thepopulation of Chicago relative to San Diego.San Diego < Chicago or 1.26 < 2.89Organic Stores in the U.S.Where Is Everyone?Source: organic.com 20080-1011-2021-3031-4041-5051-6061-7071-8081-90Number of organicgrocery storesper stateLos Angeles, CA3.80San Diego, CA 1.26Phoeniz, AZ 1.37Dallas, TX 1.21Houston, TX 2.01Chicago, IL2.89Philadelphia, PA 1.49New York City, NY* Polpulation in MillionsSource: U.S. Census Bureau8.0867. Consecutive Integers The sum of two consecutive integersis at least 583. What are the possibilities for thefirst of the two integers?At least 29169. Number Problems The sum of twice a number and six isless than ten. Find all solutions.x < 268. Consecutive Integers The sum of two consecutive integersis at most 583. What are the possibilities for the first ofthe two integers?At most 29170. Number Problems Twice the difference of a number andthree is greater than or equal to the number increased byfive. Find all solutions.x ≥ 1171. Number Problems The product of a number and four is 72. Number Problems The quotient of a number and five isgreater than the number minus eight. Find the solution less than the sum of seven and two. Find the solutionset.set._x > −​ 8 3 ​ x < 4573. Geometry Problems The length of a rectangle is 3 timesthe width. If the perimeter is to be at least 48 meters,what are the possible values for the width? (If theperimeter is at least 48 meters, then it is greater thanor equal to 48 meters.)x ≥ 6; the width is at least 6 meters.74. Geometry Problems The length of a rectangle is 3 morethan twice the width. If the perimeter is to be at least 51meters, what are the possible values for the width? (Ifthe perimeter is at least 51 meters, then it is greater thanor equal to 51 meters.)x ≥ ​_ 152 ​; the width is at least 7​ _ 1 2 ​ meters.75. Geometry Problems The numerical values of the threesides of a triangle are given by three consecutive evenintegers. If the perimeter is greater than 24 inches,what are the possibilities for the shortest side?x > 6; the shortest side is even and greater than 6 inches.77. Car Heaters If you have ever gotten in a cold car earlyin the morning you know that the heater does notwork until the engine warms up. This is because theheater relies on the heat coming off the engine. Writean equation using an inequality sign to express whenthe heater will work if the heater works only after theengine is 100°F.t ≥ 10076. Geometry Problems The numerical values of the threesides of a triangle are given by three consecutive oddintegers. If the perimeter is greater than 27 inches, whatare the possibilities for the shortest side?x > 7; the shortest side is odd and greater than 7 inches.78. Exercise When Kate exercises, she either swims or runs.She wants to spend a minimum of 8 hours a week exercising,and she wants to swim 3 times the amount sheruns. What is the minimum amount of time she mustspend doing each exercise?At least 2 hours running and 6 hours swimming.

570Chapter 8 Linear Equations and Inequalities79. Profit and Loss Movie theaters pay a certain price forthe movies that you and I see. Suppose a theater pays$1,500 for each showing of a popular movie. If theycharge $7.50 for each ticket they sell, then they willlose money if ticket sales are less than $1,500. However,they will make a profit if ticket sales are greaterthan $1,500. What is the range of tickets they can selland still lose money? What is the range of tickets theycan sell and make a profit?Lose money if they sell less than 200 tickets. Make a profit if they sellmore than 200 tickets.80. Stock Sales Suppose you purchase x shares of a stock at$12 per share. After 6 months you decide to sell all yourshares at $20 per share. Your broker charges you $15for the trade. If your profit is at least $3,985, how manyshares did you purchase in the first place?At least 500 shares.Getting Ready for the Next SectionSolve each inequality. Do not graph.81. 2x − 1 ≥ 3x ≥ 283. −2x > −8x < 485. −3 ≤ 4x + 1−1 ≤ x82. 3x + 1 ≥ 7x ≥ 284. −3x > −12x < 486. 4x + 1 ≤ 9x ≤ 2Maintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.Apply the distributive property, then simplify.87. ​_1 ​(12x + 6)688. ​_3 ​(15x − 10)589. ​_2 ​(−3x − 6)390. ​_3 ​(−4x − 12)42x + 19x − 6−2x − 4−3x − 991. 3 ​ ​ 5 _6 ​a + ​4 _9 ​ ​​_5 2 ​a + ​4 _3 ​ 92. 2 ​ ​ 3 _4 ​a − ​5 _6 ​ ​​_ 3 2 ​a − ​ _ 53 ​ 93. −3 ​ ​ 2 _3 ​a + ​5 _6 ​ ​94. −4 _​ ​ 5−2a − ​_ 5 2 ​ _−​ 1036 ​a + ​4 _9 ​ ​​a − ​16 _9 ​Apply the distributive property, then find the LCD and simplify.95. ​ 1 _2 ​x + ​1 _6 ​x​_2 3 ​x _−​ 1 4 ​x _−​ 1 6 ​x96. ​_1 2 ​x − ​3 _4 ​x97. ​_2 3 ​x − ​5 _6 ​x98. ​_199. ​ 3 _4 ​x + ​1 _6 ​x100. ​_3 2 ​x − ​2 _3 ​x101. ​_2 5 ​x + ​5 _8 ​x102. ​_312 ​x ​_ 5 6 ​x ​_ 4140 ​x​_113 ​x + ​3 _5 ​x​_ 1415 ​x5 ​x − ​3 _8 ​x​_ 940 ​x

