Chapter 3 Systems of Linear Equations Homework 3.1 1. 5. Write y = 3 ( x в€’ 1) in slope-intercept form. y y = 3 ( x в€’ 1) y = 2x +2 y = 3x в€’ 3 4 в€’2 (1,4) y в€’2 8 x 4 y = 3 x в€’3 (2, 3) y = в€’3 x + 7 в€’8 Verify that (1, 4 ) satisfies both equations. y = 2x + 2 y = в€’3 x + 7 4 = 2 (1) + 2 4 = в€’3(1) + 7 4 = 2+2 4 = в€’3 + 7 4 = 4 true 4 = 4 true 3. в€’8 y = в€’2 x + 7 Verify that ( 2,3) satisfies both equations. (в€’6, 6) y = 3 ( x в€’ 1) y = в€’2 x + 7 3 = 3 (1) 3 = в€’4 + 7 3 = в€’2 ( 2 ) + 7 3 = 3 ( 2 в€’ 1) y 8 1 y = x +8 3 8 в€’8 в€’8 3 = 3 true 3 = 3 true 7. Write 4 x + 2 y = 6 in slope-intercept form. 4x + 2 y = 6 x 1 y = в€’ x +3 2 2 y = в€’4 x + 6 y = в€’2 + 3 Verify that ( в€’6, 6 ) satisfies both equations. 1 y = в€’ x+3 2 1 6 = в€’ ( в€’6 ) + 3 2 6 = 3+3 6 = 6 true x 8 y 1 y = x+8 3 1 6 = ( в€’6 ) + 8 3 6 = в€’2 + 8 y = 3 x в€’7 2 в€’4 (2, в€’1) x в€’6 6 = 6 true y = в€’2 x + 3 y = 2x + 2 y = в€’3 x + 7 Verify that ( 2, в€’1) satisfies both equations. 4x + 2 y = 6 2x + y = 3 4 ( 2 ) + 2 ( в€’1) = 6 2 ( 2 ) + ( в€’1) = 3 8в€’2 = 6 4 в€’1 = 3 6 = 6 true 51 3 = 3 true Homework 3.1 SSM: Intermediate Algebra 13. 9. Write both equations in slope-intercept form. 5 ( y в€’ 2 ) = 21 в€’ 2 ( x + 3) y = 3 ( x в€’ 1) + 8 5 y в€’ 10 = 21 в€’ 2 x в€’ 6 The solution is (1, 4 ) . y = 3x в€’ 3 + 8 5 y = в€’2 x + 25 y = 3x + 5 15. 2 y = в€’ x+5 5 y y = 3 x +5 The solution is ( 480, 460 ) . (0, 5) 4 2 y = в€’ x +5 5 в€’4 17. Write 4 x в€’ 2 y = 2 in slope-intercept form. 4x в€’ 2 y = 2 x 4 в€’2 y = в€’4 x + 2 y = 2x в€’1 The equations are identical. Verify that (0,5 ) satisfies both equations. 5 ( y в€’ 2 ) = 21 в€’ 2 ( x + 3) y = 3 ( x в€’ 1) + 8 5 (5 в€’ 2 ) = 21 в€’ 2 (0 + 3 ) 5 = 3 (0 в€’ 1) + 8 5 (3 ) = 21 в€’ 2 (3 ) 5 = 3 ( в€’1) + 8 15 = 15 true 5 = 5 true The system is dependent. The solution set contains all ordered pairs ( x, y ) such that 11. Write both equations in slope-intercept form. 1 1 1 1 xв€’ y =1 x+ y =2 2 2 4 2 xв€’ y = 2 x + 2y = 8 в€’ y = в€’x + 2 y = xв€’2 1 y = в€’ x +4 2 y = 2x в€’1 . 19. Write 0.2 y в€’ x = 3 in slope-intercept form. 0.2 y в€’ x = 3 0.2 y = x + 3 2 y = в€’x + 8 1 y = в€’ x+4 2 y = 5 x + 15 y 4 в€’4 (4, 2) 4 x The system is inconsistent. The solution set is the empty set, в€… . в€’4 y = x в€’2 Verify that ( 4, 2 ) satisfies both equations. 1 1 xв€’ y =1 2 2 1 1 (4 ) в€’ (2 ) = 1 2 2 2 в€’1 = 1 1 в€’ 1 true 1 1 x+ y = 2 4 2 1 1 (4 ) + (2 ) = 2 4 2 1+1 = 2 2 = 2 true 52 Homework 3.1 SSM: Intermediate Algebra 21. Write both equations in slope-intercept form. 1 1 1 2 xв€’ y =1 x+ y =2 2 2 3 3 xв€’ y = 2 x + 2y = 6 в€’ y = в€’x + 2 2 y = в€’x + 6 y = xв€’2 25. a. 1 y = в€’ x+3 2 Start by plotting the data. Then find the regression lines for the data. Women: W (t ) = 0.14t + 4.52 Men: The solution is roughly (3.33,1.33) . 23. a. In 2002, t = 32 . W (32 ) = в€’0.19 (32 ) + 43.73 M (t ) = 0.083t + 4.60 = 37.65 M (32 ) = в€’0.16 (32 ) + 39.90 b. = 34.78 The womenвЂ™s time is 37.65 seconds and the menвЂ™s time is 34.78 seconds. The error for the womenвЂ™s time estimate is 0.275 seconds and the error for the menвЂ™s time estimate is 0.165 seconds. b. c. The enrollment for men and women were equal in 1971 with roughly 4.72 million each. The absolute value of the slope of W is more than the absolute value of the slope of M. This shows that the winning times of women decrease at a greater rate than the winning times of men. c. In 2009, t = 39. W (39 ) = 0.14 (39 ) + 4.52 = 9.98 M (39 ) = 0.083 (39 ) + 4.60 Since the winning time for women is decreasing at a faster rate than for men, the winning time for men may equal the winning time for women in an upcoming year since these times are only a few seconds apart in the year 2002 and they are getting closer as each year passes. = 7.837 The total enrollment will be roughly 9.98 + 7.837 = 17.817 million students. 27. a. Start by plotting the data. Then find the regression lines for the data. Punch Cards: d. f (t ) = в€’2.45t + 72.82 Men and women will have a winning time of 19.47 seconds in the year 2098. 53 Homework 3.1 SSM: Intermediate Algebra 37. Optical Scan: f ( x ) = g ( x ) when x = в€’1 . 39. Answers may vary. Possible answers: a. y 1 y =в€’ x +2 2 f (t ) = 3.08t + 12.58 y = 2 x в€’3 4 (2, 1) b. в€’4 x 4 в€’4 b. y The percentage of voters using punch cards will equal the percentage using electronic devices in 2001. This is not a national election year. The percentages will be closest in the year 2000. In 2000, t = 10. f (10 ) = в€’2.45 (10 ) + 72.82 в€’4 c. 3 x 2 1 y = (3 x + 2 ) в€’ 1 2 y 4 3 x 2 x 4 в€’4 41. Graph the equations. y = x+3 y = в€’2 x + 9 31. The slope of f ( x ) is в€’3 and the f-intercept is (0,30 ) . Therefore, f ( x ) = в€’3x + 30 . The slope of g ( x ) is 5 and the g-intercept is (0, 2 ) . Therefore, g ( x ) = 5 x + 2 . The solution for this system is (3.5,19.5 ) . You may verify your y = 3x в€’ 1 y 4 answer using the slope or substituting the values for x in the functions. f ( x ) = 3 when x = 5 . y= в€’4 29. The solution of this system is estimated to be ( в€’1.9, в€’2.8) . 35. x 4 y= 100 в€’ ( 48.32 + 43.38 ) = 8.3 In 2000, 8.3% of voters used the other two methods of voting. f ( в€’4 ) = 0 x в€’4 = 43.38 In 2000, the percentage of voters using punch cards or lever machines was 48.32% and the percentage using optical scans or other electronic devices was 43.38%. 33. y 4 = 48.32 g (10 ) = 3.08 (10 ) + 12.58 c. y = x +1 в€’2 в€’2 y = 3 x в€’1 y = x +3 (2, 5) 4 x y = в€’2 x + 9 The solution of the system is ( 2,5 ) . 54 Homework 3.2 SSM: Intermediate Algebra 5. Substitute 2 ( x в€’ 5 ) for y in 3 x в€’ 5 y = 29 . 43. Answers may vary. One possible answer: 3 y=в€’ x 4 y = x+7 3 x в€’ 5 ( 2 ( x в€’ 5 )) = 29 3 x в€’ 5 ( 2 x в€’ 10 ) = 29 3 x в€’ 10 x + 50 = 29 y = в€’2 x в€’ 5 y в€’7 x = в€’21 y = x +7 x=3 Let y = 3 in y = 2 ( x в€’ 5 ) . 4 (в€’4, 3) в€’4 в€’2 y = 2 (3 в€’ 5 ) y = в€’2 x в€’ 5 = 2 ( в€’2 ) x 2 = в€’4 The solution is (3, в€’4 ) . 3 y =в€’ x 4 45. The solution of a system of two equations is an ordered pair that satisfies both equations. This solution is the intersection point of the graphs since this is the point that satisfies both equations. 7. Substitute в€’4 y в€’ 8 for x in 3 x в€’ 2 y = 18 . 3 ( в€’4 y в€’ 8 ) в€’ 2 y = 18 в€’12 y в€’ 24 в€’ 2 y = 18 в€’14 y = 42 y = в€’3 Let y = в€’3 in x = в€’4 y в€’ 8 . Homework 3.2 1. Substitute x в€’ 5 for y in x + y = 9 . x = в€’4 ( в€’3 ) в€’ 8 x + ( x в€’ 5) = 9 = 12 в€’ 8 2x в€’ 5 = 9 =4 The solution is ( 4, в€’3) . 2 x = 14 x=7 Let x = 7 in y = x в€’ 5 . y = 7в€’5 9. Substitute 99x for y in y = 100 x . 99 x = 100 x =2 The solution is (7, 2 ) . в€’x = 0 x=0 Let x = 0 in y = 99 x . 3. Substitute 4 y + 7 for x in 2 x в€’ 3 y = 1 . y = 99 (0 ) 2 ( 4 y + 7 ) в€’ 3 y = в€’1 =0 The solution is (0, 0 ) . 8 y + 14 в€’ 3 y = в€’1 5 y + 14 = в€’1 11. Substitute 0.2 x + 0.6 for y in 2 y в€’ 3 x = в€’4 . 5 y = в€’15 2 (0.2 x + 0.6 ) в€’ 3 x = в€’4 y = в€’3 Let y = в€’3 in x = 4 y + 7 . 0.4 x + 1.2 в€’ 3 x = в€’4 x = 4 ( в€’3 ) + 7 в€’2.6 x = в€’5.2 x=2 Let x = 2 in y = 0.2 x + 0.6 . = в€’12 + 7 = в€’5 The solution is ( в€’5, в€’3) . y = 0.2 ( 2 ) + 0.6 = 0.4 + 0.6 =1 The solution is ( 2,1) . 55 Homework 3.2 SSM: Intermediate Algebra 19. Rewrite the second equation so the variables are on the left side of the equation and the constant is on the right side. 3x + 5 y = 3 1 x в€’ 5 for y in 2 x + 3 y = в€’1 . 2 пЈ«1 пЈ¶ 2 x + 3 пЈ¬ x в€’ 5 пЈ· = в€’1 пЈ2 пЈё 3 2 x + x в€’ 15 = в€’1 2 7 x = 14 2 x=4 1 Let x = 4 in y = x в€’ 5 . 2 1 y = (4 ) в€’ 5 2 = 2в€’5 13. Substitute x + 4y = 8 Multiply the second equation by в€’3 and add the equations. 3x + 5 y = 3 в€’3 x в€’ 12 y = в€’24 в€’ 7 y = в€’21 y=3 Substitute y = 3 into x = 8 в€’ 4 y and solve for x. x = 8 в€’ 4 (3) = 8 в€’ 12 = в€’3 The solution is ( 4, в€’3) . = в€’4 The solution is ( в€’4,3) . 15. Multiply the first equation by в€’1 and then add the equations. 