Complete each statement with. 6.–3 – (–3)(–4) 12 9.–1 – 2 6 – Graph the numbers on the same number line – ALGEBRA 1 LESSON – 4545 – (For help, go to Lesson 1-3.) Inequalities and Their Graphs 3-1

ALGEBRA 1 LESSON 3-1 Solutions 1.–5. 6.–3 > – < (–3)(–4) = 12 9.–1 – 2 6 – 9 –3 = – – 4545 – 10. –0.75 > – < Inequalities and Their Graphs 3-1

Is each number a solution of x 5? ALGEBRA 1 LESSON 3-1 > Yes, 5 5 is true. > No, –2 5 is not true. > Yes, 10 5 is true. > a. –2b. 10 c Inequalities and Their Graphs 3-1

ALGEBRA 1 LESSON 3-1 Is each number a solution of 3 + 2x < 8? a. –2b x < 8 –2 is a solution (–2) < 8Substitute for x. 3 – 4 < 8Simplify. –1 < 8 Compare x < 8 3 is not a solution (3) < 8 Substitute for x < 8 Simplify. 9 < 8Compare. Inequalities and Their Graphs 3-1

a. Graph d < 3. ALGEBRA 1 LESSON 3-1 b. Graph –3 ≥ g. The solutions of d < 3 are all the points to the left of 3. The solutions of –3 g are –3 and all the points to the left of –3. > Inequalities and Their Graphs 3-1

Write an inequality for each graph. ALGEBRA 1 LESSON 3-1 x < 2Numbers less than 2 are graphed. x > –2Numbers greater than –2 are graphed. x –3Numbers less than or equal to –3 are graphed. < Inequalities and Their Graphs 3-1 > x Numbers greater than or equal to are graphed

ALGEBRA 1 LESSON 3-1 Define a variable and write an inequality for each situation. a. A speed that violates the law when the speed limit is 55 miles per hour. b. A job that pays at least $500 a month. Let v = an illegal speed. The speed limit is 55, so v > 55. Let p = pay per month. The job pays $500 or more, so p 500. > Inequalities and Their Graphs 3-1

ALGEBRA 1 LESSON 3-1 Inequalities and Their Graphs pages 136–139 Exercises 1.yes 2.no 3.yes 4.yes 5.no 6.yes 7.no 8.yes 9.a. no b. no c. yes 10.a. yes b. no c. yes 11.a. no b. no c. no 12. a. yes b. no c. no 13.a. no b. yes c. no 14.a. no b. no c. yes 15.C 16.B 17.D 18.A –37. Choice of variable may vary. 27.x > –3 28.x 7 29.x 1 30.x < –6 31.x x < – > – < – > –

ALGEBRA 1 LESSON 3-1 Inequalities and Their Graphs 33.Let s = number of students. s Let a = age. a Let w = number of watts. w Let s = number of students. s Let a = number of appearances. a > n is less than b is greater than is greater than or equal to x, or x is less than or equal to z is greater than or equal to – q is less than 4, or 4 is greater than q. 43.–1 is greater than or equal to m, or m is less than or equal to – is greater than or equal to w, or w is less than or equal to g minus 2 is less than a is less than or equal to plus r is greater than – is less than or equal to h, or h is greater than or equal to is greater than k, or k is less than Use an open dot for. Use a closed dot for or. 51.Answers may vary. Sample: For x = –1, 3(–1) + 1 = –3 + 1 = –2. –2 > 0. / 3-1 < – > – < – > – > – < –

ALGEBRA 1 LESSON 3-1 Inequalities and Their Graphs 52.Put an open dot at 3 and color the rest of the number line. 53.Answers may vary. Sample: Every class has at least 18 students. 54.x > 2 55.b –5 56.r 0 57.a < “At least” is translated as. “At most” is translated as. 65.x Option A < Option B 67.“4 greater than x” means x + 4; “4 > x” means 4 is greater than x. 68.C; the inequality is true for x = 3 but not true for x = 5, so C is correct. 69.Answers may vary. Sample: a = –1, b = > – < – > – < – > –

70.a is negative, and a and b are opposites. 71.5a + 4s C 75.H 76.B 77.H 78.B 79.[2] Since 6 is half, m > 6. [1] incorrect inequality OR no diagram ALGEBRA 1 LESSON 3-1 Inequalities and Their Graphs 80.y = – x y = 5x y = x – 3 83.y = – x – I = 88. = or = – w 89.b = P – a – c 90.Associative Property of Multiplication 91.Commutative Property of Multiplication 92.Commutative Property of Addition VRVR P – 2w 2 P2P2 3-1 > –