Compound Inequalities 8.9A Solving Compound InequalitiesThe union of two sets A and B is the set of all elements that are in A or in B. Theword or is the key word in the definition. The intersection of two sets A and B isthe set of elements contained in both A and B. The key word in this definition isand. We can put the words and and or together with our methods of graphinginequalities to find the solution sets for compound inequalities.ObjectivesA Solve and graph compoundinequalities.DefinitionA compound inequality is two or more inequalities connected by the wordand or or.Example 1Graph the solution set for the compound inequalityx < −1 or x ≥ 3Solution Graphing each inequality separately, we havePractice Problems1. Graph x < −3 or x ≥ 2.x −2 and x < 3SolutionGraphing each inequality separately, we havex > –2x < 3−2 00 3Because the two inequalities are connected by the word and, we will graph theirintersection, which consists of all points that are common to both graphs; that is,we graph the region where the two graphs overlap.-2 0 38.9 Compound InequalitiesAnswer1. See Solutions to Selected PracticeProblems.2. See Solutions to Selected PracticeProblems.571

572Chapter 8 Linear Equations and Inequalities3. Solve and graph.3x + 1 ≥ 7 and −2x > −8Example 3Solve and graph the solution set for2x − 1 ≥ 3 and −3x > −12Solution Solving the two inequalities separately, we have2x − 1 ≥ 3 and −3x > −122x ≥ 4x ≥ 2 and −​ 1 _3 ​(−3x) < −​1 _3 ​(−12)x < 40 20 4Because the word and is used, their intersection is the points they have in common:0 2 4The compound inequality −2 < x and x < 3 can be written as −2 < x < 3. Theword and does not appear when an inequality is written in this form. It is implied.The solution set for −2 < x and x < 3 is-2 3It is all the numbers between −2 and 3 on the number line. It seems reasonable,then, that this graph should be the graph of −2 < x < 3. In both the graph and theinequality, x is said to be between −2 and 3.4. Solve and graph.−3 ≤ 4x + 1 ≤ 9Example 4Solve and graph −3 ≤ 2x − 1 ≤ 9.SolutionTo solve for x, we must add 1 to the center expression and thendivide the result by 2. Whatever we do to the center expression, we also must doto the two expressions on the ends. In this way we can be sure we are producingequivalent inequalities. The solution set will not be affected.−3 ≤ 2x − 1 ≤ 9−2 ≤ 2x ≤ 10 Add 1 to each expression−1 ≤ x ≤ 5 Multiply each expression by ​_1 2 ​-1 5Answer3. x ≥ 2 and x < 4 4. −1 ≤ x ≤ 2Getting Ready for ClassAfter reading through the preceding section, respond in your ownwords and in complete sentences.1. What is a compound inequality?2. Explain the shorthand notation that can be used to write two inequalitiesconnected by the word and.3. Write two inequalities connected by the word and that together areequivalent to −1 < x < 2.4. Explain in words how you would graph the compound inequalityx < 2 or x > −3