4x + y = 9 21. Multiply the first equation by в€’3 and the second equation by 2, then add the equations. в€’24 x + 27 y = 129 в€’3 x в€’ y = в€’5 24 x + 30 y = 42 x=4 Substitute x = 4 into 4 x + y = 9 and solve for y. 57 y = 171 y=3 Substitute y = 3 into 8 x в€’ 9 y = в€’43 and solve for x. 8 x в€’ 9 (3 ) = в€’43 4 (4 ) + y = 9 16 + y = 9 y = в€’7 The solution is ( 4, в€’7 ) . 8 x в€’ 27 = в€’43 8 x = в€’6 17. Multiply the first equation by 2 and add the equations. 6 x в€’ 4 y = 14 x = в€’2 The solution is ( в€’2,3) . в€’6 x в€’ 5 y = 4 23. Multiply the second equation by 2 and add the equations. 0.9 x + 0.4 y = 1.9 в€’ 9 y = 18 y = в€’2 Substitute y = в€’2 into 3 x в€’ 2 y = 7 and solve for x. 3 x в€’ 2 ( в€’2 ) = 7 0.6 x в€’ 0.4 y = 2.6 1.5 x = 4.5 x=3 Substitute x = 3 into 0.9 x + 0.4 y = 1.9 and solve for y. 0.9 (3) + 0.4 y = 1.9 3x + 4 = 7 3x = 3 x =1 The solution is (1, в€’2 ) . 2.7 + 0.4 y = 1.9 0.4 y = в€’0.8 y = в€’2 The solution is (3, в€’2 ) . 56 Homework 3.2 SSM: Intermediate Algebra 25. Use the distributive property to simplify both equations. 3 ( 2 x в€’ 1) + 4 ( y в€’ 3) = 1 29. Multiply the first equation by 3 and the second equation by в€’2 , then add the equations. 3 1 2x + y = 2 2 в€’2 x в€’ 5 y = в€’11 6 x в€’ 3 + 4 y в€’ 12 = 1 6 x + 4 y = 16 4 ( x + 5 ) в€’ 2 ( 4 y + 1) = 18 в€’ 4 x + 20 в€’ 8 y в€’ 2 = 18 4x в€’ 8 y = 0 The system can be rewritten as 6 x + 4 y = 16 7 21 y=в€’ 2 2 y=3 Substitute y = 3 into 2 1 1 x + y = and solve 3 2 6 for x. 2 1 1 x + (3) = 3 2 6 2 3 1 x+ = 3 2 6 2 4 x=в€’ 3 3 x = в€’2 The solution is ( в€’2,3) . 4x в€’ 8 y = 0 Multiply the first equation by 2 and add the equations. 12 x + 8 y = 32 4x в€’ 8 y = 0 16 x = 32 x=2 Substitute x = 2 into 4 x в€’ 8 y = 0 and solve for y. 4 (2 ) в€’ 8 y = 0 31. Substitute 2 x + 5 for y in 6 x в€’ 3 y = в€’3 . 6 x в€’ 3 ( 2 x + 5 ) = в€’3 8 в€’8y = 0 в€’8 y = в€’8 6 x в€’ 6 x в€’ 15 = в€’3 y =1 в€’15 = в€’3 false This is a contradiction. The system is inconsistent. The lines are parallel so the solution set is the empty set, в€… . The solution is ( 2,1) . 27. Multiply the first equation by 3 and add the equations. 3 9 x + y = 21 5 2 2 9 x в€’ y = в€’16 5 2 x=5 1 3 Substitute x = 5 into x + y = 7 and solve 5 2 for y. 1 3 (5 ) + y = 7 5 2 3 1+ y = 7 2 3 y=6 2 y=4 33. Multiply the first equation by 3 and the second equation by 2, then add the equations. 39 x + 30 y = в€’21 34 x в€’ 30 y = 94 73 x = 73 x =1 Substitute x = 1 into 13 x + 10 y = в€’7 and solve for y. 13 (1) + 10 y = в€’7 13 + 10 y = в€’7 10 y = в€’20 y = в€’2 The solution is (1, в€’2 ) . The solution is (5, 4 ) . 57 Homework 3.2 SSM: Intermediate Algebra 35. Multiply the first equation by 3 and add the equations. 12 x в€’ 15 y = 9 43. Substitute 1 x + 3 for y in 2 y в€’ x = 6 . 2 пЈ«1 пЈ¶ 2 пЈ¬ x + 3пЈ· в€’ x = 6 пЈ2 пЈё x+6в€’x = 6 в€’12 x + 15 y = в€’9 0 = 0 true This is an identity. The system is dependent. The solution set is the set of ordered pairs ( x, y ) 6 = 6 true This is an identity. The system is dependent. The solution set is the set of ordered pairs ( x, y ) such such that 4 x в€’ 5 y = 3 . 37. Substitute 2.3 x в€’ 7 for y in y = в€’0.6 x + 4 . 2.3 x в€’ 7 = в€’0.6 x + 4 that y = 2.9 x = 11 1 x +3. 2 45. Multiply the first equation by 2 and the second equation by 5, then add the equations. 5 1 x+ y = 6 3 2 5 25 в€’ x+ y = 20 3 2 13 y = 26 x в‰€ 3.79 Substitute x = 3.79 into y = 2.3 x в€’ 7 and solve for y. y = 2.3 (3.79 ) в€’ 7 в‰€ 1.72 The solution is roughly (3.79,1.72 ) . y=2 39. Multiply the first equation by 3 and the second equation by 4, then add the equation. в€’24 x + 27 y = в€’21 Substitute y = 2 into 5 1 x + y = 3 and solve 6 4 for x. 5 1 x + (2 ) = 3 6 4 5 1 x+ =3 6 2 5 5 x= 6 2 x=3 The solution is (3, 2 ) . 24 x в€’ 60 y = в€’12 в€’ 33 y = в€’33 y =1 Substitute y = 1 into 6 x в€’ 15 y = в€’3 and solve for x. 6 x в€’ 15 (1) = в€’3 6 x в€’ 15 = в€’3 6 x = 12 x=2 The solution is ( 2,1) . 47. Elimination: Rewrite the second equation so the variables are on the left side of the equation. 3 x + y = 11 41. Multiply the first equation by 2 and add the equations. 8x в€’ 6 y = 2 2x + y = 9 Multiply the second equation by в€’1 and add the equations. 3 x + y = 11 в€’8 x + 6 y = в€’5 0 = в€’3 false This is a contradiction. The system is inconsistent so the solution set is the empty set. в€’2 x в€’ y = в€’9 x=2 Substitute x = 2 into y = в€’2 x + 9 and solve for y. y = в€’2 ( 2 ) + 9 =5 The solution is ( 2,5 ) . 58 Homework 3.2 SSM: Intermediate Algebra Substitute 4 x + 3 for y in y = в€’6 x + 50 and solve for x. 4 x + 3 = в€’6 x + 50 Substitution: Substitute в€’2 x + 9 for y in the first equation. 3 x + ( в€’2 x + 9 ) = 11 10 x = 47 3 x в€’ 2 x + 9 = 11 x = 4.7 Substitute x = 4.7 into y = 4 x + 3 and solve for y. y = 4 ( 4.7 ) + 3 x=2 Substitute x = 2 into y = в€’2 x + 9 and solve for y. y = в€’2 ( 2 ) + 9 = 18.8 + 3 = в€’4 + 9 = 21.8 The solution is ( 4.7, 21.8 ) . =5 The solution is ( 2,5 ) . 53. The coordinate for A is (0, 0 ) since it lies at the origin. The coordinate for B is the y-intercept of l 1 , which is (0,3) . The coordinate for C is the Graphing: y 8 y = в€’3 x +11 point of intersection of l 1 and l 2 . Solve the following system: l 1 : y = 2x + 3 (2, 5) 4 y = в€’2 x + 9 4 8 l 2 : 3 y + x = 30 Substitute 2 x + 3 for y in the second equation. 3 ( 2 x + 3) + x = 30 x 49. The student is not correct. A system is inconsistent (i.e. the solution set is the empty set) when the lines are parallel. Parallel lines must have the same slope. The slopes are different yet they are close enough to where the lines look parallel around the origin. To find the correct solution, substitute 2 x + 3 for y in y = 2.01x + 1 and solve for x. 2 x + 3 = 2.01x + 1 6 x + 9 + x = 30 7 x = 21 x=3 Substitute x = 3 into the first equation and solve for y. y = 2 (3 ) + 3 = 6+3 =9 The solution is (3,9 ) so the coordinate of C is в€’0.01x = в€’2 x = 200 Substitute x = 200 into y = 2 x + 3 and solve for y. y = 2 ( 200 ) + 3 (3,9 ) . The coordinate for D is the point of intersection of l 2 and l 3 . Solve the following system: l 2 : 3 y + x = 30 = 403 The solution is ( 200, 403) . l 3 : y + 3x = 26 Multiply the second equation by в€’3 and add the equations. 3 y + x = 30 51. The function f has a slope of 4 and an f-intercept of (0,3) . Therefore, f ( x ) = 4 x + 3 . The в€’3 y в€’ 9 x = в€’78 function g has a slope of в€’6 and a g-intercept of (0,50 ) . Therefore, g ( x ) = в€’6 x + 50 . The в€’ 8 x = в€’48 x=6 Substitute x = 6 into 3 y + x = 30 and solve for y. system that describes the functions f and g is: y = 4x + 3 y = в€’6 x + 50 Solve using substitution. 59 Homework 3.2 SSM: Intermediate Algebra 3 y + (6 ) = 30 пЈ« в€’ax + c пЈ¶ dx + e пЈ¬ пЈ·= f пЈ b пЈё aex ce dx в€’ + = f b b bdx в€’ aex + ce = bf 3 y = 24 y =8 The solution is (6,8) so the coordinate of D is (6,8) . The coordinate for E is the point of (bd в€’ ae ) x + ce = bf (bd в€’ ae ) x = bf intersection of l 3 and l 4 . Solve the following system. l 3 : y + 3x = 26 bf в€’ ce ce в€’ bf or bd в€’ ae ae в€’ bd Substitute this result for x in ax + by = c and solve for y. пЈ« ce в€’ bf пЈ¶ aпЈ¬ пЈ· + by = c пЈ ae в€’ bd пЈё ace в€’ abf + by = c ae в€’ bd ace в€’ abf by = c в€’ ae в€’ bd ace в€’ bcd ace в€’ abf by = в€’ ae в€’ bd ae в€’ bd в€’bcd + abf by = ae в€’ bd abf в€’ bcd by = ae в€’ bd af в€’ cd y= ae в€’ bd Therefore, the solution is пЈ« ce в€’ bf af в€’ cd пЈ¶ , пЈ¬ пЈ· , assuming ae в€’ bd в‰ 0 . пЈ ae в€’ bd ae в€’ bd пЈё x= l 4 : y = 2 x в€’ 10 Substitute 2 x в€’ 10 in for y in y + 3 x = 26 . 