ALGEBRA 1 LESSON Is each number a solution of x > –1? a. 3b. –5 2.Is 2 a solution of 3x – 4 < 2? 3.Graph x > 4. 4.Write an inequality for the graph. 5.Graph each inequality. a. t is at most 2.b. w is at least 1. yesno p –2 < Inequalities and Their Graphs 3-1

ALGEBRA 1 LESSON 3-2 (For help, go to Lessons 1-4, 1-5, and 2-1.) Complete each statement with. 1.–3 + 4 – –3 – – – Solve each equation. 4.x – 4 = 55.n – 3 = –5 6.t + 4 = –57.k + = Solving Inequalities Using Addition and Subtraction 3-2

ALGEBRA 1 LESSON –3 + 4 – > –1 2. –3 – 6 4 – 6 –9 < –2 3. – – –1.4 > – x – 4 = 5 x = 9 5.n – 3 = –5 n = –2 6.t + 4 = –5 t = –9 Solutions Solving Inequalities Using Addition and Subtraction k + = k = – = – =

ALGEBRA 1 LESSON 3-2 Solve p – 4 < 1. Graph the solutions. p – < 1 + 4Add 4 to each side. p < 5Simplify. Solving Inequalities Using Addition and Subtraction 3-2

ALGEBRA 1 LESSON 3-2 Solve 8 d – 2. Graph and check your solution. > d – 2 + 2Add 2 to each side. > 10 d, or d 10Simplify. >< Check: 8 = d – 2Check the computation – 2Substitute 10 for d. 8 = 8 8 d – 2Check the direction of the inequality. 8 9 – 2Substitute 9 for d. 8 7 > > > Solving Inequalities Using Addition and Subtraction 3-2

ALGEBRA 1 LESSON 3-2 Solve c + 4 > 7. Graph the solutions. c + 4 – 4 > 7 – 4Subtract 4 from each side. c > 3Simplify. Solving Inequalities Using Addition and Subtraction 3-2

ALGEBRA 1 LESSON 3-2 In order to receive a B in your literature class, you must earn more than 350 points of reading credits. Last week you earned 120 points. This week you earned 90 points. How many more points must you earn to receive a B? Solving Inequalities Using Addition and Subtraction 3-2 points earnedrequired points needed Relate:plus is more than Define:Let = the number of points needed.p Write: p >

ALGEBRA 1 LESSON 3-2 (continued) You must earn 141 more points p > 350Combine like terms p > p – 210 > 350 – 210Subtract 210 from each side. p > 140Simplify. Solving Inequalities Using Addition and Subtraction 3-2

d > 3 13.y –4 14.q < 3 15.x r < m < – b < 19.n > w 5; 1212 ALGEBRA 1 LESSON 3-2 Solving Inequalities Using Addition and Subtraction pages 142–145 Exercises x > 11 5.t < 1 6.b < –4 7.d 10 8.s –4 9.r 9 10.n > w – > – < – < – 5353 > – < – > – < –

ALGEBRA 1 LESSON 3-2 Solving Inequalities Using Addition and Subtraction 36.h – ; 37.d > –1.5; 38.t < – ; 39.s , $ s , $ r , 21 reflectors 42.Add 4 to each side. 43.Subtract 9 from each side. 44.Add to each side. 45.w c 3 47.y < m > –8; 25.b > –7; 26.a –6; 27.r –6; 28.k 1; 29.x < –1; 30.p > –6; 31.z –1; 32.y < 5.5; 33.m > –1 ; 34.a –0.3; 35.p > –7; n < –5 49.z < – y < – t > 52.v k – b < 6 55.m – k 4 57.h – 58.x – w < – > – < – < – < – > – < – > – < – > – < – < – > – > – < – < – > – < –

ALGEBRA 1 LESSON 3-2 Solving Inequalities Using Addition and Subtraction 60.w –1 61.x < m > t k < 2 65.n < 6 66.m 5 67.s > a.yes b.no c.Answers may vary. Sample: The = sign indicates that each side is equal, so the two sides may be interchanged The < sign does not indicate equality. One side cannot be both greater than and less than the other side. 69.a.Let n = points scored on floor exercises n 34.0; n 9.1 b.Your sister must score 9.1 points or more to qualify for regional gymnastics competition. c.Answers may vary. Sample: 9.1, 9.2, nearly 51.2 MB 71.at least $ or more points 73.a-b. Check students’ work. 3-2 > – < – < – > – > –