8.9 Problem Set573Problem Set 8.9A Graph the following compound inequalities. [Examples 1–2]1. x < −1 or x > 5 2. x ≤ −2 or x ≥ −1−1 5 −2 −13. x < −3 or x ≥ 0 4. x < 5 and x > 1−3 0 1 55. x ≤ 6 and x > −1 6. x ≤ 7 and x > 0−1 6 0 77. x > 2 and x < 4 8. x < 2 or x > 42 4 2 49. x ≥ −2 and x ≤ 4 10. x ≤ 2 or x ≥ 4−2 4 2 411. x < 5 and x > −1 12. x > 5 or x < −1−1 5 −1 513. −1 < x < 3 14. −1 ≤ x ≤ 3−1 3 −1 315. −3 < x ≤ −2 16. −5 ≤ x ≤ 0−3 −2 −5 0A Solve the following compound inequalities. Graph the solution set in each case. [Examples 3–4]17. 3x − 1 < 5 or 5x − 5 > 10 18. x + 1 < −3 or x − 2 > 6 19. x − 2 > −5 and x + 7 < 132 3 −4 8 −3 620. 3x + 2 ≤ 11 and 2x + 2 ≥ 0 21. 11x < 22 or 12x > 36 22. −5x < 25 and −2x ≥ −12−1 3 2 3 −5 6

574Chapter 8 Linear Equations and Inequalities23. 3x − 5 < 10 and 2x + 1 > −5 24. 5x + 8 < −7 or 3x − 8 > 10 25. 2x − 3 < 8 and 3x + 1 > −10−3 5 −3 6 11−311226. 11x − 8 > 3 or 12x + 7 < −5 27. 2x − 1 < 3 and 3x − 2 > 1 28. 3x + 9 < 7 or 2x − 7 > 11−1 1 1 2 2− 9329. −1 ≤ x − 5 ≤ 2 30. 0 ≤ x + 2 ≤ 3 31. −4 ≤ 2x ≤ 64 7 −2 1 −2 332. −5 < 5x < 10 33. −3 < 2x + 1 < 5 34. −7 ≤ 2x − 3 ≤ 7−1 2 −2 2 −2 535. 0 ≤ 3x + 2 ≤ 7 36. 2 ≤ 5x − 3 ≤ 12 37. −7 < 2x + 3 < 112−3531 3 −5 438. −5 < 6x − 2 < 8 39. −1 ≤ 4x + 5 ≤ 9 40. −8 ≤ 7x − 1 ≤ 131−2533− 1−1 22For each graph below, write an inequality whose solution is the graph.41.42.-2 3 -2 3−2 < x < 3x ≤ −2 or x ≥ 343.44.-2 3 -2 3−2 ≤ x ≤ 3x < −2 or x > 3

8.9 Problem Set575Applying the Concepts45. Organic Groceries The map shows a range of how manyorganic stores are found in each state. If Oregon has24 organic grocery stores, write an inequality thatdescribes the difference, d, between the number ofstores in Oregon and the number of stores found inNevada.4 ≤ d ≤ 1346. Organic Groceries The map shows a range of howmany organic stores are found in each state. Write aninequality that describes the difference, d, in organicstores between California and the state of Washington.41 ≤ d ≤ 59Organic Stores in the U.S.Source: organic.com 20080-1011-2021-3031-4041-5051-6061-7071-8081-90Number of organicgrocery storesper stateTriangle Inequality The triangle inequality states that the sum of any two sides of a triangle must be greater than the third side.48. The following triangle ABC has sides of length x, 3x, and47. The following triangle RST has sides of length x, 2x,_​ 103 ​< x < 10 4 < x < 8and 10 as shown.16 as shown.TC2xxx16R 10 SA 3xBa. Find the three inequalities, which must be truea. Find the three inequalities, which must be true basedbased on the sides of the triangle.on the sides of the triangle.2x + x > 10; x + 10 > 2x; 2x + 10 > x3x + x > 16; 3x + 16 > x; x + 16 > 3xb. Write a compound inequality based on your resultsb. Write a compound inequality based on your resultsabove.above.49. Engine Temperature The engine in a car gives off a lot ofheat due to the combustion in the cylinders. The waterused to cool the engine keeps the temperature withinthe range 50 ≤ F ≤ 266 where F is in degrees Fahrenheit.Graph this inequality on the number line.50. Engine Temperature To find the engine temperature rangefrom Problem 47 in degrees Celsius, we use thefact that F = ​_9 ​C + 32 to rewrite the inequality as550 ≤ ​_9 ​C + 32 ≤ 2665Solve this inequality and graph the solution set.10 ≤ C ≤ 13010 13050 26651. Number Problem The difference of twice a number and3 is between 5 and 7. Find the number.4 < x < 552. Number Problem The sum of twice a number and 5 isbetween 7 and 13. Find the number.1 < x < 453. Perimeter The length of a rectangle is 4 inches longer 54. Perimeter The length of a rectangle is 6 feet longer thanthan the width. If the perimeter is between 20 inches the width. If the perimeter is between 24 feet and 36 feet,and 30 inches, find all possible values for the width. find all possible values for the width.The width is between 3 inches and ​_ 112 ​ inches. The width is between 3 feet and 6 feet.