2 x в€’ 10 + 3 x = 26 5 x = 36 x = 7.2 Substitute x = 7.2 into y = 2 x в€’ 10 and solve for x. y = 2 ( 7.2 ) в€’ 10 = 4.4 The solution is (7.2, 4.4 ) so the coordinate of E is (7.2, 4.4 ) . The coordinate of F is the x- intercept of l 4 . Let y = 0 in l 4 and solve for x. y = 2 x в€’ 10 0 = 2 x в€’ 10 2 x = 10 x=5 The coordinate of F is (5, 0 ) . y C (3 , 9 ) D (6 , 8 ) 8 b. E (7 .2, 4 .4 ) 4 B (0 , 3 ) F (5 , 0 ) A (0 , 0 ) 55. a. 4 8 в€’ ce x Substitute 3 for a, 5 for b, 2 for c, 4 for d, 3 for e, and 4 for f in the solution from part a. 2 (3 ) в€’ 5 ( 4 ) в€’14 14 x= = = 3 (3) в€’ 5 ( 4 ) в€’11 11 y= You may use more than one method to solve the system. For example, you may use substitution as follows: Solve the first equation for y. ax + by = c 3 (4 ) в€’ 2 (4 ) 3 ( 3) в€’ 5 ( 4 ) = 4 4 =в€’ 11 в€’11 пЈ« 14 4 пЈ¶ The solution is пЈ¬ , в€’ пЈ· . пЈ 11 11 пЈё 57. Written response. Answers may vary. by = в€’ ax + c в€’ax + c b Substitute this result for y in the second equation and solve for x. y= 60 Homework 3.3 SSM: Intermediate Algebra 7. a. Homework 3.3 1. Solve the system y = в€’0.19t + 43.73 Solve the system y = 13.5t + 229 y = в€’5t + 365 Substitute 13.5t + 229 for y in the second equation and solve for t. 13.5t + 229 = в€’5t + 365 y = в€’0.16t + 39.90 Substitute в€’0.19t + 43.73 for y in the second equation and solve for t. в€’0.19t + 43.73 = в€’0.16t + 39.90 18.5t = 136 t в‰€ 7.35 Substitute this result into the first equation and solve for y. y = 13.5 (7.35 ) + 229 в€’0.03t = в€’3.83 t в‰€ 127.67 Substitute this result into the first equation and solve for y. y = в€’0.19 (127.67 ) + 43.73 в‰€ 328.23 According to the models, the two newspapers had equal circulations of roughly 328 thousand in 1997. в‰€ 19.47 According to the models, the winning times for women and men will both be 19.47 seconds in the year 2098. b. Since the circulations were roughly equal in 1997, competition heated up as each newspaper tried to overtake the other. c. D (10 ) = 13.5 (10 ) + 229 3. Solve the system E = 0.14t + 4.52 E = 0.083t + 4.60 Substitute 0.14t + 4.52 for E in the second equation and solve for t. 0.14t + 4.52 = 0.083t + 4.60 = 135 + 229 = 364 R (10 ) = в€’5 (10 ) + 365 0.057t = 0.08 = в€’50 + 365 t в‰€ 1.40 Substitute this result into the first equation and solve for E. E = 0.14 (1.4 ) + 4.52 в‰€ 4.71 According to the models, the enrollments for men and women were approximately equal in 1971 (roughly 4.71 million). = 315 826 в€’ (364 + 315 ) = 826 в€’ 679 = 147 According to the models, the combined increase due to bonus copies was roughly 147 thousand bonus copies. d. 5. Solve the system p = в€’2.45t + 72.82 D (11) = 13.5 (11) + 229 = 377.5 R (11) = в€’5 (11) + 365 p = 3.08t + 12.58 Substitute в€’2.45t + 72.82 for p in the second equation and solve for t. в€’2.45t + 72.82 = 3.08t + 12.58 = 310 377.5 + 310 = 687.5 According to the models, the combined circulation from the two newspapers was roughly 688 thousand copies in 2001. в€’5.53t = в€’60.24 t в‰€ 10.89 Substitute this result into the first equation and solve for p. p = в€’2.45 (10.89 ) + 72.82 в‰€ 46.14 According to the models, the percentage of voters using optical scan or other electronic devices equaled the percentage of voters using punch cards or lever machines in 2001. This was not a national election year. The percentages were closest in 2000. e. 61 The estimate in part d. was an overestimate. After joining revenue streams, the competition for subscribers ceased (or at least reduced if there were other competitors). The end of bonus copies, or just the merger in general, may have caused some subscribers to cancel subscriptions. Homework 3.3 9. a. SSM: Intermediate Algebra c. Start by plotting the data, then compute the regression line for each data set. Women: Women: W (t ) = в€’1.01t + 323.57 Men: W (t ) = в€’1.22t + 338.47 M (t ) = в€’0.36t + 240.73 Men: Solve the system y = в€’1.01t + 323.57 y = в€’0.36t + 240.73 Substitute в€’1.01t + 323.57 for y in the second equation and solve for t. в€’1.01t + 323.57 = в€’0.36t + 240.73 в€’0.65t = в€’82.84 t в‰€ 127.45 Substitute this result into the first equation and solve for y. y = в€’1.01(127.45 ) + 323.57 M (t ) = в€’0.36t + 240.44 b. в‰€ 194.85 Now the models predict that the record times for men and women will be the same in 2027. Solve the system y = в€’1.22t + 338.47 y = в€’0.36t + 240.44 Substitute в€’1.22t + 338.47 for y in the second equation and solve for t. в€’1.22t + 338.47 = в€’0.36t + 240.44 в€’0.86t = в€’98.03 t в‰€ 113.99 Substitute this result into the first equation and solve for y. y = в€’1.22 (113.99 ) + 338.47 в‰€ 199.40 According to the models, the record times for men and women will both be approximately 199.40 seconds in 2014. d. For the most part the data appear to be linear. However, the data value for women in 1926 seems to deviate somewhat from the linear pattern. Removing this value makes the linear model a better fit. 11. a. Since a 2001 TaurusвЂ™ value decreases by a constant $1582 each year, the function T (t ) is linear and its slope is вЂ“1582. The Tintercept is (0,12939 ) , since the car is worth $12,939 at year t = 0 (2002). Doing similar work for E (t ) , we get the following equations: T (t ) = в€’1582t + 12,939 E (t ) = в€’1024t + 9330 62 Homework 3.3 SSM: Intermediate Algebra b. Solve the system V = в€’1582t + 12,939 c. Find the intersection point using a graphing utility. 15. a. Since Jenny CraigвЂ™s program fees increase by a constant $72 each week, the function J is linear and its slope is 72. The J-intercept is (0,19 ) , since the start-up fee is $19 at V = в€’1024t + 9330 Substitute в€’1582t + 12,939 in for V in the second equation and solve for t. в€’1582t + 12,939 = в€’1024t + 9330 в€’558t = в€’3609 t в‰€ 6.468 Substitute this result into the first equation and solve for V. V = в€’1582 (6.468 ) + 12,939 c. в‰€ 2707 The cars will both be worth roughly $2707 in 2008. t = 0 . So, an equation for J (t ) is: Find the intersection point using a graphing utility. Since Weight WatchersвЂ™ program fees increase by a constant $77 each week ($17 fee + $60 food), the function W is linear and its slope is 77. The W-intercept is (0, 0 ) J (t ) = 72t + 19 . since there is no start-up fee at t = 0 . An equation for W (t ) is: W (t ) = 77t . 13. a. b. Since college AвЂ™s tuition increases by a constant $670 each year, the function A (t ) y = 77t Substitute 72t + 19 for y into the second equation. 72t + 19 = 77t is linear and its slope is 670. The A-intercept is (0, 6100 ) , since tuition is $6100 in year t = 0 (2000). Similar work for college B gives the following equations: A (t ) = 670t + 6100 5t = 19 t = 3.8 Substitute this result into the first equation and solve for y. y = 72 (3.8 ) + 19 B (t ) = 440t + 8500 b. Solve the system y = 72t + 19 Solve the system y = 670t + 6100 = 292.6 The total cost at both Jenny Craig and Weight Watchers is approximately $293 in 4 weeks. y = 440t + 8500 Substitute 670t + 6100 for y in the second equation and solve for t. 670t + 6100 = 440t + 8500 c. Find the intersection point using a graphing utility. 230t = 2400 t в‰€ 10.435 Substitute this result in the first equation and solve for y. y = 670 (10.435 ) + 6100 = 13091.45 The tuition at both colleges will be $13,091 in approximately 10 years (in 2010). 63 Homework 3.4 SSM: Intermediate Algebra 17. Let P (t ) represent the average price (in dollars) order to predict when the family will be able to pay a 10% down payment on an average-priced house, solve the following system for t when y = 0.1( P (t )) = 0.1(8000t + 214000 ) of a home in a community and S (t ) represent the amount of money (in dollars) a family has saved at t years since 2000. Since the average price of a home increases at a constant $8000 per year, the function P is linear and its slope is 8000. the P-intercept is (0, 214000 ) since the y = S (t ) = 3600t + 10000 Substitute 0.1(8000t + 214000 ) for y in the second equation and solve for t. 0.1(8000t + 214000 ) = 3600t + 10000 price of a home is $214,000 in year t = 0 . So, an equation for P is : P (t ) = 8000t + 214000 . 800t + 21400 = 3600t + 10000 Similar work in finding the equation for the function S gives: S (t ) = 3600t + 10000 . (The slope of S is 3600 since the family plans to save $300 each month which is $3600 each year.) In в€’2800t = в€’11400 t в‰€ 4.07 The family will be able to pay a 10% down payment in 4 years (2004). Homework 3.4 1. 