ALGEBRA 1 LESSON 3-2 Solving Inequalities Using Addition and Subtraction 74.a.No; the solution is z 13.8, so z 14 is not correct. b.Answers may vary. Sample: Substituting values does not work because there is always the possibility that the solution lies between values that make the inequality true and a value that does not. 75.x 1 76.n < 5 77.t –3 78.k > r < –2 80.r –13 81.a – m 83.d > –8 84.y < 6 85.a a < –8 87.a.a > c – b b.b > a – c c.c > b – a d.The length of the third side must be greater than the difference of the lengths of the other two sides. 88.not true; sample counter example: for a = 5 and b = –6, 5 –(–6) < 5 +(–6) 89.true 90.true 91.not true; sample counter example: for a = 0, b = 1, and c = –2, 0 < 1 and 0 < 1 + (–2) 92.Answers may vary. Sample: For x = 2, y = 1, z = 4, and w = 3, 2 > 1 and 4 > 3, but 2 – 4 > 1 – 3. / / / 3-2 > – < – > – > – > – > – < – < –

ALGEBRA 1 LESSON 3-2 Solving Inequalities Using Addition and Subtraction 93.B 94.G 95.A 96.I 97.D 98.[2] at least 33 points; 24(19.5) + x > 25(20) x > 500 x > 32 (OR equivalent explanation) [1] incorrect answer OR no work or explanation 99.Let c = length of octopus in feet. c Let h = distance in miles a hummingbird migrates. h > Let a = average. a Let p = number of pages to read. p –1 105.– – – > – < – > –

ALGEBRA 1 LESSON 3-2 Solve each inequality. Graph the solutions. 1.p – 7 –52.w – 3 < –9 3.x + 6 > h > > p 2 > w < –6 x > –2 < 4 h, or h 4 > Solving Inequalities Using Addition and Subtraction 3-2

(For help, go to Lessons 2-1 and 3-1.) ALGEBRA 1 LESSON 3-3 Solve each equation. 1.8 = t2.14 = –21x3. = –1 4.5d = 325. x = – n = 9 Write an inequality for each graph x6x Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-3 Solutions 1. 8 = t t = 8 t = 8(2) = = –21x –21x = 14 x = – = – = –1 x = –1(6) = –6 x6x6 4. 5d = 32 d = = x = –12 x = –12 x = – n = 9 = n = n x –1 8.x > 3 < Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-3 Solve > –2. Graph and check the solutions. z3z3 z > –6Simplify each side. 3 > 3(–2)Multiply each side by 3. Do not reverse the inequality symbol. z3z3 ( ) z3z3 > –2Check the direction of the inequality. Check: = –2 Check the computation. z3z3 – = –2Substitute –6 for z –2 = –2Simplify. – > –2Substitute –3 for z –1 > –2Simplify. Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-3 Solve 3 – x. Graph and check the solutions. < 3535 ( ) 5353 – 5353 – (3) > ( ) 3535 x Multiply each side by the reciprocal of –, which is –, and reverse the inequality symbol –5 x, or x –5Simplify. <> Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-3 (continued) Check: 3 = – xCheck the computation. 3 = – (–5) Substitute –5 for x. 3 = 3 3 – xCheck the direction of the inequality. 3 – (–10)Substitute –10 for x < < < Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-3 Solve –4c < 24. Graph the solutions. c > –6Simplify. Divide each side by –4. Reverse the inequality symbol. –4c –4 > 24 –4 Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-3 Your family budgets $160 to spend on fuel for a trip. How many times can they fill the car’s gas tank if it cost $25 each time? Your family can fill the car’s tank at most 6 times. cost per total fuel tankbudget Relate:times number of tanks is at most Define:Let = the number of tanks of gas. t Write:25160t < 25t 160 < Solving Inequalities Using Multiplication and Division 3-3 Divide each side by 25. < 25t t 6.4Simplify. <

ALGEBRA 1 LESSON 3-3 Solving Inequalities Using Multiplication and Division pages 149–151 Exercises 1.t –4; 2.s < 6; 3.w –2; 4.p < –8; 5.y > –4; 6.v –1.5; 7.x < 6; 8.k –2; 9.x < 0; 10.y 0; 11.x < 7; 12.d –4; 13.c > – ; 14.b –1 ; 15.u < – ; 16.n > ; 17.t < –3; 18.m ≥ 2; 19.w –5; 20.c > 4; 21.z –9; 22.b < –6; 23.d < – ; < – < – > – < – < – > – > – > – > –