576Chapter 8 Linear Equations and InequalitiesMaintaining Your SkillsThe problems that follow review some of the more important skills you have learned in previous sections and chapters.You can consider the time you spend working these problems as time spent studying for exams.Answer the following percent problems.55. What number is 25% of 32?856. What number is 15% of 75?11.2557. What number is 20% of 120?2458. What number is 125% of 300?37559. What percent of 36 is 9?25%60. What percent of 16 is 9?56.25%61. What percent of 50 is 5?10%62. What percent of 140 is 35?25%63. 16 is 20% of what number?8064. 6 is 3% of what number?20065. 8 is 2% of what number?40066. 70 is 175% of what number?40Simplify each expression.67. −​| −5 |​ −5 68. −​ _​ 2 33 ​ ​​−​ 8 _27 ​69. −3 − 4(−2) 5 70. 2 4 + 3 3 ÷ 9 − 4 2 371. 5 ​| 3 − 8 |​−6 ​| 2 − 5 |​ 7 72. 7 − 3(2 − 6) 195 + 3(7 − 2)73. 5 − 2[−3(5 − 7) − 8] 9 74. ​__2(−3) − 4 ​ −275. Find the difference of −3 and −9. 6 76. If you add −4 to the product of −3 and 5, what numberresults? −1977. Apply the distributive property to ​ 1 _2 ​(4x − 6). 2x − 3 78. Use the associative property to simplify −6 ​ ​ 1 _3 ​x ​. −2xFor the set {−3, −​ 4 _5 ​, 0, ​5 _8 ​, 2, ​√— 5 ​}, which numbers are79. Integers −3, 0, 2 80. Rational numbers−3, −​ 4 _5 ​, 0, ​ 5 _8 ​, 2

Chapter 8 SummarySimilar Terms [8.1]A term is a number or a number and one or more variables multiplied together.Similar terms are terms with the same variable part.EXAMPLEs1. The terms 2x, 5x, and −7x areall similar because their variableparts are the same.Simplifying Expressions [8.1]In this chapter we simplified expressions that contained variables by using thedistributive property to combine similar terms.2. Simplify 3x + 4x.3x + 4x = (3 + 4)x= 7xSolution Set [8.2]The solution set for an equation (or inequality) is all the numbers that, when usedin place of the variable, make the equation (or inequality) a true statement.3. The solution set for the equationx + 2 = 5 is {3} becausewhen x is 3 the equation is3 + 2 = 5, or 5 = 5.Equivalent Equations [8.2]Two equations are called equivalent if they have the same solution set.4. The equation a − 4 = 3 anda − 2 = 5 are equivalentbecause both have solution set{7}.Addition Property of Equality [8.2]When the same quantity is added to both sides of an equation, the solution set forthe equation is unchanged. Adding the same amount to both sides of an equationproduces an equivalent equation.5. Solve x − 5 = 12.x − 5 + 5 = 12 + 5x + 0 = 17x = 17Multiplication Property of Equality [8.3]If both sides of an equation are multiplied by the same nonzero number, the solutionset is unchanged. Multiplying both sides of an equation by a nonzero quantityproduces an equivalent equation.6. Solve 3x = 18.​ 1 _3 ​(3x) = ​1 _3 ​(18)x = 6Strategy for Solving Linear Equations in One Variable [8.4]Step 1a: Use the distributive property to separate terms, if necessary.1b: If fractions are present, consider multiplying both sides by the LCD toeliminate the fractions. If decimals are present, consider multiplyingboth sides by a power of 10 to clear the equation of decimals.1c: Combine similar terms on each side of the equation.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 side. A variableterm is a term that contains the variable (for example, 5x). A constantterm is a term that does not contain the variable (the number 3, forexample.)Chapter 8 Summary7. Solve 2(x + 3) = 10.2x + 6 = 102x + 6 + (−6) = 10 + (−6)2x = 4​ 1 _2 ​(2x) = ​1 _2 ​(4)x = 2577