3. Words Inequality Notation Numbers greater than 3 x>3 Numbers less than or equal to 4 xв‰¤4 Numbers less than 5 x<5 Numbers greater than or equal to в€’1 x в‰Ґ в€’1 0 ( в€’в€ћ,5) 5 в€’1 7. ( в€’в€ћ, в€’4] 0 в€’4 x + 2в€’ 2 в‰Ґ 5в€’ 2 [в€’1, в€ћ ) 0 2 x + 7 < 11 2 x + 7 в€’ 7 < 11 в€’ 7 xв‰Ґ3 Interval: [3, в€ћ ) 5. (3,в€ћ ) 3 0 x+2в‰Ґ5 0 Interval Notation Graph 2x < 4 2x 4 < 2 2 x<2 Interval: ( в€’в€ћ, 2 ) 3 в€’x + 2 в‰Ґ 5 в€’x + 2 в€’ 2 в‰Ґ 5 в€’ 2 0 в€’x в‰Ґ 3 в€’x 3 в‰¤ в€’1 в€’1 x в‰¤ в€’3 Interval: ( в€’в€ћ, в€’3] в€’3 0 64 2 Homework 3.4 SSM: Intermediate Algebra 9x < 4 + 5x 9. 17. 9x в€’ 5x < 4 + 5x в€’ 5x 3 в€’ 2x + 8 > 4x +1 4x < 4 в€’2 x + 11 > 4 x + 1 в€’2 x + 11 в€’ 11 > 4 x + 1 в€’ 11 4x 4 < 4 4 x <1 Interval: ( в€’в€ћ,1) в€’2 x > 4 x в€’ 10 в€’2 x в€’ 4 x > 4 x в€’ 10 в€’ 4 x в€’6 x > в€’10 0 в€’6 x в€’10 < в€’6 в€’6 5 x< 3 5пЈ¶ пЈ« Interval: пЈ¬ в€’в€ћ, пЈ· 3пЈё пЈ 1 2.1x в€’ 7.4 в‰¤ 10.4 11. 2.1x в€’ 7.4 + 7.4 в‰¤ 10.4 + 7.4 2.1x в‰¤ 17.8 2.1x 17.8 в‰¤ 2.1 2.1 x в‰¤ 8.48 Interval: ( в€’в€ћ,8.48] 0 13. в€’3 19. 0 3 в€’6.23 x + 2.35 < 1.76 (3 в€’ 2.73 x ) в€’6.23 x + 2.35 < 5.28 в€’ 4.8048 x 8 4 в€’6.23 x + 2.35 в€’ 2.35 < 5.28 в€’ 4.8048 x в€’ 2.35 2 x в€’ 3 > 7 x + 22 в€’6.23 x < 2.93 в€’ 4.8048 x 2 x в€’ 3 + 3 > 7 x + 22 + 3 в€’6.23 x + 4.8048 x < 2.93 в€’ 4.8048 x + 4.8048 x 2 x > 7 x + 25 в€’1.4252 x < 2.93 2 x в€’ 7 x > 7 x + 25 в€’ 7 x в€’1.4252 x 2.93 > в€’1.4252 в€’1.4252 x > в€’2.06 Interval: ( в€’2.06, в€ћ ) в€’5 x > 25 в€’5 x 25 < в€’5 в€’5 x < в€’5 Interval: ( в€’в€ћ, в€’5 ) в€’3 в€’5 15. 3 в€’ 2 ( x в€’ 4) > 4x + 1 0 3 21. 7 ( x + 1) в€’ 8 ( x в€’ 2 ) в‰¤ 0 0 7 x + 7 в€’ 8 x + 16 в‰¤ 0 7 x в€’ 3 в‰¤ в€’2 x в€’ 21 в€’ x + 23 в‰¤ 0 7 x в€’ 3 + 3 в‰¤ в€’2 x в€’ 21 + 3 в€’ x в‰¤ в€’23 7 x в‰¤ в€’2 x в€’ 18 в€’ x в€’23 в‰Ґ в€’1 в€’1 x в‰Ґ 23 7 x + 2 x в‰¤ в€’2 x в€’ 18 + 2 x 9 x в‰¤ в€’18 Interval: [23, в€ћ ) 9 x в€’18 в‰¤ 9 9 x в‰¤ в€’2 Intervals: ( в€’в€ћ, в€’2] 23 в€’2 0 65 26 Homework 3.4 23. SSM: Intermediate Algebra 2 в€’ x>4 3 3 2 3 в€’ в‹…в€’ x < в€’ в‹…4 2 3 2 x < в€’6 Interval: ( в€’в€ћ, в€’6 ) 29. 0 в€’6 25. 1 xв€’2 <3 4 1 x в€’2+ 2 < 3+ 2 4 1 x<5 4 1 4в‹… x < 4в‹…5 4 x < 20 Interval: ( в€’в€ћ, 20 ) в€’1 31. T >E в€’1582t + 12, 939 в€’ 12, 939 > в€’1024t + 9330 в€’ 12, 939 в€’1582t > в€’1024t в€’ 3609 в€’1582t + 1024t > в€’1024t в€’ 3609 + 1024t в€’558t > в€’3609 в€’558t в€’3609 < в€’558 в€’558 t < 6.47 0 20 2 3 5 в€’ xв‰¤ 3 4 2 2 3 2 5 2 в€’ xв€’ в‰¤ в€’ 3 4 3 2 3 3 11 в€’ xв‰¤ 4 6 4 3 4 11 в€’ в‹…в€’ x в‰Ґ в€’ в‹… 3 4 3 6 22 xв‰Ґв€’ 9 пЈ® 22 пЈ¶ Interval: пЈЇ в€’ , в€ћ пЈ· пЈ° 9 пЈё в€’ 0 в€’1582t + 12, 939 > в€’1024t + 9330 15 27. 1 1 1 в€’ xв€’ в‰Ґ 2 6 3 1 1 1 1 1 в€’ xв€’ + в‰Ґ + 2 6 6 3 6 1 1 в€’ xв‰Ґ 2 2 1 1 в€’2 в‹… в€’ x в‰¤ в€’2 в‹… 2 2 x в‰¤ в€’1 Interval: ( в€’в€ћ, в€’1] 22 9 The value of the Taurus is more than the value of the Escort for years up until 2008 ( t < 6 ). 33. a. Since U-HaulвЂ™s charge increases at a constant rate of $0.69 per mile, the equation is linear with slope 0.69 . The U-intercept is 19.95 since U-Haul charges a flat fee of $19.95. An equation for U-HaulвЂ™s charge is: U ( x ) = 0.69 x + 19.95 . After similar work for Penske, an equation for PenskeвЂ™s charge is: P ( x ) = 0.39 x + 29.95 . 0 b. U<P 0.69 x + 19.95 < 0.39 x + 29.95 0.69 x + 19.95 в€’ 19.95 < 0.39 x + 29.95 в€’ 19.95 0.69 x < 0.39 x + 10 0.69 x в€’ 0.39 x < 0.39 + 10 в€’ 0.39 x 0.3 x < 10 0.3x 10 < 0.3 0.3 100 x< 3 U-Haul will be cheaper for miles driven less than 33.3 miles. 66 Homework 3.4 SSM: Intermediate Algebra 35. a. W ( 28 ) = 0.098 ( 28 ) + 77.58 37. The student made a mistake. It is not necessary to switch the direction of the inequality when dividing by a positive number. в‰€ 80.3 M ( 28 ) = 0.192 ( 28 ) + 69.98 39. a. в‰€ 75.4 80.3 в€’ 75.4 = 4.9 Women born in 2008 will live roughly 5 years longer, on average, than men born in 2008. b. Solve the inequality for x. 3( x в€’ 2) + 1 в‰Ґ 7 в€’ 4x 3x в€’ 6 + 1 в‰Ґ 7 в€’ 4 x 3x в€’ 5 в‰Ґ 7 в€’ 4 x 3x в€’ 5 + 5 в‰Ґ 7 в€’ 4 x + 5 W <M 3 x в‰Ґ 12 в€’ 4 x 0.098t + 77.58 < 0.192t + 69.98 3 x + 4 x в‰Ґ 12 в€’ 4 x + 4 x 0.098t + 77.58 в€’ 77.58 < 0.192t + 69.98 в€’ 77.58 7 x в‰Ґ 12 0.098t < 0.192t в€’ 7.6 7 x 12 в‰Ґ 7 7 12 xв‰Ґ 7 Any three numbers that are greater than or 12 equal to are possible solutions. 7 0.098t в€’ 0.192t < 0.192t в€’ 7.6 в€’ 0.192t в€’0.094t < в€’7.6 в€’0.094t в€’7.6 > в€’0.094 в€’0.094 t > 80.9 Men will have a longer average life expectancy than women for birth years starting with 2061. c. i. ii. b. The y-intercept represents the average life expectancy for 1980. Since this value is larger for women than men, she should marry a younger man. From part a., any number less than 12 is 7 not a solution. 41. We have mx < c and x > 2 . This implies that m must be negative so that dividing both sides by m produces: mx < c The average life expectancy for a woman born in 1980 is 77.58 years. She wants to marry a younger man, but each year later decreases her average life expectancy. So, t years after 1980, the average life expectancy of a born in 1980 is 77.58 в€’ t . This value needs to be less than or equal to the life expectancy of the man she will marry. Thus, we have 77.58 в€’ t в‰¤ 0.192t + 69.98 mx c > m m c x> m c = 2 , this means c must also be m negative. Thus, c and m can be any negative c numbers such that = 2. m Since we want 77.58 в€’ t в€’ 77.58 в‰¤ 0.192t + 69.98 в€’ 77.58 в€’t < 0.192t в€’ 7.6 в€’t в€’ 0.192t < 0.192t в€’ 7.6 в€’ 0.192t 43. A solution of the form 2 < x < 5 requires a compound inequality and is not possible for a single linear inequality. в€’1.192t < в€’7.6 в€’1.192t в€’7.6 > в€’1.192 в€’1.192 t > 6.4 45. True. When x = в€’4 , the graph of f is above the graph of g. 47. False. When x = в€’1 , the graph of f is above the graph of g. She must marry a man who is born in 1986 or later. 67 Chapter 3 Review Exercises SSM: Intermediate Algebra y = 2 (1) + 5 49. The graph of f is above the graph of g for all values of x such that x < 2.8 . So, f ( x ) > g ( x ) = 2+5 when x < 2.8 . =7 The solution is (1, 7 ) . 51. Written response. Answers may vary. 4. Solve using elimination. Multiply the first equation by 3 and the second equation by в€’4 , then add the equations. 12 x в€’ 15 y = в€’66 в€’12 x в€’ 8 y = 20 Chapter 3 Review Exercises 3 1. y = в€’ x + 1 2 1 y = xв€’6 4 y в€’4 4 в€’ 23 y = в€’46 y=2 Substitute y = 2 into 3 x + 2 y = в€’5 and solve for x. 3 x + 2 ( 2 ) = в€’5 x в€’4 3 x + 4 = в€’5 (4, в€’5) 3 x = в€’9 The solution is ( 4, в€’5 ) . y = в€’2 ( x в€’ 4 ) 2. 3 x в€’ 5 y = в€’1 в€’5 y = в€’3 x в€’ 1 y x = в€’3 The solution is ( в€’3, 2 ) . = в€’2 x + 8 5. Solve using elimination. Rewrite the second equation so all the variables are on the left side of the equation and the constant is on the right side. в€’2 x + 3 y = 7 3 1 y = x+ 5 5 8 4 x в€’ 6 y = в€’14 Multiply the first equation by 2 and add the equations. в€’4 x + 6 y = 14 4 (3, 2) 4 8 4 x в€’ 6 y = в€’14 x 0 = 0 true This is an identity. The system is dependent. The solution set contains all ordered pairs ( x, y ) The solution is (3, 2 ) . 3. Solve using substitution. Substitute 2 x + 5 for y in the second equation. y = в€’3 x + 10 such that y = 2 x + 5 = в€’3 x + 10 2 7 x+ . 3 3 6. Solve using elimination. Multiply the first equation by вЂ“2 and add the equations. в€’6 x + 14 y = в€’10 2 x + 5 в€’ 5 = в€’3 x + 10 в€’ 5 2 x = в€’3 x + 5 2 x + 3 x = в€’3 x + 5 + 3 x 6 x в€’ 14 y = в€’1 5x = 5 x =1 Substitute this result into the first equation and solve for y. 0 = в€’11 false This is a contradiction. The system is inconsistent. The solution set is the empty set. 68 Chapter 3 Review Exercises SSM: Intermediate Algebra 7. Solve using elimination. Rewrite the second equation so all the variables are on the left side of the equation and the constant is on the right side. в€’4 x в€’ 5 y = 3 10. Solve using elimination. Multiply the first equation by 2 and add the equations. 