ALGEBRA 1 LESSON 3-3 Solving Inequalities Using Multiplication and Division 34., 0, –1, –2 35.–6, –7, –8, –9 36.–2, –3, –4, –5 37.5, 4, 3, 2 38.–10.4, –11, –12, –13 39.Multiply each side by –4 and reverse the inequality symbol. 40.Multiply each side by Divide each side by Multiply each side by. 43.Divide each side by Multiply each side by –1 and reverse the inequality symbol. 45.–2 46.– – x and y are equal x –5 ; 25.q > –3 ; 26.h < 5; 27.d > 4; 28.m –4.5; c 300, 67 cars h 100, 17 hours 31–38. Answers may vary. Samples are given. 31.–2, –3, –4, –5 32.–12, –10, –8, 0 33.–3, –4, –5, – > – < – > – > –

59.d –7 60.u > s < – 62.k –30 63.y < 9 64.t – h –4 66.x > 2 67.x < –9 68.b < 0 69.p > –2 70.m –47 71.g < – 72.n 2 73.m < z –2 75.Yes; in each case, y is greater than a.She should have divided each side by –15. b.150 satisfies the original inequality –15q 135. ALGEBRA 1 LESSON 3-3 Solving Inequalities Using Multiplication and Division –55. Estimates may vary. 52.r > –2 53.j > –6 54.p s concrete blocks 57.For both the equation and the inequality, you multiply each side by –3. For the inequality, you must reverse the inequality symbol. 58.Answers may vary. Sample: 2x > 6, x > 4, –x < –3, – < – 4343 x5x > – < – < – > – < – < – < – > – > – < –

ALGEBRA 1 LESSON 3-3 Solving Inequalities Using Multiplication and Division 77. (continued) c.Answers may vary. Sample: –15 78.a > 0 79.a > 0 80.a 0 81.a < d > 15, 5-in. box or 6-in. box 83. x 18(15), 480 tiles , x –7 91.w < – 92.t d > 4 94.–3.9 m 95.y < > a 97.k 3 98.t = 2 99.n = –2 100.x = Prop. of Opposites 102.Prop. of Reciprocals 103.Ident. Prop. of Add. 104.Ident. Prop. of Mult. 105.Assoc. Prop. of Mult. 106.Comm. Prop. of Mult /= 9 16 > – > – < – > – < –

ALGEBRA 1 LESSON 3-3 Solve each inequality. Graph the solution. 1. –3 2.– < –1 3.6x < –12h y2y2 > p3p3 > y –6 > p > 3 x < 5 > < –4 h, or h –4 Solving Inequalities Using Multiplication and Division 3-3

ALGEBRA 1 LESSON 3-4 (For help, go to Lessons 2-2 and 2-3.) Solve each equation, if possible. If the equation is an identity or if it has no solution, write identity or no solution. 1.3(c + 4) = 62.3t + 6 = 3(t – 2) 3.5p + 9 = 2p – 14.7n + 4 – 5n = 2(n + 2) 5. k – + k = 6.2t – 32 = 5t + 1 Find the missing dimension of each rectangle Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON (c + 4) = 62.3t + 6 = 3(t – 2) c + 4 = 23t + 6 = 3t – 6 c = –2 6 = –6 no solution 3. 5p + 9 = 2p – 14.7n + 4 – 5n = 2(n + 2) 3p = –10 2n + 4 = 2n + 4 p = –3identity 1313 Solving Multi-Step Inequalities Solutions 3-4

ALGEBRA 1 LESSON k – + k = 6.2t – 32 = 5t + 1 k = –3t = 33 k = = 1 t = –11 7. P = 2( + w)8. P = 2( + w) 110 = 2( + 15)78 = 2(26 + w) 55 = = 26 + w 40 = 13 = w length = 40 cmwidth = 13 in Solutions Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-4 Solve 5 + 4b < b – 5 < 21 – 5Subtract 5 from each side. 4b < 16Simplify. b < 4Simplify. < Divide each side by 4. 4b44b Check: 5 + 4b = 21Check the computation b < 21Check the direction of the inequality (3) < 21Substitute 3 for b (4) 21Substitute 4 for b. 21 = < 21 Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-4 The band is making a rectangular banner that is 20 feet long with trim around the edges. What are the possible widths the banner can be if there is no more than 48 feet of trim? twice the the length lengthof trim Relate:plus twice the width can be no more than Write:2(20) + 2w 48 < Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-4 (continued) The banner’s width must be 4 feet or less. 2(20) + 2w 48 < w 48Simplify 2(20). < w – – 40Subtract 40 from each side. < 2w 8Simplify. < w 4Simplify. < Divide each side by 2. < 2w22w Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-4 Solve 3x + 4(6 – x) < 2. 3x + 24 – 4x < 2Use the Distributive Property. –x + 24 < 2Combine like terms. –x + 24 – 24 < 2 – 24Subtract 24 from each side. –x < –22Simplify. x > 22Simplify. > Divide each side by –1. Reverse the inequality symbol. –x –1 –22 –1 Solving Multi-Step Inequalities 3-4