578Chapter 8 Linear Equations and InequalitiesStep 3: Use the multiplication property of equality to get x (that is, 1x) by itselfon one side of the equation.Step 4: Check your solution in the original equation to be sure that you havenot made a mistake in the solution process.Formulas [8.5]8. Solving P = 2l + 2w for l, we haveP − 2w = 2l​_P − 2w ​= l2A formula is an equation with more than one variable. To solve a formula for oneof its variables, we use the addition and multiplication properties of equality tomove everything except the variable in question to one side of the equal sign sothe variable in question is alone on the other side.Blueprint for Problem Solving [8.6, 8.7]Step 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 cases thiswill amount to letting x = the item that is asked for in the problem.)Then translate the other information in the problem to expressionsinvolving 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 original wordsin the problem.Addition Property for Inequalities [8.8]9. Solve x + 5 < 7.x + 5 + (−5) < 7 + (−5)x < 2Adding the same quantity to both sides of an inequality produces an equivalentinequality, one with the same solution set.Multiplication Property for Inequalities [8.8]10. Solve −3a ≤ 18.−​ 1 _3 ​(−3a) ≥ −​ 1 _3 ​(18)a ≥ −611. Solve 3(x − 4) ≥ −2.3x − 12 ≥ −23x − 12 + 12 ≥ −2 + 123x ≥ 10​ 1 _3 ​(3x) ≥ ​1 _3 ​(10)x ≥ ​ 10 _3 ​0 103Multiplying both sides of an inequality by a positive number never changes thesolution set. If both sides are multiplied by a negative number, the sense of theinequality must be reversed to produce an equivalent inequality.Strategy for Solving Linear Inequalities in One Variable [8.8]Step 1a: Use the distributive property to separate terms, if necessary.1b: If fractions are present, consider multiplying both sides by the LCD toeliminate the fractions. If decimals are present, consider multiplyingboth sides by a power of 10 to clear the inequality of decimals.1c: Combine similar terms on each side of the inequality.Step 2: Use the addition property for inequalities to get all variable terms onone side of the inequality and all constant terms on the other side.

Chapter 8Summary579Step 3: Use the multiplication property for inequalities to get x by itself on oneside of the inequality.Step 4: Graph the solution set.Compound Inequalities [8.9]Two inequalities connected by the word and or or form a compound inequality. Ifthe connecting word is or, we graph all points that are on either graph. If the connectingword is and, we graph only those points that are common to both graphs.The inequality −2 ≤ x ≤ 3 is equivalent to the compound inequality −2 ≤ x andx ≤ 3.Common Mistakes12. x < −3 or x > 1−2 ≤ x ≤ 3−3 0 1−2 0 31. Trying to subtract away coefficients (the number in front of variables)when solving equations. For example:4x = 124x − 4 = 12 − 4x = 8 9 MistakeIt is not incorrect to add (−4) to both sides; it’s just that 4x − 4 is notequal to x. Both sides should be multiplied by ​_1 ​to solve for x.42. Forgetting to reverse the direction of the inequality symbol whenmultiplying both sides of an inequality by a negative number. For instance:−3x < 12_−​ 1 3 ​(−3x) < −​1 _​(12) 9 Mistake3x < −4_It is not incorrect to multiply both sides by −​ 1 ​. But if we do, we must3also reverse the sense of the inequality.

