51 Chapter 3 Systems of Linear EquationsHomework 3.1 1. 4yx422(1,4)2 2 y x = +3 7 y x = + Verify that( ) 1, 4satisfies both equations. ( )2 24 2 1 24 2 24 4truey x = += += +=3 74 3(1) 74 3 74 4 truey x = += += += 3. 888yx8(6, 6)183y x = +132y x = + Verify that( ) 6, 6 satisfies both equations. ( )13216 6 326 3 36 6truey x = += += +=

( )18316 6 836 2 86 6 truey x = += += += 5.Write( ) 3 1 y x = in slope-intercept form. ( ) 3 13 3y xy x= = 888yx8(2, 3)3 3 y x = 2 7 y x = + Verify that( ) 2, 3satisfies both equations. ( )( )( )3 13 3 2 13 3 13 3 truey x = = == ( )2 73 2 2 73 4 73 3 truey x = += += += 7.Write4 2 6 x y + =in slope-intercept form. 4 2 62 4 62 3x yy xy+ == += + 624yx(2, 1)2 3 y x = +3 7 y x = 2 23 7y xy x= += + Verify that( ) 2, 1 satisfies both equations. ( ) ( )4 2 64 2 2 1 68 2 66 6truex y + =+ = == ( ) ( )2 32 2 1 34 1 33 3truex y + =+ = == Homework 3.1SSM: Intermediate Algebra 52 9.Write both equations in slope-intercept form. ( ) ( ) 5 2 21 2 35 10 21 2 65 2 25255y xy xy xy x = + = = += +( ) 3 1 83 3 83 5y xy xy x= += += + 4 4yx4(0, 5)255y x = +3 5 y x = + Verify that( ) 0, 5satisfies both equations. ( ) ( )( ) ( )( ) ( )5 2 21 2 35 5 2 21 2 0 35 3 21 2 315 15 truey x = + = += = ( )( )( )3 1 85 3 0 1 85 3 1 85 5 truey x = += += += 11.Write both equations in slope-intercept form. 1 112 2222x yx yy xy x = = = +=

1 124 22 82 8142x yx yy xy x+ =+ == += + 444yx4(4, 2)142y x = +2 y x = Verify that( ) 4, 2satisfies both equations. ( ) ( )1 112 21 14 2 12 22 1 11 1truex y = = =

( ) ( )1 124 21 14 2 24 21 1 22 2 truex y + =+ =+ == 13.

The solution is( ) 1, 4 . 15.

The solution is( ) 480, 460 . 17.Write4 2 2 x y =in slope-intercept form. 4 2 22 4 22 1x yy xy x = = += The equations are identical.

The system is dependent. The solution set contains all ordered pairs( ) , xysuch that 2 1 y x = . 19.Write0.2 3 y x =in slope-intercept form. 0.2 30.2 35 15y xy xy x == += +

The system is inconsistent. The solution set is the empty set,. SSM: Intermediate AlgebraHomework 3.1 53 21.Write both equations in slope-intercept form. 1 112 2222x yx yy xy x = = = +=