Solve 8z – 6 < 3z ALGEBRA 1 LESSON 3-4 8z – 6 – 3z < 3z + 12 – 3zSubtract 3z from each side. 5z – 6 < 12Combine like terms. 5z – < Add 6 to each side. 5z < 18Simplify. < Divide each side by 5. 5z55z z < 3Simplify Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-4 Solve 5(–3 + d) 3(3d – 2). < –15 – 4d + 15 –6 + 15Add 15 to each side. < –4d 9Simplify. < –15 + 5d – 9d 9d – 6 – 9dSubtract 9d from each side. < –15 – 4d –6Combine like terms. < –15 + 5d 9d – 6Use the Distributive Property. < Divide each side by –4. Reverse the inequality symbol. –4d –4 9 –4 > d –2Simplify. > 1414 Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-4 Solving Multi-Step Inequalities 10.8t 420 and t 52.5, so the average rate of speed must be at least 52.5 mi/h s + 8 and s 9.5, so the two equal sides must be no longer than 9.5 cm. 12.–6x j 1 14.b < 15.h > 5 16.x 3 17.y > –8 18.w 4 19.c < –7 20.r n 9 22.w < 3 23.t –1 24.d 6 25.n 2 26.k –4 27.s < p > 3 29.x > 2 30.m > – 31.d < –1 32.y – pages 155–159 Exercises 1.d 4 2.m > –3 3.x > –2 4.n 3 5.q 4 6.h –2 7.a 3 8.b > 6 9.c < c1c 1212 > – < – > – < – < – > – > – > – < – > – > – < – < – < – < – < – > – > – < –

ALGEBRA 1 LESSON 3-4 Solving Multi-Step Inequalities k 34.v –4 35.q –2 36.r 1 37.x < 1 38.m > –8 39.v 2 40.Subtract 8 from each side. 41.Subtract 7 from each side. 42.Subtract y from each side and add 5 to each side Add 2 to each side, then multiply each side by –5, and reverse the inequality sign. 44.Subtract 3j from each side and subtract 5 from each side. 45.Answers may vary. Sample: Multiply q – 3 by 2, add 3q to each side, add 6 to each side, and divide each side by a.–3t –9, t 3 b.9 3t, t 3 c.The results are the same – (r + 3) – (t – 6) 4, t > – < – < – < – > – < – > – < – > – > – > –

ALGEBRA 1 LESSON 3-4 Solving Multi-Step Inequalities 49.3(z + 2) > 12, z > 2 50.Answers may vary. Sample: To solve 2.5(p – 4) > 3(p + 2), first use the Distributive Property to simplify each side. The result is 2.5p – 10 > 3p + 6. Then use the Subtraction Property of Inequality. Subtract 6 from each side and 2.5p from each side. The result is –16 > 0.5p. Then use the Division Property of Inequality. Divide each side by 0.5. The result is –32 > p. So the solution is –32 > p or p < – a.i. always true ii. always true iii. never true b.If the coefficients of the only variable on each side of an inequality are the same, then the inequality will either be always true or never true. 52.For x = the number of guests, (0.75) x 250; x 80; at least 80 guests must attend. 53.a.maximum b. no more than B 55.E 56.F 3-4 > – > –

ALGEBRA 1 LESSON 3-4 Solving Multi-Step Inequalities 57.A 58.D 59.C 60.Answers may vary. Samples: – x – 5 > 0, x < –15; – x – 5 10, x –45 61.r < 5 62.m –2 63.s n –2 65.s < 8 66.k For x = number of hours of work each month, 15x – ( ) 600, so to make a profit he must work at least 80 h per month. 78.For x = amount of sales, x 460, so to reach her goal, she must have at least $7000 in sales. 79.Add 2x to each side rather than subtract 2x, so x. 80.Distribute 4 to 2 as well as n, so n > –7. 81.a.x > b.x < n > –2 68.r 1 69.x > 1 70.a 71.m 1 72.k 2 73.n –23 74.x 0 75.a < –6 76.c c – b a c – b a > – < – > – < – < – < – < – > – > – < – > – < – > – > – > – < –