Chapter 8 ReviewSimplify each expression as much as possible. [8.1]1. 5x − 8x −3x 2. 6x − 3 − 8x −2x − 3 3. −a + 2 + 5a − 9 4a − 74. 5(2a − 1) − 4(3a − 2) −2a + 3 5. 6 − 2(3y + 1) − 4 −6y 6. 4 − 2(3x − 1) − 5 −6x + 1Find the value of each expression when x is 3. [8.1]7. 7x − 2 19 8. −4x − 5 + 2x −11 9. −x − 2x − 3x −18Find the value of each expression when x is −2. [8.1]10. 5x − 3 −13 11. −3x + 2 8 12. 7 − x − 3 6Solve each equation. [8.2, 8.3]13. x + 2 = −6 −8 14. x − ​_1 2 ​= ​4 _7 ​ ​15 _​ 15. 10 − 3y + 4y = 12 21416. −3 − 4 = −y − 2 + 2y −5 17. 2x = −10 −5 18. 3x = 0 019. ​_x 3 ​= 4 12 20. −​x _​= 2 −8 21. 3a − 2 = 5a −1422. ​_710 ​a = ​1 _5 ​a + ​1 _​ 1 23. 3x + 2 = 5x − 8 5 24. 6x − 3 = x + 7 2225. 0.7x − 0.1 = 0.5x − 0.1 0 26. 0.2x − 0.3 = 0.8x − 0.3 0Solve each equation. Be sure to simplify each side first. [8.4]27. 2(x − 5) = 10 10 28. 12 = 2(5x − 4) 2 29. ​ 1 _2 ​(3t − 2) + ​1 _2 ​= ​5 _2 ​ 230. ​_3 5 ​(5x − 10) = ​2 _3 ​(9x + 3) −​8 _​ 31. 2(3x + 7) = 4(5x − 1) + 18 0 32. 7 − 3(y + 4) = 10 −53Use the formula 4x − 5y = 20 to find y if [8.5]33. x is 5 0 34. x is 0 −4 35. x is −5 −8 36. x is 10 4Solve each of the following formulas for the indicated variable. [8.5]37. 2x − 5y = 10 for y 38. 5x − 2y = 10 for y 39. V = πr 2 h for hy = ​_2 5 2 y = ​_ 5 2 5 h = ​_ Vπr 2​40. P = 2l + 2w for ww = ​_ P − 2l ​241. What number is 86% of 240? [8.5] 206.4 42. What percent of 2,000 is 180? [8.5] 9%Chapter 8Review581

582Chapter 8 Linear Equations and InequalitiesSolve each of the following word problems. In each case, be sure to show the equation that describes the situation.[8.6, 8.7]43. Number Problem The sum of twice a number and 6 is28. Find the number. 1144. Geometry The length of a rectangle is 5 times as long asthe width. If the perimeter is 60 meters, find the lengthand the width. 5 meters; 25 meters45. Investing A man invests a certain amount of money inan account that pays 9% annual interest. He invests$300 more than that in an account that pays 10%annual interest. If his total interest after a year is $125,how much does he have invested in each account?$500 at 9%; $800 at 10%46. Coin Problem A collection of 15 coins is worth $1.00. Ifthe coins are dimes and nickels, how many of each coinare there? 10 nickels; 5 dimesSolve each inequality. [8.8]_47. −2x < 4 x > −2 48. −5x > −10 x < 2 49. −​ a 2 ​≤ −3 a ≥ 6 50. −​a _​> 5 a < −153Solve each inequality, and graph the solution. [8.8]51. −4x + 5 > 37 52. 2x + 10 < 5x − 11 53. 2(3t + 1) + 6 ≥ 5(2t + 4)Graph the solution to each of the following compound inequalities. [8.9]54. x < −2 or x > 5 55. −5x ≥ 25 or 2x − 3 ≥ 9 56. −1 < 2x + 1 < 9