1 223 32 62 6132x yx yy xy x+ =+ == += +

The solution is roughly( ) 3.33,1.33 . 23.a.In 2002,32 t = . ( ) ( ) 32 0.19 32 43.7337.65W = += ( ) ( ) 32 0.16 32 39.9034.78M = += The womens time is 37.65 seconds and the mens time is 34.78 seconds. The error for the womens time estimate is 0.275 seconds and the error for the mens time estimate is 0.165 seconds. b.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.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. d. Men and women will have a winning time of 19.47 seconds in the year 2098. 25.a.Start by plotting the data. Then find the regression lines for the data. Women: ( ) 0.14 4.52 Wt t = + Men: ( ) 0.083 4.60 Mt t = +b. The enrollment for men and women were equal in 1971 with roughly 4.72 million each. c.In 2009,39. t =( ) ( ) 39 0.14 39 4.529.98W = += ( ) ( ) 39 0.083 39 4.607.837M = += 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: ( ) 2.45 72.82 f t t = + Homework 3.1SSM: Intermediate Algebra 54 Optical Scan: ( ) 3.08 12.58 f t t = +b. 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 2000,10. t =( ) ( ) 10 2.45 10 72.8248.32f = += ( ) ( ) 10 3.08 10 12.5843.38g = += 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%. c.( ) 100 48.32 43.38 8.3 + =In 2000, 8.3% of voters used the other two methods of voting. 29.The solution of this system is estimated to be ( ) 1.9, 2.8 . 31.The slope of( ) f xis3 and the f-intercept is ( ) 0, 30 . Therefore, ( ) 3 30 f x x = + . The slope of( ) g xis 5 and the g-intercept is( ) 0, 2 . Therefore,( ) 5 2 g x x = + . The solution for this system is( ) 3.5,19.5 . You may verify your answer using the slope or substituting the values for x in the functions. 33.( ) 4 0 f =35.( ) 3 f x =when5 x = . 37.( ) ( ) f x g x =when1 x = . 39.Answers may vary. Possible answers: a. 444yx4(2, 1)122y x = +2 3 y x = b. 444yx4y x1 y x = + c. ( )3213 2 12y xy x== + 444yx432y x = 41.Graph the equations. 32 93 1y xy xy x= += += 4yx422(2, 5)3 y x = +2 9 y x = +3 1 y x = The solution of the system is( ) 2, 5 . SSM: Intermediate AlgebraHomework 3.2 55 43.Answers may vary. One possible answer: 3472 5y xy xy x= = += 42yx24(4, 3)34y x =7 y x = +2 5 y x = 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. Homework 3.2 1.Substitute5 x for y in9 x y + = . ( ) 5 92 5 92 147x xxxx+ = === Let7 x =in5 y x = . 7 52y = = The solution is( ) 7, 2 . 3.Substitute4 7 y +for x in2 3 1 x y = . ( ) 2 4 7 3 18 14 3 15 14 15 153y yy yyyy+ = + = + = = = Let3 y = in4 7 x y = + . ( ) 4 3 712 75x = += += The solution is( ) 5, 3 . 5.Substitute( ) 2 5 x for y in3 5 29 x y = . ( ) ( )( )3 5 2 5 293 5 2 10 293 10 50 297 213x xx xx xxx = = + = = = Let3 y =in( ) 2 5 y x = . ( )( )2 3 52 24y = = = The solution is( ) 3, 4 . 7.Substitute4 8 y for x in3 2 18 x y = . ( ) 3 4 8 2 1812 24 2 1814 423y yy yyy = = == Let3 y = in4 8 x y = . ( ) 4 3 812 84x = = = The solution is( ) 4, 3 . 9.Substitute99xfor y in100 y x = . 99 10000x xxx= == Let0 x =in99 y x = . ( ) 99 00y == The solution is( ) 0, 0 . 11.Substitute0.2 0.6 x +for y in2 3 4 y x = . ( ) 2 0.2 0.6 3 40.4 1.2 3 42.6 5.22x xx xxx+ = + = = = Let2 x =in0.2 0.6 y x = + . ( ) 0.2 2 0.60.4 0.61y = += += The solution is( ) 2,1 . Homework 3.2SSM: Intermediate Algebra 56 13.Substitute 152 x for y in2 3 1 x y + = . 12 3 5 1232 15 1271424x xx xxx + = + = == Let4 x =in 152y x = . ( )14 522 53y = = = The solution is( ) 4, 3 . 15.Multiply the first equation by1 and then add the equations. 4 93 5 4x yx yx+ = = = Substitute4 x =into4 9 x y + =and solve for y. ( ) 4 4 916 97yyy+ =+ == The solution is( ) 4, 7 . 17.Multiply the first equation by 2 and add the equations. 6 4 146 5 4 9 18 2x yx yyy = = == Substitute2 y = into3 2 7 x y =and solve for x. ( ) 3 2 2 73 4 73 31xxxx =+ === The solution is( ) 1, 2 . 19.Rewrite the second equation so the variables are on the left side of the equation and the constant is on the right side. 3 5 34 8x yx y+ =+ = Multiply the second equation by3 and add the equations. 3 5 33 12 24 7 21 3x yx yyy+ = = = = Substitute3 y =into8 4 x y = and solve for x. ( ) 8 4 38 124x = = = The solution is( ) 4, 3 . 21.Multiply the first equation by3 and the second equation by 2, then add the equations. 24 27 129 24 30 4257 171 3x yx yyy + =+ === Substitute3 y =into8 9 43 x y = and solvefor x. ( ) 8 9 3 438 27 438 62xxxx = = = = The solution is( ) 2, 3 . 23.Multiply the second equation by 2 and add the equations. 0.9 0.4 1.90.6 0.4 2.6 1.5 4.5 3x yx yxx+ = === Substitute3 x =into0.9 0.4 1.9 x y + =and solve for y. ( ) 0.9 3 0.4 1.92.7 0.4 1.90.4 0.82yyyy+ =+ == = The solution is( ) 3, 2 . SSM: Intermediate AlgebraHomework 3.2 57 25.Use the distributive property to simplify both equations. ( ) ( ) 3 2 1 4 3 16 3 4 12 16 4 16x yx yx y + = + =+ = ( ) ( ) 4 5 2 4 1 184 20 8 2 184 8 0x yx yx y+ + =+ = = The system can be rewritten as 6 4 164 8 0x yx y+ = = Multiply the first equation by 2 and add the equations. 