ALGEBRA 1 LESSON 3-4 Solving Multi-Step Inequalities 82.x is an integer between –3 and 7, inclusive in.  14.2 in. 84.– h 86.a.74 boxes b.5 trips 87.C 88.F 89.A 90.F 91.[2]Maxwell should continue ordering from the current supplier. Let x = number of bottles Maxwell orders per month. The cost from the current supplier is 3x The cost from the competitor is 4x + 5. The inequality 4x + 5 3x + 25 represents the number of bottles of shampoo for which the competitor will be less expensive. Since x = 20, the competitor will not be less expensive for orders of 30 bottles or more. (OR equivalent explanation) [1]incorrect answer OR insufficient explanation 92.m –4 93.y –8 94.x > –12 95.t –3 96.b < w > –98 98.p > 1 99.d h 101.– – > – < – > – < – > –

ALGEBRA 1 LESSON 3-4 Solve each inequality a 232.– p < p – 6 3.3(x – 4) > 4x (3c + 2) 2(3c – 2) > < a 3 > p > x < –19 < c – Solving Multi-Step Inequalities 3-4

ALGEBRA 1 LESSON 3-5 (For help, go to Lessons 1-1 and 3-1.) Graph each pair of inequalities on one number line. 1.c 12 Use the given value of the variable to evaluate each expression. 4.3n – 6; 45.7 – 2b; 5 6. ; 177. ; 9 > < > < y 3 2d – 3 5 Compound Inequalities 3-5

ALGEBRA 1 LESSON c < 8; c t –2; t –5 3. m 7; m > n – 6 for n = 4: 3(4) – 6 = 12 – 6 = – 2b for b = 5: 7 – 2(5) = 7 – 10 = –3 6. for y = 17: = = for d = 9: = = = 3 > >< < y 3 2d – (9) – – Compound Inequalities Solutions 3-5

ALGEBRA 1 LESSON 3-5 Write two compound inequalities that represent each situation. Graph the solutions. a. all real numbers that are at least –1 and at most 3. b –1 and b 3 –1 b 3 ><<< b. all real numbers that are at less than 31, but greater than > n and n > > n > 25 Compound Inequalities 3-5

Solve 5 > 5 – f > 2. Graph the solutions. ALGEBRA 1 LESSON – f – 5 > 2 – 5 –f > –3 –3 –1 –f –1 < 5 – 5 > 5 – f – 5 0 > –f 0 –1 –f –1 < 5 – f > 25 > 5 – f and f < 30 < f and 0 < f < 3 Write the compound inequality as two inequalities joined by and. Compound Inequalities 3-5

ALGEBRA 1 LESSON 3-5 Your test grades in science so far are 83 and 87. What possible grades g can you make on your next test to have an average between 85 and 90? Relate:the average which is less than or equal to is less than or equal to 8590 Write:8590 < g 3 g < Compound Inequalities 3-5

ALGEBRA 1 LESSON 3-5 (continued) 85 < g 3 < 90 Multiply by 3. 3(85) << 3(90) g 3 3 ( ) << g270Simplify. << 255 – g – – 170Subtract 170. << 85g100Simplify. The third test grade must be between 85 and 100, inclusive. Compound Inequalities 3-5

ALGEBRA 1 LESSON 3-5 Write an inequality that represents each situation. Graph the solutions. a. all real numbers that are less than 0 or greater than 3. n 3 b. Discounted tickets are available to children under 7 years old or to adults 65 and older. a 0. > Compound Inequalities 3-5

ALGEBRA 1 LESSON 3-5 Solve the compound inequality 3x + 2 < –7 or –4x + 5 < 1. Graph the solution. 3x + 2 – 2 < –7 – 2 3x < –9 < 3x33x3 –9 3 –4x + 5 – 5 < 1 – 5 –4x < –4 > –4x –4 3x + 2 < –7–4x + 5 < 1or x < –3x > 1 or Compound Inequalities 3-5

ALGEBRA 1 LESSON 3-5 Compound Inequalities pages 163–166 Exercises 1.–4 < x and x < 6 or –4 < x < 6; 2.2 n and n 9 or 2 n 9; 3.23 < c < 23.5; 4.40 w 74; 5.–5 < j < 5; 6.1 w 5; 7.2 < n 6; 8.–4 < p –2; 9.1 < x < 3; 10.–1.5 < w 3.5; 11.5 < x < 8; 12.–2 < m 1; 13.3 s < 4; 14.1 < t < 3 ; 15.1 r 2; 16.–2 < r 1.5; 17.5 a 17; 18.1 < x < 7; 19.–1 x 11; 20.n –3 or n 5, 21.x 7; 22.h 3; < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – < – > –

ALGEBRA 1 LESSON 3-5 Compound Inequalities 23.b 300; 24.b 2; 25.k –1; 26.c < 2 or c 3; 27.a 4 or a > 5; 28.c 2 or c 3; 29.y –2 or y 5; 30.d –3 or d 0; 31.z 2; 32.x < –3 or x 5; 33.n 3; 34.–2 < x < 3 35.x < –3 or x 2 36.x ≤ 0 or x > 2 37.–4 x 3 38.q ≤ –2 or q > 4 39.h ≤ t ≤14 41.r –6 ≤ t < 1 43.x ≤ d ≤ all real numbers except > – < – < – > – < – < – < – < – > – > – > – > –

ALGEBRA 1 LESSON 3-5 Compound Inequalities 46.The word “and” means both statements must be true. The word “or” means that at least one of the statements must be true < x < < x < < x < < x < C D Charlotte: 29 C 90 Detroit: 15 D Answers may vary. Sample: Elevation near a coastline varies between 2 m below and 8 m above sea level –2 < x < 0 or 0 < x < 3 57.n = 0 or n 3 58.a.130 R 157 b.20 years old 59.16, 18, , 12, B 62.G 3-5 > – < – < – < – < – < – < – < – < – < – < –

ALGEBRA 1 LESSON 3-5 Compound Inequalities 63.[2] g g < g < 6712 minimum consumption: 4169 gal maximum consumption: 6712 gal (OR equivalent explanation) [1] incorrect answer OR insufficient explanation 64.b > 65.n 3 66.r 7 67.x = –1 68.w = 2 69.identity < – < – < – < – < – < –

ALGEBRA 1 LESSON Write two compound inequalities that represent the given situation. Graph the solution. all real numbers that are at least 2 and at most 5 2.Write an inequality that represents the given situation. Graph the solution. all real numbers that are less than –3 or greater than –1 3.Solve –2 2x – 4 < 6. Graph the solution. 4.Solve 3x – 2 < –8 or –2x Graph the solution. < < b 2 and b 5, 2 b 5 ><<< n –1 1 x < 5 < > x < –2 or x 1 Compound Inequalities 3-5

11.|–4 | |–3 – 4 | 7 ALGEBRA 1 LESSON 3-6 (For help, go to Lessons 1-3 and 1-4.) Simplify. 1.|15|2.|–3|3.|18 – 12| 4.–|–7|5.|12 – (–12)|6.|–10 + 8| Complete each statement with. 7.|3 – 7| 48.|–5| |7| – |6 – 2 | Absolute Value Equations and Inequalities

ALGEBRA 1 LESSON |15| = 152.|–3| = 3 3. |18 – 12| = |6| = 64.–|–7| = –(7) = –7 5. |12 – (–12)| = | | = |24| = 24 6.|–10 + 8| = |–2| = 2 7.|3 – 7| 4 |–4| 4 4 = 4 8.|–5| > 6 Absolute Value Equations and Inequalities Solutions 3-6

ALGEBRA 1 LESSON 3-6 Solutions 10.|6 – 2 | 3 |3 | 3 3 > |7| – – < 8 11.|–4 | > |–3 – 4 | 7 |–7 | 7 7 = Absolute Value Equations and Inequalities 3-6

ALGEBRA 1 LESSON 3-6 Solve and check |a| – 3 = 5. |a| – = 5 + 3Add 3 to each side. |a| = 8Simplify. a = 8 or a = –8Definition of absolute value. Check: |a| – 3 = 5 |8| – 3 5Substitute 8 and –8 for a.|–8| – – 3 = 5 8 – 3 = 5 Absolute Value Equations and Inequalities 3-6

ALGEBRA 1 LESSON 3-6 Solve |3c – 6| = 9. The value of c is 5 or –1. Absolute Value Equations and Inequalities 3-6 3c – 6 = 9Write two equations. 3c – 6 = –9 3c – = 9 + 6Add 6 to each side.3c – = – c = 153c = –3 Divide each side by 3. 3c33c3 = c33c3 = –3 3 c = 5c = –1

ALGEBRA 1 LESSON 3-6 Solve |y – 5| 2. Graph the solutions. < 3 y 7 << Write a compound inequality.y – 5 –2 > y – 5 2 < and Add 5 to each side.y – –2 + 5 > y – < Simplify.y 3 > y 7 < and Absolute Value Equations and Inequalities 3-6

The ideal diameter of a piston for one type of car is mm. The actual diameter can vary from the ideal diameter by at most mm. Find the range of acceptable diameters for the piston. ALGEBRA 1 LESSON 3-6 greatest difference between actual and ideal Relate:0.007 mm is at most Define:Let d = actual diameter in millimeters of the piston. Write:| d – |0.007 < Absolute Value Equations and Inequalities 3-6