Chapter 8 Cumulative ReviewSimplify.1. 7 − 9 − 12 −14 2. −11 + 17 + (−13) −73. 8 − 4 ⋅ 5 −12 4. 30 ÷ 3 ⋅ 10 1005. 6 + 3(6 + 2) 30 6. −6(5 − 11) − 4(13 − 6) 87. ​ _−​ 2 33 ​ ​​ −​ 8_ 27 ​ 8. ​ _​ 1 22 ​(−8) ​​ 16_9. ​ −30120 ​ −​1_ 4​ 10. ​234_312 ​ ​3_ 4 ​_11. −​ 3 _​÷ ​154 16 ​ −​4_ 5 ​ 12. ​2 _5 ​⋅ ​4 _7 ​ ​8_ 35 ​_13. ​ −4(−6) ​ −​ 8_ __ + 3(6)−93​ 14. ​2(−9) ​ 0−7 − 815. ​ 5 _9 ​+ ​1 _3 ​ ​8_ 9 ​ 16. ​4 _21 ​− ​9 _35 ​ −​1_ 15 ​_17. ​ 1 ​(10x) 2x 18. −8(9x) −72x519. ​_1 ​(8x − 4) 2x − 1 20. 3(2x − 1) + 5(x + 2)411x + 7Solve each equation.21. 7x = 6x + 4 4 22. x + 12 = −4 −16_23. −​ 3 5 ​x = 30 −50 24. ​x _5 ​= −​ _ 310 ​ −​3_ 2 ​25. 3x − 4 = 11 5 26. 5x − 7 = x − 1 ​ 3_ 2 ​27. 4(3x − 8) + 5(2x + 7) = 25 128. 15 − 3(2t + 4) = 1 ​ 1_ 3 ​29. 3(2a − 7) − 4(a − 3) = 15 12_30. ​ 1 3 ​(x − 6) = ​1 _​(x + 8) 48431. Solve P = a + b + c for c. c = P − a − b32. Solve 3x + 4y = 12 for y. y = −​ 3_ 4 ​x + 3Solve each inequality, and graph the solution.33. 3(x − 4) ≤ 6x ≤ 635. x − 1 < 1 or 2x + 1 > 7x < 2 or x > 3Translate into symbols.34. −5x + 9 < −6x > 336. −2 < x + 1 < 5−3 < x < 4For the set {−5, −3, −1.7, 2.3, ​ _ 127 ​, π} list all the39. Integers−5, −340. Rational numbers−5, −3, −1.7, 2.3, ​ 12 _ 7 ​41. Evaluate x 2 − 8x − 9 when x = −2. 1142. Evaluate a 2 − 2ab + b 2 when a = 3, b = −2. 2543. What is 30% of 50? 1544. What percent of 36 is 27? 75%45. Number Problem The sum of a number and 9 is 23.Find the number. 1446. Number Problem Twice a number increased by 7 is 31.Find the number. 1247. Geometry A right triangle has one 42° angle. Find theother two angles. 48°, 90°48. Geometry Two angles are complementary. If one is25°, find the other one. 65°49. Hourly Pay Carol tutors in the math lab. She gets paid$8 per hour for the first 15 hours and $10 per hour foreach hour after that. She made $150 one week. Howmany hours did she work? 18 hours50. Cost of a Letter The cost of mailing a letter was 32¢ forthe first ounce and 29¢ for each additional ounce. If thecost of mailing a letter was 90¢, how many ounces didthe letter weigh? 3 ozThe illustration shows the annual proceeds (in millions ofdollars) from sales of the top five magazines. Use the illustrationto answer the following questions.Top Five Magazine Sales1TV Guide2$548.9M3People$494.6MTime The Weekly Magazine4$313.8M5Sports Illustrated$306.9MReader’s Digest$288.4MSource: Magazine Publishers of America 200451. Find the total annual proceeds from sales of TV Guideand Reader’s Digest. $837.3 million52. How much more are the annual proceeds from sales ofPeople than from sales of Time? $180.8 million37. The difference of x and 5 is 12. x − 5 = 1238. The sum of x and 7 is 4. x + 7 = 4Chapter 8Cumulative Review583