0.8 x + 0.6 y = 0.8 в€’0.8 x + 1.2 y = 6.4 8 x + 10 y = в€’6 Multiply the first equation by 2 and add the equations. в€’8 x в€’ 10 y = 6 1.8 y = 7.2 y=4 Substitute y = 4 into 0.4 x + 0.3 y = 0.4 and solve for x. 0.4 x + 0.3 ( 4 ) = 0.4 8 x + 10 y = в€’6 0 = 0 true This is an identity. The system is dependent. The solution set contains all ordered pairs ( x, y ) 0.4 x + 1.2 = 0.4 0.4 x = в€’0.8 x = в€’2 The solution is ( в€’2, 4 ) . 4 3 such that y = в€’ x в€’ . 5 5 11. Solve using substitution. 1 Substitute x в€’ 4 for y in the first equation. 2 пЈ«1 пЈ¶ 3 x в€’ 5 пЈ¬ x в€’ 4 пЈ· = 21 2 пЈ пЈё 5 3 x в€’ x + 20 = 21 2 1 x + 20 = 21 2 1 x =1 2 x=2 Substitute x = 2 into the second equation and solve for y. 1 y = (2 ) в€’ 4 2 = 1в€’ 4 = в€’3 The solution is ( 2, в€’3) . 8. Solve using substitution. Substitute 4.2 x в€’ 7.9 for y in the second equation. y = в€’2.8 x + 1.1 4.2 x в€’ 7.9 = в€’2.8 x + 1.1 4.2 x в€’ 7.9 + 7.9 = в€’2.8 x + 1.1 + 7.9 4.2 x = в€’2.8 x + 9 4.2 x + 2.8 x = в€’2.8 x + 9 + 2.8 x 7x = 9 x в‰€ 1.29 Substitute this result into the first equation and solve for y. y = 4.2 (1.29 ) в€’ 7.9 в‰€ в€’2.5 The solution is approximately (1.29, в€’2.5) . 9. Solve using substitution. Substitute 4.9x for y in the second equation and solve for x. в€’3.2 y = x в€’3.2 ( 4.9 x ) = x 12. Solve using elimination. 6 4 xв€’ y =8 5 3 6 8 в€’ x + y = в€’4 5 3 4 y=4 3 y=3 в€’15.68 x = x в€’16.68 x = 0 x=0 Substitute this result into the first equation and solve for y. y = 4.9 (0 ) =0 The solution is (0, 0 ) . Substitute y = 3 into for x. 69 3 2 x в€’ y = 4 and solve 5 3 Chapter 3 Review Exercises SSM: Intermediate Algebra 2 x в€’ 5 y = 15 3 2 x в€’ (3) = 4 5 3 3 xв€’2= 4 5 3 x=6 5 x = 10 The solution is (10,3) . в€’2 x в€’ y = в€’9 в€’ 6y = 6 y = в€’1 Substitute y = в€’1 into 2 x в€’ 5 y = 15 and solve for x. 2 x в€’ 5 ( в€’1) = 15 2 x + 5 = 15 2 x = 10 13. First use the distributive property to simplify each equation. 2 (3 x в€’ 4 ) + 3 ( 2 y в€’ 1) = в€’5 x=5 The solution is (5, в€’1) . 6 x в€’ 8 + 6 y в€’ 3 = в€’5 Substitution: Substitute в€’2 x + 9 for y in the first equation and solve for x. 2 x в€’ 5 y = 15 6 x + 6 y в€’ 11 = в€’5 6x + 6 y = 6 в€’3 ( 2 x + 1) + 4 ( y + 3 ) = в€’7 2 x в€’ 5 ( в€’2 x + 9 ) = 15 в€’6 x в€’ 3 + 4 y + 12 = в€’7 в€’6 x + 4 y + 9 = в€’7 2 x + 10 x в€’ 45 = 15 в€’6 x + 4 y = в€’16 The system is: 6x + 6 y = 6 12 x = 60 x=5 Substitute this result into the second equation and solve for y. y = в€’2 x + 9 в€’6 x + 4 y = в€’16 Solve using elimination. Add the two equations together. 6x + 6 y = 6 = в€’2 (5 ) + 9 = в€’10 + 9 = в€’1 The solution is (5, в€’1) . в€’6 x + 4 y = в€’16 10 y = в€’10 y = в€’1 Substitute y = в€’1 into 6 x + 6 y = 6 and solve for x. 6x + 6 y = 6 Graphically: 2 x в€’ 5 y = 15 y = в€’2 x + 9 в€’5 y = в€’2 x + 15 2 y = xв€’3 5 6 x + 6 ( в€’1) = 6 6x в€’ 6 = 6 y 6 x = 12 x=2 The solution is ( 2, в€’1) . 4 8 14. Elimination: Rewrite the second equation so that all the variables are on the left side of the equation and the constant is on the right side. 2 x в€’ 5 y = 15 в€’4 2 y = x в€’3 5 x (5, в€’1) y = в€’2 x + 9 The solution is (5, в€’1) . 2x + y = 9 Multiply the second equation by в€’1 and add the equations. 70 Chapter 3 Review Exercises SSM: Intermediate Algebra 2 x + 3 y = 19 15. Answers may vary. Possible answers: a. 3 y = в€’2 x + 19 x + 2y = 4 2 19 y = в€’ x+ 3 3 3 x + 6 y = 12 Solving this system, you will encounter an identity such as 0 = 0 . The solution set is the set of ordered pairs ( x, y ) such that 6 x в€’ 4 y = 18 в€’4 y = в€’6 x + 18 y= 3 9 xв€’ 2 2 x + 2y = 4 . b. x + 2y = 4 18. The coordinate for A is (0, 0 ) since it lies at the 2 x + 6 y = 10 Solving this system, you will encounter a contradiction such as 0 = в€’2 . The solution set is the empty set, в€… . c. origin. The coordinate for B is the y-intercept of l 1 , which is (0, 4 ) . The coordinate for C is the point of intersection of l 1 and l 2 . Solve the following system: l 1 : y = 3x + 4 x + y = 10 3 x в€’ 3 y = в€’6 l 2 : 3 y + 2 x = 34 Substitute 3 x + 4 for y in the second equation. 3 (3 x + 4 ) + 2 x = 34 The point ( 4, 6 ) satisfies both equations. 16. The approximate solutions for this system are ( 2.6, 21.9 ) and (0.4,15.1) . Although answers 9 x + 12 + 2 x = 34 11x = 22 may vary, your answers should be close to these points. x=2 Substitute x = 2 into the first equation and solve for y. y = 3 (2 ) + 4 17. Begin by substituting 5 for x and 3 for y in the equations and solve for a and b. 2 (5 ) + 3 ( 3 ) = a 6 (5 ) в€’ 4 (3 ) = b 10 + 9 = a = 6+ 4 30 в€’ 12 = b = 10 The solution is ( 2,10 ) so the coordinate of C is 19 = a 18 = b Substitute 19 for a and 18 for b in the original system. This gives: 2 x + 3 y = 19 6 x в€’ 4 y = 18 Verify: 2 (5 ) + 3 (3 ) = 19 6 (5 ) в€’ 4 (3 ) = 18 10 + 9 = 19 30 в€’ 12 = 18 ( 2,10 ) . The coordinate for D is the point of intersection of l 2 and l 3 . Solve the following system: l 2 : 3 y + 2 x = 34 l 3 : y + 4 x = 28 Multiply the second equation by в€’3 and add the equations. 3 y + 2 x = 34 19 = 19 true 18 = 18 To verify graphically, put each equation in slopeintercept form and then find the intersection point using a graphing utility. в€’3 y в€’ 12 x = в€’84 в€’ 10 x = в€’50 x=5 Substitute x = 5 into 3 y + 2 x = 34 and solve for y. 3 y + 2 (5 ) = 34 3 y = 24 y =8 71 Chapter 3 Review Exercises SSM: Intermediate Algebra y The solution is (5,8 ) so the coordinate of D is y = 3 x в€’6 (5,8) . The coordinate for E is the point of 4 intersection of l 3 and l 4 . Solve the following system. l 3 : y + 4 x = 28 в€’4 l 4 : y = 3 x в€’ 14 Substitute 3 x в€’ 14 in for y in y + 4 x = 28 . y = в€’2 x + 7 (3x в€’ 14 ) + 4 x = 28 20. в€’2 x в‰¤ 18 в€’2 x 18 в‰Ґ в€’2 в€’2 x в‰Ґ в€’9 Interval: [в€’9, в€ћ ) x=6 Substitute x = 6 into y = 3 x в€’ 14 and solve for x. y = 3 ( 6 ) в€’ 14 = 28 в€’ 14 =4 The solution is (6, 4 ) so the coordinate of E is 21. 3 x в€’ 8 в‰¤ 13 3 x в€’ 8 + 8 в‰¤ 13 + 8 l 4 . Let y = 0 in l 4 and solve for x. y = 3 x в€’ 14 3 x в‰¤ 21 3 x 21 в‰¤ 3 3 xв‰¤7 Interval: ( в€’в€ћ, 7 ] 0 = 3 x в€’ 14 3 x = 14 14 3 пЈ« 14 пЈ¶ The coordinate of F is пЈ¬ , 0 пЈ· . пЈ 3 пЈё 4 C (2 , 1 0) D (5 , 8 ) в€’8 x + 5 x > в€’5 x + 15 + 5 x в€’3 x > 15 E (6 , 4 ) F пЈ« 14 пЈ¶ пЈ¬ ,0пЈ· пЈ3 пЈё 8 10 в€’8 x > в€’3 x + 6 в€’ 2 x + 9 в€’8 x > в€’5 x + 15 8 B (0 , 4 ) 7 в€’8 x > в€’3 ( x в€’ 2 ) в€’ 2 x + 9 22. 16 A (0 , 0 ) в€’4 в€’9 (6, 4) . The coordinate of F is the x-intercept of y y = в€’x + 2 The three graphs do not share a common point. Thus, the solution set is the empty set. 7 x = 42 x= x 8 в€’3 x 15 < в€’3 в€’3 x < в€’5 Interval: ( в€’в€ћ, в€’5 ) x 19. Graph each of the equations. y = в€’x + 2 в€’10 y = в€’2 x + 7 y = 3x в€’ 6 72 в€’5 0 Chapter 3 Review Exercises SSM: Intermediate Algebra 23. 4.2 в€’ 3.6 x в‰Ґ 3.9 ( x + 2.1) 26. 4.2 в€’ 3.6 x в‰Ґ 3.9 x + 8.19 4.2 в€’ 3.6 x в€’ 4.2 в‰Ґ 3.9 x + 8.19 в€’ 4.2 в€’3.6 x в‰Ґ 3.9 x + 4.01 в€’3.6 x в€’ 3.9 x в‰Ґ 3.9 x + 4.01 в€’ 3.9 x в€’7.5 x в‰Ґ 3.99 в€’7.5 x 3.99 в‰¤ в€’7.5 в€’7.5 x в‰¤ в€’0.532 Interval: ( в€’в€ћ, в€’0.532] в€’4 в€’ 27. в€’0.532 0 24. в€’5 ( 2 x + 3) в‰Ґ 2 (3 x в€’ 4 ) в€’10 x в€’ 15 в‰Ґ 6 x в€’ 8 в€’10 x в€’ 15 + 15 в‰Ґ 6 x в€’ 8 + 15 в€’10 x в‰Ґ 6 x + 7 в€’10 x в€’ 6 x в‰Ґ 6 x + 7 в€’ 6 x в€’16 x в‰Ґ 7 7 в€’16 x в‰¤ в€’16 в€’16 7 xв‰¤в€’ 16 7пЈ№ пЈ« Interval: пЈ¬ в€’в€ћ, в€’ пЈє 16 пЈ» пЈ в€’4 3 5 в€’ xв‰¤ 4 2 4 3 4 5 в€’ в‹…в€’ x в‰Ґ в€’ в‹… 3 4 3 2 10 xв‰Ґв€’ 3 пЈ® 10 пЈ¶ Interval: пЈЇ в€’ , в€ћ пЈ· пЈ° 3 пЈё в€’ 7 16 0 10 3 в€’2 2 3 x > xв€’ 5 2 2 3 3в€’ x в€’3 > x в€’ в€’3 5 2 2 9 в€’ x > xв€’ 5 2 2 9 в€’ xв€’x > xв€’ в€’x 5 2 7 9 в€’ x>в€’ 5 2 5 7 5 9 в€’ в‹…в€’ x < в€’ в‹…в€’ 7 5 7 2 45 x< 14 45 пЈ¶ пЈ« Interval: пЈ¬ в€’в€ћ, пЈ· 14 пЈ пЈё 3в€’ 0 2 5 7 25. в€’ x+ < 3 2 3 2 5 5 7 5 в€’ x+ в€’ < в€’ 3 2 2 3 2 2 14 15 в€’ x< в€’ 3 6 6 2 1 в€’ x<в€’ 3 6 3 2 3 1 в€’ в‹…в€’ x > в€’ в‹…в€’ 2 3 2 6 1 x> 4 пЈ«1 пЈ¶ Interval: пЈ¬ , в€ћ пЈ· пЈ4 пЈё 0 1 4 2 0 28. a. 45 14 Solve the inequality for x. 7 в€’ 2 (3 x + 5 ) < 4 x + 1 7 в€’ 6 x в€’ 10 < 4 x + 1 в€’6 x в€’ 3 < 4 x + 1 в€’6 x в€’ 3 + 3 < 4 x + 1 + 3 в€’6 x < 4 x + 4 в€’6 x в€’ 4 x < 4 x + 4 в€’ 4 x в€’10 x < 4 4 в€’10 x > в€’10 в€’10 2 x>в€’ 5 Any three numbers that are greater than в€’ 1 are possible solutions. 