(continued) ALGEBRA 1 LESSON 3-6 The actual diameter must be between mm and mm, inclusive d Simplify. << Add – d – << |d – | < –0.007 d – Write a compound inequality. << Absolute Value Equations and Inequalities 3-6

ALGEBRA 1 LESSON 3-6 Absolute Value Equations and Inequalities pages 169–172 Exercises 1.–2, 2 2.–4, 4 3.–, 4.–6, 6 5.–3, 3 6.–7, 7 7.–5, 5 8.–2, no solution 11.–3, –4, , –8, 4 15.–3, 1 16.–5, 1 17.–5, 9 18.no solution 19.–7, 1 20.–1, 1 21.–0.6, a.less than b.greater than k 2.5; 24.–2 < w < 2; 25.–8 < x < 2; 26.n –11 or n –5; 27.1 y 3; 28.1 p 7; 29.–2 < c < 7; 30.y –2 or y 5; 31.t 2 ; 32.x 2.5; 1313 > – < – < – < – < – < – < – > –

ALGEBRA 1 LESSON 3-6 Absolute Value Equations and Inequalities 33.t –2.4 or t 4; 34.–2 < r < 8; 35.between mm and mm, inclusive 36.between 49 in. and 50 in., inclusive 37.–9, 9 38.–3, 3 39.–1, 1 40.–1.8, –5, , 8 43.d –2 or d 2 44.n c 2 46.–12.6, –6 < x < 6 48.–8, 8 49.–8, 8 50.–3, 3 51.–8 < m < 4 52.|n| < 3 53.|n| > |n – 6| > 2 55.|n + 1| 3 56.|w – 16| 0.05, w > – < – > – < – > – < – < – < –

ALGEBRA 1 LESSON 3-6 Absolute Value Equations and Inequalities 57.39%, 45% 58.a.|w – 454| 5 b.between 449 g and 459 g, inclusive 59.|g – 6.27| The absolute value of a number cannot be less than zero. 61.Sample: |5x – 12| = 3; 1, 3 62.|x – 4| = 2 63.|x – 2| = 4 64.|x – 3| = 6 65.|x – 12 | = |x – 3| = 4 67.|x – 5 | = 2 68.|x + 9| = 6 69.|x – 6| = 4 70.a.between oz and oz, inclusive b.No; the excess weight of some coins may be balanced by the lower weight of other coins. 71.a.11°F, 39°F b.|t – 25| = , = 78.= 79.|x + 1| > 3 80.|x – 2| 4 81.A 82.G 83.D 84.I 85.A 3-6 < – < – > – < – < –

86.[2]a. b.The overlap of the three graphs is from 20 to 25. [1]no graph OR insufficient explanation 87.–282 e 20, Let t = body temperature (°C), 36.0 t – –8 93. ALGEBRA 1 LESSON 3-6 Absolute Value Equations and Inequalities –2.5, –2, 0, 3, 96.–2, –1.5, –, 7, , 0.009, 0.01, –, –3, –2.5, 3, < – < – < – < –

Solve. 1.|a| + 6 = 92.|2x + 3| = 7 3.|p + 6| 14.3|x + 4| > 15 < a = 3 or a = –3x = 2 or x = –5 –7 p –5 << x > 1 or x < –9 ALGEBRA 1 LESSON 3-6 Absolute Value Equations and Inequalities 3-6

ALGEBRA 1 CHAPTER 3 Solving Inequalities 1.a. no b. yes c. yes 2.a. yes b. no c. no 3.c 7 4.r 5 5.g 40 6.h > 32 7.n < –7 8.n n –5 10.n > z 2; 12.–4 y; 13.x > –6; 14.u < 4; 15.t 2; 16.w 1; 17.m > –4; 18.y < 2.5; 19.x 2 or x 8; 20.–3 < h < 2; 21.–6 < b –3; 22.–2.5 < q < 3; 23.n < –5 or n –1; 24.k 4 and k < 7.5; 3-A > – < – > – < – < – < – > – < – > – < – > – < – > – > –

Solving Inequalities 25.d > – 26.b < 6 27.m 28.x – Answers may vary. Sample: x –6 or x > 2 30.Answers may vary. Sample: –3 < x < 2 31.–2.25 or –33 or –6 or or Answers may vary. Sample: Depending on the sign of a, the inequality symbol may need to be reversed. a = 2:2x – 1< 3 2x< 4 x< 2 a = –2:–2x – 1 < 3 –2x< 4 x> –2 36.Answers may vary. Sample: |x| < 6 37.Let c = number of cans c c boxes 39.Let = length w = width | – | |w – 7.625| w ALGEBRA 1 CHAPTER 3 3-A > – < – < – > – > – < – < – < – < – < – < –