Chapter 8 TestSimplify each of the following expressions. [8.1]1. 3x + 2 − 7x + 3−4x + 53. 7 − 3(y + 5) − 4−3y − 122. 4a − 5 − a + 13a − 44. 8(2x + 1) − 5(x − 4)11x + 285. Find the value of 2x − 3 − 7x when x = −5. [8.1] 226. Find the value of x 2 + 2xy + y 2 when x = 2 andy = 3. [8.1] 257. Fill in the tables below to find the sequences formedby substituting the first four counting numbers into theexpressions (n + 1) 2 and n 2 + 1. [8.1]a.b.n (n + 1) 2n n 2 + 11 42 93 164 251 22 53 104 17Solve the following equations. [8.2, 8.3, 8.4]8. 2x − 5 = 7 6 9. 2y + 4 = 5y ​ 4_ 3 ​10. ​ 1 _2 ​x − ​1 _10 ​= ​1 _5 ​x + ​1 _2 ​ 2 11. ​2 _5 ​(5x − 10) = −5 −​1_ 2 ​12. −5(2x + 1) − 6 = 19 −313. 0.04x + 0.06(100 − x) = 4.6 7014. 2(t − 4) + 3(t + 5) = 2t − 2 −315. 2x − 4(5x + 1) = 3x + 17 −116. What number is 15% of 38? [8.5] 5.717. 240 is 12% of what number? [8.5] 2,00018. If 2x − 3y = 12, find x when y = −2. [8.5] 319. If 2x − 3y = 12, find y when x = −6. [8.5] −820. Solve 2x + 5y = 20 for y. [8.5] y = −​ 2_ 5 ​x + 421. Solve h = x + vt + 16t 2 for v. [8.5] ​__v = h − x − 16t 2​tSolve each word problem. [8.6, 8.7]22. Age Problem Dave is twice as old as Rick. Ten yearsago the sum of their ages was 40. How old are theynow? Rick is 20, Dave is 4023. Geometry A rectangle is twice as long as it is wide. Theperimeter is 60 inches. What are the length and width?l = 20 in., w = 10 in.24. Coin Problem A man has a collection of dimes andquarters with a total value of $3.50. If he has 7 moredimes than quarters, how many of each coin does hehave? 8 quarters, 15 dimes25. Investing A woman has money in two accounts. Oneaccount pays 7% annual interest, whereas the otherpays 9% annual interest. If she has $600 more investedat 9% than she does at 7% and her total interest fora year is $182, how much does she have in eachaccount? $800 at 7% and $1,400 at 9%Solve each inequality, and graph the solution. [8.8]26. 2x + 3 < 5 x < 1 27. −5a > 20 a < −428. 0.4 − 0.2x ≥ 1 x ≤ −3 29. 4 − 5(m + 1) ≤ 9 m ≥ −2Solve each inequality, and graph the solution. [8.9]30. 3 − 4x ≥ −5 or 2x ≥ 10x ≤ 2 or x ≥ 531. −7 < 2x − 1 < 9−3 < x < 5Write an inequality whose solution is the given graph. [8.8,8.9]32.33.34.–4 0–3 1–4 1x > −4−3 ≤ x ≤ 1x < −4 or x > 1Use the illustration below to find the percent of the hoursin one week (168 hours) that a person from the givencountry spends watching television. Round your answersto the nearest tenth of a percent. [8.5]Watching the Tubeweekly hours spent watching TVUnited States 19Turkey 20.2Egypt 20.9Thailand 22.4Philippines 21Indonesia 19.7Source: NOP World35. United States 11.3% 36. Thailand 13.3%584Chapter 8 Linear Equations and Inequalities

Chapter 8 ProjectsLinear Equations and Inequalitiesgroup PROJECTTables and GraphsNumber of PeopleTime NeededEquipmentBackground2-35–10 minutesPencil and graph paperBuilding tables is a method of visualizing information.We can build a table from a situation(as below) or from an equation. In this project,we will first build a table and then write anequation from the information in the table.ProcedureA parking meter, which accepts only dimes and quarters, is emptied at the end of each day. Theamount of money in the meter at the end of one particular day is $3.15.1. Complete the following table so that all possible combinations of dimes and quarters, alongwith the total number of coins, is shown. Remember, although the number of coins will vary,the value of the dimes and quarters must total $3.15.number ofNumber of Dimes Quarters Total Coins Value29 1 30 $3.1524 $3.15$3.15$3.15$3.15$3.152. From the information in the table, answer the following questions.a. What is the maximum possible number of coins taken from the meter?b. What is the minimum possible number of coins taken from the meter?c. When is the number of dimes equal to the number of quarters?3. Let x = the number of dimes and y = the number of quarters. Write an equation in two variablessuch that the value of the dimes added to the value of the quarters is $3.15.Chapter 8Projects585

RESEARCH PROJECTStand and DeliverThe 1988 film Stand and Deliver starring EdwardJames Olmos and Lou Diamond Phillips is basedon a true story. Olmos, in his portrayal of highschoolmath teacher Jaime Escalante, earnedan academy award nomination for best actor.Watch the movie Stand and Deliver. After brieflydescribing the movie, explain how Escalante’sstudents became successful in math. Make a listof specific things you observe that the studentshad to do to become successful. Indicate whichitems on this list you think will also help youbecome successful.The Kobal Collection586Chapter 8 Linear Equations and Inequalities