73 2 5 Chapter 3 Review Exercises b. SSM: Intermediate Algebra From part a., any number less than or equal 2 to в€’ is not a solution. 5 c. 0.22 x + 75 < 0.69 x + 29.95 0.22 x + 75 в€’ 75 < 0.69 x + 29.95 в€’ 75 0.22 x < 0.69 x в€’ 45.05 29. The studentвЂ™s work is incorrect. When dividing both sides of an inequality by a negative number, you must switch the direction of the inequality. 30. 0.22 x в€’ 0.69 x < 0.69 в€’ 45.05 в€’ 0.69 x в€’0.47 x < в€’45.05 в€’0.47 x в€’45.05 > в€’0.47 в€’0.47 x > 95.9 Rent A Wreck will be cheaper for miles driven more than 95.9 miles. f (4) = 1 31. g ( 4 ) = в€’3 32. f ( x ) = 0 when x = 1 . 38. Let P (t ) represent the average price (in dollars) 33. g ( x ) = 0 when x = в€’5 . 34. of a home in a community and S (t ) represent the amount of money (in dollars) a family has saved at t years since 2000. f ( x ) = g ( x ) when x = в€’2 . 35. The graph of f ( x ) is above the graph of g ( x ) Since the average price of a home increases at a constant $9000 per year, the function P is linear and its slope is 9000. the P-intercept is (0, 250000) since the price of a home is for values of x that are greater than в€’2 . Thus, f ( x ) > g ( x ) when x > в€’2 . $250,000 in year t = 0 . So, an equation for P is : P (t ) = 9000t + 250000 . cв€’b =5, a a < 0 , and b > c . For example, a = в€’1 , b = 6 , and c = 1 will work. 36. Any values for a, b, and c so that 37. a. b. R <U Similar work in finding the equation for the function S gives: S (t ) = 4800t + 12000 . (The Since U-HaulвЂ™s charg e increases at a constant rate of $0.69 per mile, the equation is linear with slope 0.69 . The U-intercept is 29.95 since U-Haul charges a flat fee of $29.95. An equation for U-HaulвЂ™s charge is: U ( x ) = 0.69 x + 29.95 . After similar work for Rent A Wreck, an equation for Rent A WreckвЂ™s charge is: R ( x ) = 0.22 x + 75.00 . slope of S is 4800 since the family plans to save $400 each month which is $4800 each year.) In order to predict when the family will be able to pay a 10% down payment on an averagepriced house, solve the following system for t when y = 0.1( P (t )) = 0.1(9000t + 250000 ) y = S (t ) = 4800t + 12000 Substitute 0.1(9000t + 250000 ) for y in the U (x) = R (x) second equation and solve for t. 0.1(9000t + 250000 ) = 4800t + 12000 0.69 x + 29.95 = 0.22 x + 75 0.69 x + 29.95 в€’ 29.95 = 0.22 x + 75 в€’ 29.95 900t + 25000 = 4800t + 12000 0.69 x = 0.22 x + 45.05 в€’3900t = в€’13000 0.69 x в€’ 0.22 x = 0.22 x + 45.05 в€’ 0.22 x t в‰€ 3.33 The family will be able to pay a 10% down payment in 3 years (2003). 0.47 x = 45.05 x в‰€ 95.9 The two charges will be the same when the number of miles driven is roughly 95.9 miles. 74 Chapter 3 Test SSM: Intermediate Algebra 39. a. 40. a. Start by plotting the data sets, then find the regression line for each region. Solve the system y = H (t ) = 21.14t + 100.00 y = A (t ) = в€’15.63t + 220.68 Solve using substitution. Substitute 21.14t + 100.00 for y in the second equation and solve for t. y = в€’15.63t + 220.68 North America: 21.14t + 100.00 = в€’15.63t + 220.68 n (t ) = 0.52t + 35.04 21.14t + 100 в€’ 100 = в€’15.63t + 220.68 в€’ 100 21.14t = в€’15.63t + 120.68 Far East: 21.14t + 15.63t = в€’15.63t + 120.68 + 15.63t 36.77t = 120.68 t в‰€ 3.3 Substitute t = 3.3 into y = 21.14t + 100.00 and solve for y. y = 21.14t + 100.00 f (t ) = 1.50t + 13.89 b. Both slopes are positive indicating an increase in consumption each year. The slope for the Far East is larger than the slope for North America. The consumption of petroleum is increasing at a faster rate in the Far East than in North America. c. n (t ) = f (t ) = 21.14 (3.3) + 100.00 в‰€ 169.8 According to the models, there were the same number of jobs in Hollywood as in the aerospace industry in the year 1993 ( t = 3 ). b. inequality is true when t > 3.3 . Therefore, there are more jobs in Hollywood than in the aerospace industry in years after 1993 ( t > 3 ). 0.52t + 35.04 = 1.50t + 13.89 0.52t + 35.04 в€’ 35.04 = 1.50t + 13.89 в€’ 35.04 0.52t = 1.50t в€’ 21.15 0.52t в€’ 1.50t = 1.50t в€’ 21.15 в€’ 1.50t в€’0.98t = в€’21.15 c. t в‰€ 21.6 The consumption of petroleum was the same for the Far East and North America in 2002. d. Solve for t when H (t ) > A (t ) . This n ( t ) < f (t ) Hollywood has experienced the most change in terms of employment. This is evident from the greater difference in the number of jobs in Hollywood compared to the difference in the number of jobs in the aerospace industry between the years 1992 and 1996. Chapter 3 Test 0.52t + 35.04 < 1.50t + 13.89 0.52t + 35.04 в€’ 35.04 < 1.50t + 13.89 в€’ 35.04 1. Solve using substitution. Substitute 3 x в€’ 1 for y in the second equation and solve for x. 3 x в€’ 2 y = в€’1 0.52t < 1.50t в€’ 21.15 0.52t в€’ 1.50t < 1.50t в€’ 21.15 в€’ 1.50t в€’0.98t < в€’21.15 3 x в€’ 2 (3 x в€’ 1) = в€’1 в€’0.98t в€’21.15 > в€’0.98 в€’0.98 t > 21.6 The consumption of petroleum in North America will be less than the consumption of petroleum in the Far East for years after 2002. 3 x в€’ 6 x + 2 = в€’1 в€’3 x + 2 = в€’1 в€’3 x = в€’3 (cont.) 75 x =1 Chapter 3 Test SSM: Intermediate Algebra Substitute x = 1 into the first equation and solve for y. y = 3 (1) в€’ 1 3 1 (5 ) + 5 4 1 3+ 4 1 4 1 4 = 3 в€’1 =2 The solution is (1, 2 ) . 2. Solve using elimination. First write the second equation so that all the variables are on the left side of the equation and the constant is on the right side. 2x в€’ 5 y = 3 y =1 y =1 y = 1в€’ 3 y = в€’2 y = в€’8 The solution is (5, в€’8) . 5. First use the distributive property to simplify each equation. в€’4 ( x + 2 ) + 3 ( 2 y в€’ 1) = 21 6 x в€’ 15 y = 9 Multiply the first equation by в€’3 and add the equations. в€’6 x + 15 y = в€’9 в€’4 x в€’ 8 + 6 y в€’ 3 = 21 в€’4 x + 6 y в€’ 11 = 21 6 x в€’ 15 y = 9 в€’4 x + 6 y = 32 2 x в€’ 3 y = в€’16 0 = 0 true This is an identity. The system is dependent. The solution set is the set of ordered pairs ( x, y ) 5 (3 x в€’ 2 ) в€’ ( 4 y + 3) = в€’59 such that 2 x в€’ 5 y = 3 . 15 x в€’ 10 в€’ 4 y в€’ 3 = в€’59 15 x в€’ 4 y в€’ 13 = в€’59 3. Solve using elimination. Multiply the first equation by 3 and the second equation by в€’2 , then add the equations. 12 x в€’ 18 y = 15 15 x в€’ 4 y = в€’46 The system can be rewritten as: 2 x в€’ 3 y = в€’16 15 x в€’ 4 y = в€’46 Solve using elimination. Multiply the first equation by 4 and the second equation by в€’3 , then add the equations. 8 x в€’ 12 y = в€’64 в€’12 x + 18 y = 4 0 = 19 false This is a contradiction. The system is inconsistent. The solution set is the empty set. 4. Solve using elimination. Multiply the second equation by 3 and add the equations. 2 3 xв€’ y =8 5 4 9 3 x+ y =3 5 4 11 x = 11 5 x=5 9 3 Substitute x = 5 into x + y = 3 and solve 5 4 for y. в€’45 x + 12 y = 138 в€’ 37 x = 74 x = в€’2 Substitute x = в€’2 into 2 x в€’ 3 y = в€’16 and solve for y. 2 ( в€’2 ) в€’ 3 y = в€’16 в€’4 в€’ 3 y = в€’16 в€’3 y = в€’12 y=4 The solution is ( в€’2, 4 ) . 6. Answers may vary. One possible answer: x в€’ 2y =1 3 x + y = 17 76 Chapter 3 Test SSM: Intermediate Algebra 7. Elimination: First rewrite the second equation so that all the variables are on the left side and the constant is on the right side. 4x в€’ 3y = 9 8. If the solution set is the empty set, the system is two parallel lines. Therefore, m in y = mx + b is 5 since this is the slope in y = 5 x в€’ 13 and parallel lines have the same slope. The y-intercept in y = mx + b is any number other than в€’13 since y = 5 x в€’ 13 and в€’2 x + y = в€’5 Multiply the second equation by 2 and add the equations. 4x в€’ 3 y = 9 в€’4 x + 2 y = в€’10 y = mx + b intersect the y-axis at different points ( b в‰ в€’13 ). 9. в€’ y = в€’1 2 в€’ 10 x в€’ 2 в‰Ґ 3 x + 14 в€’ 2 y =1 Substitute y = 1 into 4 x в€’ 3 y = 9 and solve for x. 4 x в€’ 3 (1) = 9 в€’10 x в‰Ґ 3 x + 12 в€’10 x в€’ 3 x в‰Ґ 3 x + 12 в€’ 3 x в€’13 x в‰Ґ 12 в€’13 x 12 в‰¤ в€’13 в€’13 12 xв‰¤в€’ 13 12 пЈ№ пЈ« Interval: пЈ¬ в€’в€ћ, в€’ пЈє 13 пЈ» пЈ 4x в€’ 3 = 9 4 x = 12 x=3 The solution is (3,1) . Substitution: Substitute 2 x в€’ 5 for y in the first equation. 