17. 8x and 4x 2 > x 1 < 7 and 6x x or 2x x 7 < 3 and 8x x 9 9 and 5x > x + 3 < 3 or 8x 2

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1 Section 1.4 Compound Inequalities Exercises In Exercises 1-12, solve the inequality. Express your answer in both interval and set notations, and shade the solution on a number line. 1. 8x 16x x 6 > x +. 12x + 5 x x + 2x x 9 < x x 8 4x x 5 > 5x x + 4 > 6x x 1 > 7x x 2 > 4x x + < 11x 12. 6x + < 8x + 8 In Exercises 1-50, solve the compound inequality. Express your answer in both interval and set notations, and shade the solution on a number line. 1. 2x 1 < 4 or 7x x + 9 < and 7x + 1 > 15. 6x 4 < 4 and x x + 8 and x 6 > x and 4x 2 > x 1 < 7 and 6x x or 2x x 7 < and 8x 21. 9x 9 9 and 5x > x + < or 8x 2 2. x 5 < 4 and x + 9 > 24. 8x 6 < 5 or 4x x + 5 or 2x x + 6 < 4 or 7x 5 > x 2 2 or x x + 5 < 4 or 5x x + 1 < 6 and x + 9 > x + 2 < 5 or 6x x 7 < 2 and x 2. 4x + 1 < 0 or 8x + 6 > 9. 7x + 8 < and 8x x < 2 and 7x x and 6x x 1 8 or x 9 > x 5 1 and 4x + 7 > 7 8. x + 1 < 0 or 5x + 5 > 8 1 Copyrighted material. See:

2 64 Chapter 1 Preliminaries 9. 8x or 5x + 6 > x 6 5 and 6x 2 > 41. 4x 8 < 4 or 4x + 2 > 42. 9x 5 < 2 or 8x x 5 or x + 1 > 44. 5x 6 and 2x < x x < 4 x < 5 2x x < 2x x < x x < x x < 5x 5 < < 9x < 7x < 7x + 6 < < 2x In Exercises 51-62, solve the given inequality for x. Graph the solution set on a number line, then use interval and setbuilder notation to describe the solution set < x < < x < < 1 x 2 < Aeron has arranged for a demonstration of How to make a Comet by Professor O Commel. The wise professor has asked Aeron to make sure the auditorium stays between 15 and 20 degrees Celsius (C). Aeron knows the thermostat is in Fahrenheit (F) and he also knows that the conversion formula between the two temperature scales is C = (5/9)(F 2). a) Setting up the compound inequality for the requested temperature range in Celsius, we get 15 C 20. Using the conversion formula above, set up the corresponding compound inequality in Fahrenheit. b) Solve the compound inequality in part (a) for F. Write your answer in set notation. c) What are the possible temperatures (integers only) that Aeron can set the thermostat to in Fahrenheit? x < x x < x 2x < x < 2 < 4 x + 1 2

3 Section 1.4 Compound Inequalities 1.4 Solutions 1. 8x 16x 1 = 8x + 16x 1 + = 8x 2 = x 1 4 Thus, the solution interval is (, 1 4 ] = {x x 1 4 } x + 5 x 4 = 12x + x 4 5 = 9x 9 = x 1 Thus, the solution interval is [1, ) = {x x 1} x 9 < x + 1 = 11x + x < = 8x < 10 = x > 5 4 Thus, the solution interval is ( 5 4, ) = {x x > 5 4 }. 5 4

4 Chapter 1 Preliminaries 7. 4x 5 > 5x 7 = 4x 5x > = x > 2 = x < 2 Thus, the solution interval is (, 2) = {x x < 2} x 1 > 7x + 2 = 2x 7x > = 5x > = x < 5 Thus, the solution interval is (, 5 ) = {x x < 5 } x + < 11x = x + 11x < = 8x < 6 = x < 4 Thus, the solution interval is (, 4 ) = {x x < 4 } x 1 < 4 or 7x = 2x < 5 or 7x 5 = x < 5 2 or x 5 7

5 Section 1.4 Compound Inequalities 5/2 5/7 For the union, shade anything shaded in either graph. The solution is the set of all real numbers (, ) x 4 < 4 and x = 6x < 0 and x 12 = x > 0 and x 4 = 0 < x 4 0 The intersection is all points shaded in both graphs, so the solution is (0, 4] = {x 0 < x 4} x and 4x 2 > 1 = 8x 6 and 4x > 1 = x 4 and x > 1 4 /4 1/4 Shade all numbers that are shaded in both graphs. There are no such numbers, so the solution set is empty. No solution.

6 Chapter 1 Preliminaries 19. x or 2x 4 = x 1 or 2x 1 = x 1 or x 1 2 1/ 1/2 For the union, shade all points that are shaded in either graph: (, 1 ] [ ) 1 2, = {x x 1 1 or x 2 } x 9 9 and 5x > 1 = 9x 18 and 5x > 1 = x 2 and x > 1 5 1/5 For the intersection, shade any points that are shaded in both graphs. The solution set is ( 1 5, 2] = {x 1 5 < x 2} x 5 < 4 and x + 9 > = x < 9 and x > 6 = x < and x < 6

7 Section 1.4 Compound Inequalities 6 For the intersection, shade all points shaded in both graphs. (, ) = {x x < }. The solution set is 25. 9x + 5 or 2x 4 9 = 9x 8 or 2x 1 = x 8 9 8/9 or x 1 2 1/2 Note that 8 9 > 1 2. For the union, shade any points that are shaded in either graph. The solution set is (, 8 9 ] = {x x 8 9 } x 2 2 or x 9 = 4x 4 or x 12 = x 1 or x For the union, shade any points that are shaded in either graph: (, 1] [4, ) = {x x 1 or x 4} 1 4

