Section 1.4Solving Absolute Value Equations33In Exercises 31−34, write an absolute value equation that has the given solutions.31. x 8 and x 18 32. x 6 and x 1033. x 2 and x 9 34. x 10 and x 5In Exercises 35−44, solve the equation. Check your solutions. (See Examples 4, 5, and 6.)35. 4n 15 n 36. 2c + 8 10c 37. 2b 9 b 6 38. 3k 2 2 k + 2 39. 4 p 3 2p + 8 40. 2 4w1 3 4w + 2 41. 3h + 1 7h 42. 6a 5 4a43. f 6 f + 8 44. 3x 4 3x 5 45. MODELING WITH MATHEMATICS Starting from 300 feet away, a car drives toward you. It then passes by you at a speed of 48 feet per second. The distance d (in feet) of the car from you after t seconds is given by the equation d 300 48t . At what times is the car 60 feet from you?46. MAKING AN ARGUMENT Your friend says that the absolute value equation 3x + 8 9 5 has no solution because the constant on the right side of the equation is negative. Is your friend correct? Explain.47. MODELING WITH MATHEMATICS You randomly survey students about year-round school. The results are shown in the graph.Year-Round SchoolOpposeFavor0%20%40%60%80%32%Error: ±5%68% The error given in the graph means that the actual percent could be 5% more or 5% less than the percent reported by the survey. a. Write and solve an absolute value equation to nd the least and greatest percents of students who could be in favor of year-round school. b. A classmate claims that 1 — 3 of the student body is actually in favor of year-round school. Does this con ict with the survey data? Explain. 48. MODELING WITH MATHEMATICS The recommended weight of a soccer ball is 430 grams. The actual weight is allowed to vary by up to 20 grams. a. Write and solve an absolute value equation to nd the minimum and maximum acceptable soccer ball weights. b. A soccer ball weighs 423 grams. Due to wear and tear, the weight of the ball decreases by 16 grams. Is the weight acceptable? Explain.ERROR ANALYSIS In Exercises 49 and 50, describe and correct the error in solving the equation. 49. 2x − 1 = −9 2x − 1 = −9 or 2x − 1 = −(−9) 2x = −8 2x = 10 x = −4 x = 5The solutions are x = −4 and x = 5.
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50. 5x 8 = x 5x 8 = x or 5x 8 = −x 4x 8 = 0 6x 8 = 0 4x = −8 6x = −8 x = −2 x = − 4 — 3 The solutions are x = −2 and x = 4 — 3 .
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51. ANALYZING EQUATIONS Without solving completely, place each equation into one of the three categories. No solutionOne solutionTwo solutions x 2 + 6 0 x + 3 1 0 x + 8 + 2 7 x 1 + 4 4 x 6 5 9 x + 5 8 8