4x в€’ 3 y = 9 в€’2 3 x + 13 < 5 x в€’ 10 3 x + 13 в€’ 13 < 5 x в€’ 10 в€’ 13 в€’2 x + 15 = 9 в€’2 x = в€’6 3 x < 5 x в€’ 23 x=3 Substitute x = 3 into y = 2 x в€’ 5 and solve for y. 3 x в€’ 5 x < 5 x в€’ 23 в€’ 5 x в€’2 x < в€’23 y = 2 (3 ) в€’ 5 в€’2 x в€’23 > в€’2 в€’2 23 x> 2 пЈ« 23 пЈ¶ Interval: пЈ¬ , в€ћ пЈ· пЈ 2 пЈё =1 The solution is (3,1) . Graphically: 4x в€’ 3 y = 9 y = 2x в€’ 5 4 y = xв€’3 3 0 y = 2 x в€’5 4 (3, 1) y= в€’2 4 2 3 x + 12 + 1 < 5 x в€’ 10 4 x в€’ 6 x + 15 = 9 в€’4 0 10. 3 ( x + 4 ) + 1 < 5 ( x в€’ 2 ) 4 x в€’ 3 (2 x в€’ 5) = 9 y 2 в€’ 10 x в‰Ґ 3 x + 14 x 4 x в€’3 3 The solution is (3,1) . 77 6 12 Chapter 3 Test SSM: Intermediate Algebra 2.6 ( x в€’ 3.1) > 4.7 x в€’ 5.9 11. 18. a. 2.6 x в€’ 8.06 > 4.7 x в€’ 5.9 2.6 x в€’ 8.06 + 8.06 > 4.7 x в€’ 5.9 + 8.06 3 x в€’ 11 + 11 < 7 в€’ 6 x + 11 2.6 x > 4.7 x + 2.16 3 x < 18 в€’ 6 x 3 x + 6 x < 18 в€’ 6 x + 6 x 2.6 x в€’ 4.7 x > 4.7 x + 2.16 в€’ 4.7 x в€’2.1x > 2.16 в€’2.1x 2.16 < в€’2.1 в€’2.1 x < в€’1.03 Interval: ( в€’в€ћ, в€’1.03) в€’2 в€’1 9 x < 18 9 x 18 < 9 9 x<2 Any number less than 2 will satisfy the inequality. 0 b. 5 1 7 12. в€’ x+ в‰¤ x 3 6 4 5 1 5 7 5 в€’ x+ + x в‰¤ x+ x 3 6 3 4 3 1 41 x в‰¤ 6 12 12 1 12 41 в‹… в‰¤ в‹… x 41 6 41 12 2 в‰¤x 41 2 xв‰Ґ 41 пЈ®2 пЈ¶ Interval: пЈЇ , в€ћ пЈ· пЈ° 41 пЈё 0 13. 1 19. a. Answers may vary. From part a., any number greater than or equal to 2 does not satisfy the inequality. In the year 2000, t = 100 . Compare A (100 ) and B (100 ) . A (100 ) = в€’0.12 (100 ) + 31.75 = 19.75 B (100 ) = 0.08 (100 ) + 9.43 = 17.43 Since 19.75 в€’ 17.43 = 2.32 , College A had an enrollment of 2320 more than College B. b. The slopes of A and B are в€’0.12 and 0.08 , respectively. The enrollment at college A is decreasing by 120 students per year while the enrollment at college B is increasing by 80 students per year. c. Solve for t when A (t ) = B (t ) . в€’0.12t + 31.75 = 0.08t + 9.43 2 f (5 ) = 3 14. g ( x ) = 3 when x = в€’4 . 15. Answers may vary. Solve the inequality for x. 3 x в€’ 11 < 7 в€’ 6 x в€’0.12t + 31.75 в€’ 31.75 = 0.08t + 9.43 в€’ 31.75 в€’0.12t = 0.08t в€’ 22.32 f ( x ) = g ( x ) when x = 2 . в€’0.12t в€’ 0.08t = 0.08t в€’ 22.32 в€’ 0.08t в€’0.2t = в€’22.32 t = 111.6 16. The graph of f is below the graph of g for x < 2 and the two graphs are the same for x = 2 . Therefore, f ( x ) в‰¤ g ( x ) when x в‰¤ 2 . According to the models, the colleges will have the same enrollment in the year 2012 ( t = 112 ). 17. The graph of f is lower than the graph of g for all values of x that are greater than 13. Thus, f ( x ) < g ( x ) for x > 13 . 78 Cumulative Review Chapters 1-3 SSM: Intermediate Algebra d. Solve for t when A (t ) > B (t ) . d. в€’0.12t + 31.75 > 0.08t + 9.43 в€’0.12t + 31.75 в€’ 31.75 > 0.08t + 9.43 в€’ 31.75 в€’0.12t > 0.08t в€’ 22.32 в€’0.12t в€’ 0.08t > 0.08t в€’ 22.32 в€’ 0.08t в€’0.2t > в€’22.32 в€’0.2t в€’22.32 < в€’0.2 в€’0.2 t < 111.6 According to the models, college A will have a larger enrollment than college B for years before 2012 ( t < 112 ). 20. a. The slope for the two-year colleges is larger than for the four-year colleges. This means that the number of two-year colleges is growing at a faster rate. However, the slopes are not very different so it will take many years for the number of two-year colleges to overcome the difference in numbers. Cumulative Review Chapters 1-3 1. T O rig in a l te m p . Start by plotting each set of data, then find the regression lines for the data. R e frig e ra to r te m p . 2-Year Colleges: t 2 2. y = в€’ x + 4 3 y f ( x ) = 25.59t + 929.18 4 4-Year Colleges: в€’4 4 x в€’4 3. 5 x в€’ 3 y = 15 g (t ) = 23.14t + 1685.30 b. в€’3 y = в€’5 x + 15 5 y = xв€’5 3 In 2008, t = 38 . f (38 ) = 25.59 (38 ) + 929.18 y = 1901.6 g (38 ) = 23.14 (38 ) + 1685.30 4 = 2564.62 The models predict that there will be a total of 4466 two-year and four-year colleges in 2008. c. в€’4 4 в€’4 Solve for t when f (t ) = g (t ) . 25.59t + 929.18 = 23.14t + 1685.30 25.59t = 23.14t + 756.12 2.45t = 756.12 t в‰€ 308.62 The number of two-year and four-year colleges will be the same in 2279 ( t = 309 ). 79 x Cumulative Review Chapters 1-3 SSM: Intermediate Algebra 4. 3 ( x в€’ 4 ) = в€’2 ( y + 5 ) + 4 8. First find the slope: 3 в€’ ( в€’2 ) 3+ 2 5 m= = = в€’2 в€’ ( в€’5 ) в€’2 + 5 3 3 x в€’ 12 = в€’2 y в€’ 10 + 4 3 x в€’ 12 = в€’2 y в€’ 6 Next use the slope and either point (we will use the first point) to find b. 5 m = ; ( в€’5, в€’2 ) 3 y = mx + b 2 y = в€’3 x + 6 y 3 y = в€’ x+3 2 3 5 ( в€’5 ) + b 3 25 в€’2 = в€’ + b 3 19 b= 3 The equation of the line is: 5 19 y = x+ 3 3 в€’2 = 2 1 в€’1 1 в€’1 2 3 x 5. x + 2 = 0 x = в€’2 y 9. 2 x в€’ 5 y = 20 4 в€’4 4 в€’5 y = в€’2 x + 20 2 y = xв€’4 5 x в€’4 6. m = = Since the slope of the given line is 2 , our line 5 5 since perpendicular lines 2 have opposite-reciprocal slopes. 5 m = в€’ ; ( в€’5,3) 2 y = mx + b y2 в€’ y1 x2 в€’ x1 must have slope в€’ в€’1 в€’ 2 3 в€’ ( в€’4 ) в€’3 7 3 =в€’ 7 = 5 ( в€’5 ) + b 2 25 3= +b 2 19 b=в€’ 2 The equation of the line is: 5 19 y = в€’ xв€’ or y = в€’2.5 x в€’ 9.5 2 2 3=в€’ 3 7. m = в€’ ; ( 2, в€’3) 5 y = mx + b 3 (2 ) + b 5 6 в€’3 = в€’ + b 5 9 b=в€’ 5 The equation of the line is: 3 9 or y = в€’0.6 x в€’ 1.8 y = в€’ xв€’ 5 5 в€’3 = в€’ 10. Answers may vary. The lines y = 2 x + 3 and y = 2 x + 3.1 are parallel. A line between these lines and parallel to both is y = 2 x + 3.05 . We can select any three points on this line. Some possible points: (0,3.05) , (1,5.05) , and ( 2, 7.05) 80 Cumulative Review Chapters 1-3 SSM: Intermediate Algebra 11. Determine the slopes from the given data and use them to obtain the remaining values. k (x) : f (x) : g (x) : h (x) : m= 58 в€’ 97 в€’39 = = в€’13 3в€’0 3 Equation 1 x 12. f (x) Equation 2 m= 43 в€’ 4 39 = = 13 9в€’6 3 m= Equation 3 x g (x) x в€’22 в€’ 23 в€’45 = = в€’9 6 в€’1 5 15. в€’16 в€’ ( в€’28 ) 14 в€’ 10 = 12 =3 4 Equation 4 h (x) x k (x) 0 97 4 -22 1 23 10 -28 1 84 5 -9 2 14 11 -25 2 71 6 4 3 5 12 -22 3 58 7 17 4 -4 13 -19 4 45 8 30 5 -13 14 -16 5 32 9 43 6 -22 15 -13 f (5) = 32 17. The x-intercept is found by solving f ( x ) = 0 . 3 0= в€’ x+7 2 13. g ( x ) = 30 when x = 8 . 14. m= 3 x=7 2 2 3 2 в‹… x = в‹…7 3 2 3 14 x= 3 3 f (x) = в€’ x + 7 2 3 f ( в€’4 ) = в€’ ( в€’4 ) + 7 2 = 6+7 = 13 пЈ« 14 пЈ¶ The x-intercept is пЈ¬ , 0 пЈ· . пЈ 3 пЈё 3 f (x) = в€’ x + 7 2 3 в€’4 = в€’ x + 7 2 3 в€’11 = в€’ x 2 2 2 3 в€’ в‹… в€’11 = в€’ в‹… в€’ x 3 3 2 22 x= 3 18. Since the equation is in the form y = mx + b , the y-intercept is (0,b ) or (0, 7 ) . 19. 3 f (x) = в€’ x + 7 2 y 8 4 16. Since the equation is in the form y = mx + b , the slope is the coefficient on x. 3 m=в€’ 2 4 20. 81 f ( в€’6 ) = в€’3 . 8 x Cumulative Review Chapters 1-3 SSM: Intermediate Algebra 21. g ( x ) = 0 when x = в€’6 . 28. 22. The y-intercept of f is (0,1) . 23. To find the slope of g, pick two points on the graph. Two possible points are the intercepts ( в€’6, 0 ) and (0, в€’2 ) . m= в€’2 в€’ 0 в€’2 1 = =в€’ 0 в€’ ( в€’6 ) 6 3 1 The slope of g is в€’ . 3 24. To find an equation for f, first determine the slope by finding two points on the graph. Two possible points are (0,1) and (3,3) . Use the 29. Yes, the graph is that of a function. The graph passes the vertical line test. points to find the slope of the line. 3 в€’1 2 m= = 3в€’ 0 3 Since the y-intercept is (0,1) , the equation of the line is y = 25. 7 1 1 3 в€’ x= + x 8 4 2 8 7 1 7 1 3 7 в€’ xв€’ = + xв€’ 8 4 8 2 8 8 1 3 3 в€’ x = xв€’ 4 8 8 1 3 3 3 3 в€’ xв€’ x = xв€’ в€’ x 4 8 8 8 8 5 3 в€’ x=в€’ 8 8 3 x= 5 30. x = y y 8 2 x +1. 3 4 f ( x ) = g ( x ) when x = в€’3 . 2 x The relation is a function. It passes the vertical line test. For each input value, there is only one output value. 26. The graph of f is at or below the graph of g for all values of x that are less than or equal to в€’3 . Therefore, f ( x ) в‰¤ g ( x ) when x в‰¤ в€’3 . 