12 8 32 4 8 0 16 322x yx yxx+ = === Substitute2 x =into4 8 0 x y =and solvefor y. ( ) 4 2 8 08 8 08 81yyyy = = = = The solution is( ) 2,1 . 27.Multiply the first equation by 3 and add the equations. 3 9215 22 9165 2 5x yx yx+ = = = Substitute5 x =into 1 375 2x y + =and solvefor y. ( )1 35 75 231 723624yyyy+ =+ === The solution is( ) 5, 4 . 29.Multiply the first equation by 3 and the second equation by2 , then add the equations. 3 1 22 22 5 117 21 2 2 3x yx yyy+ = = = = Substitute3 y =into 2 1 13 2 6x y + =and solvefor x. ( )2 1 133 2 62 3 13 2 62 43 32xxxx+ =+ == = The solution is( ) 2, 3 . 31.Substitute2 5 x +for y in6 3 3 x y = . ( ) 6 3 2 5 36 6 15 315 3 falsex xx x + = = = This is a contradiction. The system is inconsistent. The lines are parallel so the solution set is the empty set,. 33.Multiply the first equation by 3 and the second equation by 2, then add the equations. 39 30 2134 30 9473 73 1x yx yxx+ = === Substitute1 x =into13 10 7 x y + = and solve for y. ( ) 13 1 10 713 10 710 202yyyy+ = + = = = The solution is( ) 1, 2 . Homework 3.2SSM: Intermediate Algebra 58 35.Multiply the first equation by 3 and add the equations. 12 15 912 15 9 0 0truex yx y = + = = This is an identity. The system is dependent. The solution set is the set of ordered pairs( ) , xysuch that4 5 3 x y = . 37.Substitute2.3 7 x for y in0.6 4 y x = + . 2.3 7 0.6 42.9 113.79x xxx = += Substitute3.79 x =into2.3 7 y x = and solve for y. ( ) 2.3 3.79 71.72y = The solution is roughly( ) 3.79,1.72 . 39.Multiply the first equation by 3 and the second equation by 4, then add the equation. 24 27 2124 60 12 33 33 1x yx yyy + = = = = Substitute1 y =into6 15 3 x y = and solve for x. ( ) 6 15 1 36 15 36 122xxxx = = == The solution is( ) 2,1 . 41.Multiply the first equation by 2 and add the equations. 8 6 28 6 50 3 falsex yx y = + = = This is a contradiction. The system is inconsistent so the solution set is the empty set. 43.Substitute 132 x +for y in2 6 y x = . 12 3 626 66 6 truex xx x + = + == This is an identity. The system is dependent. The solution set is the set of ordered pairs( ) , xy such that 132y x = + . 45.Multiply the first equation by 2 and the second equation by 5, then add the equations. 5 1 63 25 25203 213 26 2x yx yyy+ = + === Substitute2 y =into 5 136 4x y + =and solvefor x. ( )5 12 36 45 136 25 56 23xxxx+ =+ === The solution is( ) 3, 2 . 47.Elimination: Rewrite the second equation so the variables are on the left side of the equation. 3 112 9x yx y+ =+ = Multiply the second equation by1 and add the equations. 3 112 9 2x yx yx+ = = = Substitute2 x =into2 9 y x = +and solvefor y. ( ) 2 2 95y = += The solution is( ) 2, 5 . SSM: Intermediate AlgebraHomework 3.2 59 Substitution: Substitute2 9 x +for y in the first equation. ( ) 3 2 9 113 2 9 112x xx xx+ + = + == Substitute2 x =into2 9 y x = +and solve for y. ( ) 2 2 94 95y = += += The solution is( ) 2, 5 . Graphing: 48yx4 8(2, 5)2 9 y x = +3 11 y 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, substitute2 3 x +for y in2.01 1 y x = +and solve for x. 2 3 2.01 10.01 2200x xxx+ = + = = Substitute200 x =into2 3 y x = +and solvefor y. ( ) 2 200 3403y = += The solution is( ) 200, 403 . 51.The function f has a slope of 4 and an f-intercept of( ) 0, 3 . Therefore,( ) 4 3 f x x = + . The function g has a slope of6 and a g-intercept of ( ) 0, 50 . Therefore,( ) 6 50 g x x = + . The system that describes the functions f and g is: 4 36 50y xy x= += + Solve using substitution. Substitute4 3 x +for y in6 50 y x = +and solve for x. 4 3 6 5010 474.7x xxx+ = +== Substitute4.7 x =into4 3 y x = +and solvefor y. ( ) 4 4.7 318.8 321.8y = += += The solution is( ) 4.7, 21.8 . 53.The coordinate for A is( ) 0, 0since it lies at the origin. The coordinate for B is the y-intercept of 1l , which is( ) 0, 3 . The coordinate for C is the point of intersection of 1land 2l . Solve the following system: 12: 2 3: 3 30y xy x= ++ =ll Substitute2 3 x +for y in the second equation. ( ) 3 2 3 306 9 307 213x xx xxx+ + =+ + === Substitute3 x =into the first equation and solve for y. ( ) 2 3 36 39y = += += The solution is( ) 3, 9so the coordinate of C is ( ) 3, 9 . The coordinate for D is the point of intersection of 2land 3l . Solve the following system: 23: 3 30: 3 26y xy x+ =+ =ll Multiply the second equation by3 and add the equations. 3 303 9 78 8 48 6y xy xxx+ = = = = Substitute6 x =into3 30 y x + =and solve for y. Homework 3.2SSM: Intermediate Algebra 60 ( ) 3 6 303 248yyy+ === The solution is( ) 6, 8so the coordinate of D is ( ) 6, 8 . The coordinate for E is the point of intersection of 3land 4l . Solve the following system. 34: 3 26: 2 10y xy x+ == ll Substitute2 10 x in for y in3 26 y x + = . 2 10 3 265 367.2x xxx + === Substitute7.2 x =into2 10 y x = and solve for x. ( ) 2 7.2 104.4y = = The solution is( ) 7.2, 4.4so the coordinate of E is( ) 7.2, 4.4 . The coordinate of F is the x-intercept of 4l . Let0 y =in 4land solve for x. 2 100 2 102 105y xxxx= = == The coordinate of F is( ) 5, 0 . 48yx4 8B(0,3)C(3,9)D(6,8)F(5,0)E(7. 2,4.4)A(0,0)