8 Chapter 1 Preliminaries 29. 5x + 1 < 6 and x + 9 > 4 = 5x < 7 and x > 1 = x < 7 5 and x > 1 7/5 1/ For the intersection, shade any points that are shaded in both graphs. The solution set is ( 1, 7 5 ) = {x 1 < x < 7 5 } x 7 < 2 and x = 7x < 5 and x = x > 5 7 and x 1 5/7 For the intersection, shade any points that are shaded in both graphs. The solution set is [1, ) = {x x 1} x + 8 < and 8x + 9 = 7x < 11 and 8x 12 = x < /7 and x 2

9 Section 1.4 Compound Inequalities /2 For the intersection, shade all points that are shaded in both graphs. There are no such points, so there is no intersection. 5. 5x and 6x + 2 = 5x 4 and 6x 1 = x 4 5 and x 1 6 4/5 1/6 For the intersection, shade all points that are shaded in both graphs. There are no such points, so there is no solution. 7. 2x 5 1 and 4x + 7 > 7 = 2x 6 and 4x > 0 = x and x > 0 0 For the intersection, shade all points that are shaded in both graphs. Thus, the solution set is (0, ] = {x 0 < x }. 0

10 Chapter 1 Preliminaries 9. 8x or 5x + 6 > 2 = 8x 2 or 5x > 8 = x 1 4 or x < 8 5 1/4 For the union, shade all points that are shaded in either graph. Every number is shaded in one graph or the other, so the solution is the set of all real numbers (, ). 8/ x 8 < 4 or 4x + 2 > = 4x < 12 or 4x > 1 = x > or x < 1 4 1/4 For the union, shade all numbers that are shaded in either graph. Every number is shaded in one of the graphs, so the solution is the set of all real numbers (, ). 4. 9x 5 or x + 1 > = 9x 2 or x > 2 = x 2 9 or x > 2 2/9 2

11 Section 1.4 Compound Inequalities For the union, shade all numbers that shaded in either graph. The solution interval is [ 2 9, ) = {x x 2 9 } x 2 = 2 7x 5 = 2 7 x 5 7 = 5 7 x 2 7 Thus, the solution interval is [ 5 7, 2 7 ] = {x 5 7 x 2 7 } < 9x 6 = 8 < 9x 9 = 8 9 < x 1 Thus, the solution interval is ( 8 9, 1] = {x 8 9 < x 1} < 7x + 6 < 6 = 8 < 7x < 0 = 8 7 > x > 0 = 0 < x < 8 7 Thus, the solution set is (0, 8 7 ){x 0 < x < 8 7 }

12 Chapter 1 Preliminaries 51. Multiply by 12 to clear the fractions. 1 < x < 1 ( 12 1 ) ( x < ) ( ) 1 < < 6x + < 4 Subtract from all three members, then divide all three members of the resulting inequality by 6. 7 < 6x < < x < 1 6 Thus, the solution interval is ( 7/6, 1/6), or equivalently, {x : 7/6 < x < 1/6}. 7/6 1/6 5. Multiply by 6 to clear the fractions. 1 2 < 1 x 2 < 1 ( ) ( 1 < 6 2 x ) ( ) 1 < < 2 x < Subtract 2 from all three members, then divide all three members of the resulting inequality by. Remember to reverse the inequality symbols. 5 < x < 1 5 > x > 1 It is conventional to change the order of this solution to match the order of the shaded solution on the number line. So, equivalently, 1 < x < 5. Thus, the solution interval is ( 1/, 5/), or equivalently, {x : 1/ < x < 5/}. 1/ 5/

13 Section 1.4 Compound Inequalities 55. Multiply by 5 to clear the fractions. 1 < x x + 1 < 2 ( 5 5( 1) < 5 x x + 1 ) < 5(2) 5 5 < 5x (x + 1) < 10 5 < 5x x 1 < 10 5 < 4x 1 < 10 Add 1 to all three members, then divide all three members of the resulting inequality by 4. 4 < 4x < 11 1 < x < 11 4 Thus, the solution interval is ( 1, 11/4), or equivalently, {x : 1 < x < 11/4}. 1 11/4 57. Multiply by 6 to clear the fractions. 2 < x + 1 x ( 2 x + 1 6( 2) < 6 x + 1 ) 6(2) 2 12 < (x + 1) 2(x + 1) < x + 2x < x Subtract 1 from all three members. 1 < x 11 Thus, the solution interval is ( 1, 11], or equivalently, {x : 1 < x 11} We ll need to split the compound inequality x < 4 x < 5 and write it using and. Then we can solve each part independently.

14 Chapter 1 Preliminaries x < 4 x and 4 x < 5 2x < 4 and x < 1 x < 2 and x > 1 Thus, the solution interval is ( 1, 2), or equivalently, {x : 1 < x < 2} We ll want to split the compound inequality x < x and write it using and. Then we can solve each part independently. x < x + 5 and x x < 5 and x 6 x > 5/2 Thus, the solution interval is ( 5/2, 6], or equivalently, {x : 5/2 < x 6}. 5/ a) (F 2) 20 b) 15 5 (F 2) 20 9 = 9(15) (9) 5 (F 2) (9)20 9 = 15 5(F 2) 180 = 15 5F = 295 5F 40 = The solution is {F 59 F 68} F = 59 F 68 c) In roster form, the solutions are {59, 60, 61, 62, 6, 64, 65, 66, 67, 68}.

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