31. Answers may vary. One possible answer: Equation: y = 3x + 4 27. в€’5 x в€’ 3 ( 2 x + 4 ) = 8 в€’ 2 x в€’5 x в€’ 6 x в€’ 12 = 8 в€’ 2 x в€’11x в€’ 12 = 8 в€’ 2 x Table: x y в€’2 в€’2 в€’1 1 0 4 1 7 2 10 в€’11x в€’ 12 + 12 = 8 в€’ 2 x + 12 в€’11x = 20 в€’ 2 x в€’11x + 2 x = 20 в€’ 2 x + 2 x в€’9 x = 20 x=в€’ 4 20 9 Graph: y 4 в€’4 4 в€’4 82 x Cumulative Review Chapters 1-3 SSM: Intermediate Algebra 32. Solve using elimination. Multiply the first equation by 3 and the second equation by 4, then add the equations. 6 x + 12 y = в€’24 Substitution: x + 3y = 9 x = в€’3 y + 9 Substitute в€’3 y + 9 for x in the first equation and solve for y. 2 ( в€’3 y + 9 ) в€’ 5 y = в€’4 20 x в€’ 12 y = 76 26 x = 52 x=2 Substitute x = 2 into 2 x + 4 y = в€’8 and solve for y. 2 ( 2 ) + 4 y = в€’8 в€’6 y + 18 в€’ 5 y = в€’4 в€’11 y + 18 = в€’4 в€’11 y = в€’22 y=2 Substitute y = 2 in x = в€’3 y + 9 and solve for x. 4 + 4 y = в€’8 4 y = в€’12 x = в€’3 ( 2 ) + 9 y = в€’3 = в€’6 + 9 The solution is ( 2, в€’3) . =3 The solution is (3, 2 ) . 33. Solve using substitution. 3 Substitute x в€’ 2 for y in the first equation. 7 3 x в€’ 7 y = 14 Graphically: 2 x в€’ 5 y = в€’4 в€’5 y = в€’2 x в€’ 4 пЈ«3 пЈ¶ 3 x в€’ 7 пЈ¬ x в€’ 2 пЈ· = 14 7 пЈ пЈё 3 x в€’ 3 x + 14 = 14 3y = в€’x + 9 2 4 y = x+ 5 5 1 y = в€’ x+3 3 y 14 = 14 true This is an identity. The system is dependent. The solution set consists of all ordered pairs ( x, y ) such that y = x + 3y = 9 4 3 xв€’2. 7 (3, 2) в€’2 4 x в€’4 34. Elimination: Multiply the second equation by в€’2 and add the equations. 2 x в€’ 5 y = в€’4 в€’2 x в€’ 6 y = в€’18 35. в€’2 ( 4 x + 5 ) в‰Ґ 3 ( x в€’ 7 ) + 1 в€’8 x в€’ 10 в‰Ґ 3 x в€’ 21 + 1 в€’8 x в€’ 10 в‰Ґ 3 x в€’ 20 в€’8 x в€’ 10 + 10 в‰Ґ 3x в€’ 20 + 10 в€’ 11 y = в€’22 в€’8 x в‰Ґ 3 x в€’ 10 y=2 Substitute y = 2 into 2 x в€’ 5 y = в€’4 and solve for x. 2 x в€’ 5 ( 2 ) = в€’4 в€’8 x в€’ 3 x в‰Ґ 3x в€’ 10 в€’ 3x в€’11x в‰Ґ в€’10 в€’11x в€’10 в‰¤ в€’11 в€’11 10 xв‰¤ 11 10 пЈ№ пЈ« Interval: пЈ¬ в€’в€ћ, пЈє 11 пЈ» пЈ 2 x в€’ 10 = в€’4 2x = 6 x=3 The solution is (3, 2 ) . 0 83 10 11 2 Cumulative Review Chapters 1-3 36. SSM: Intermediate Algebra d. 4 1 5 xв€’ < 2в€’ x 3 6 6 4 1 1 5 1 xв€’ + < 2в€’ x+ 3 6 6 6 6 4 13 5 x< в€’ x 3 6 6 4 5 13 5 5 x+ x < в€’ x+ x 3 6 6 6 6 13 13 x< 6 6 6 13 6 13 в‹… x< в‹… 13 6 13 6 x <1 Interval: ( в€’в€ћ,1) в€’3 x + 6 = 4 x в€’ 1 в€’3 x = 4 x в€’ 7 в€’7 x = в€’7 x =1 Substitute this result into the first equation and solve for y. y = в€’3 x + 6 y = в€’3 (1) + 6 =3 The solution is (1,3) . 39. Answers may vary. Possible answers: 1 0 Solve using substitution. Substitute в€’3 x + 6 for y in the second equation and solve for x. y = 4x в€’ 1 a. 37. Decrease m to make the slope more negative. Increase b to raise the y-intercept. For example: 3x + 2 = 8 2x = 6 y x=3 3 b. 2 y = в€’ x+2 3 y x 38. a. 4 в€’3 x + 6 = 0 в€’3 x = в€’6 x=2 b. в€’4 4 в€’3 x + 6 < 0 x в€’4 в€’3 x < в€’6 в€’3 x в€’6 > в€’3 в€’3 x>2 Interval: ( 2, в€ћ ) c. y = 2x в€’1 y = в€’x + 2 y y = 2 x в€’1 4 2 c. (1, 1) в€’4 y 4 в€’4 4 в€’2 в€’2 d. 4 y = в€’x + 2 4 x + 2 в‰¤ 10 4x в‰¤ 8 x xв‰¤2 0 84 x 2 Cumulative Review Chapters 1-3 SSM: Intermediate Algebra 40. a. The grass height is declining at a constant rate. Thus, h (t ) is linear and has the form e. h (t ) = m в‹… t + b . Selecting two ordered pairs allows us to find 1пЈ¶ пЈ« пЈ« 1пЈ¶ the slope. We will use пЈ¬ 55, пЈ· and пЈ¬ 95, пЈ· . 4 пЈ пЈё пЈ 8пЈё 1 1 в€’ 1 в€’1/ 8 m= 8 4 = =в€’ 95 в€’ 55 40 320 Using the slope and the first point, we can find the y-intercept. y = mx + b 1 1 =в€’ (55) + b 4 320 1 11 = в€’ +b 4 64 27 b= 64 1 27 Therefore, h (t ) = в€’ t+ . 320 64 b. c. d. 1 27 t+ 320 64 1 27 t= 320 64 t = 135 The t-intercept is (135, 0 ) . 0=в€’ The putting greens will have no grass in 2035. 41. a. Since the number of bicyclists younger than 13 hit by motorists declines at a constant rate, f (t ) is linear. The slope is в€’12 since the number hit decreases by 12 each year. The n-intercept is (0, 446 ) since 446 bicyclists younger than 13 were hit by motorists in 1975. Thus, an equation for f (t ) is: f (t ) = в€’12t + 446 1 27 (105 ) + в‰€ 0.094 320 64 The grass on the green will have a height of about 0.094 inches in 2005. h (105 ) = в€’ b. (see a.) The slope of f is в€’12 . The number of bicyclists younger than 13 hit by motorists decreases by 12 each year. c. (see a.) The n-intercept is (0, 446 ) . There were 466 bicyclists younger than 13 hit by motorists in 1975. 1 1 27 t+ =в€’ 16 320 64 23 1 t в€’ =в€’ 64 320 t = 115 The grass on the green will have a height 1 that is of an inch in 2015. 16 d. Find the t-intercept by solving f (t ) = 0 . 0 = в€’12t + 446 12t = 446 t в‰€ 37.17 The t-intercept is (37.17,0 ) . The number of bicyclists younger than 13 hit by motorists will decrease to 0 in 2012. 42. a. 1 1 в€’ 1 в€’1/ 8 m= 8 4 = =в€’ 95 в€’ 55 40 320 1 The slope of h is в€’ . The height of the 320 grass on the green gets shorter each year by 1 1 of an inch. Note: в‰€ 0.0031 320 320 Since Company AвЂ™s sales increase by a constant 1.3 million dollars each year, the slope for A (t ) is 1.3. The A-intercept is (0,9.5) since the total sales are 9.5 million dollars in year t = 0 . Therefore, the equation for A (t ) is A (t ) = 1.3t + 9.5 . Similar work for B (t ) gives the equation B (t ) = 1.8t + 5.2 . 85 Cumulative Review Chapters 1-3 b. M illio n s o f d olla rs y 30 О’ SSM: Intermediate Algebra 43. a. О‘ 24 Women: 18 12 6 t 3 6 9 Ye ars W (t ) = в€’0.064t + 27 12 15 The intersection of A and B is (8.6, 20.68 ) . Men: According to the sketch, sales at the companies will both equal $20.68 million dollars in 2006 ( t = 9 ). c. Solve for t when A (t ) = B (t ) . M (t ) = в€’0.029t + 22.10 A (t ) = B (t ) 1.3t + 9.5 = 1.8t + 5.2 b. 1.3t + 9.5 в€’ 9.5 = 1.8t + 5.2 в€’ 9.5 W (107 ) = в€’0.064 (107 ) + 27 = 20.15 M (107 ) = в€’0.029 (107 ) + 22.10 1.3t = 1.8t в€’ 4.3 1.3t в€’ 1.8t = 1.8t в€’ 4.3 в€’ 1.8t = 19 The models predict that 200-Meter Run record time in 2007 will be 20.15 seconds for women and 19 seconds for men. в€’0.5t = в€’4.3 t = 8.6 Substitute 8.6 for t in A (t ) = 1.3t + 9.5 . A (8.6 ) = 1.3 (8.6 ) + 9.5 = 20.68 In 2009 ( t = 9 ), the sales at both companies will equal $20.68 million dollars. This is the same result as in part b. d. Start by plotting each set of data and then find the regression lines. Solve the inequality A (t ) < B (t ) . c. The absolute value of the slope for the women is larger than for the men. This means that the record time for women is decreasing at a faster rate than for men. d. Since the slopes are not equal, the graphs of the two equations will cross eventually. The record times are getting closer which indicates that the times will eventually be equal sometime in the future. e. Solve for t in W (t ) = M (t ) . A (t ) < B (t ) 1.3t + 9.5 < 1.8t + 5.2 1.3t + 9.5 в€’ 9.5 < 1.8t + 5.2 в€’ 9.5 1.3t < 1.8t в€’ 4.3 1.3t в€’ 1.8t < 1.8t в€’ 4.3 в€’ 1.8t W ( t ) = M (t ) в€’0.5t < в€’4.3 в€’0.064t + 27 = в€’0.029t + 22.1 в€’0.5t в€’4.3 > в€’0.5 в€’0.5 t > 8.6 According to the models, Company AвЂ™s sales will be less than Company BвЂ™s sales after 2009 ( t = 9 ). в€’0.064t = в€’0.029t в€’ 4.9 в€’0.035t = в€’4.9 t = 140 The models predict that the record times will be equal in 2040. 86 Cumulative Review Chapters 1-3 SSM: Intermediate Algebra W ( t ) > M (t ) f. в€’0.064t + 27 > в€’0.029t + 22.1 в€’0.064t > в€’0.029t в€’ 4.9 в€’0.035t > в€’4.9 в€’0.035t в€’4.9 < в€’0.035 в€’0.035 t < 140 WomenвЂ™s record times are greater than menвЂ™s record times for years before 2040. g. Women: 0 = в€’0.064t + 27 0.064t = 27 t в‰€ 421.88 Men: 0 = в€’0.029t + 22.1 0.029t = 22.1 t в‰€ 762.07 The t-intercept for women is ( 421.88, 0 ) and the t-intercept for men is (762.07, 0 ) . According to the models, the womenвЂ™s record time will be 0 seconds in 2322 and the menвЂ™s record time will be 0 seconds in 2662. This is not feasible so model breakdown has occurred. h. Answers may vary. One possible answer: r W M t 87

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