55.a.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 cby ax cax cyb+ == + += Substitute this result for y in the second equation and solve for x. ( )( ) orax cdx e fbaex cedx fb bbdx aex ce bfbd ae x ce bfbd ae x bf cebf ce ce bfxbd ae ae bd + + = + = + = + = = = Substitute this result for x inax by c + =and solve for y. ce bfa by cae bdace abfby cae bdace abfby cae bdace bcd ace abfbyae bd ae bdbcd abfbyae bdabf bcdbyae bdaf cdyae bd + = + == = +=== Therefore, the solution is ,ce bf af cdae bd ae bd , assuming0 ae bd . b.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 5 414 143 3 5 4 11 11x= = = ( ) ( )( ) ( )3 4 2 44 43 3 5 4 11 11y= = = The solution is 14 4,11 11 . 57.Written response. Answers may vary. SSM: Intermediate AlgebraHomework 3.3 61 Homework 3.3 1.Solve the system 0.19 43.730.16 39.90y ty t= += + Substitute0.19 43.73 t +for y in the second equation and solve for t. 0.19 43.73 0.16 39.900.03 3.83127.67t ttt + = + = Substitute this result into the first equation and solve for y. ( ) 0.19 127.67 43.7319.47y = + According to the models, the winning times for women and men will both be 19.47 seconds in the year 2098. 3.Solve the system 0.14 4.520.083 4.60E tE t= += + Substitute0.14 4.52 t +for E in the second equation and solve for t. 0.14 4.52 0.083 4.600.057 0.081.40t ttt+ = += Substitute this result into the first equation and solve for E. ( ) 0.14 1.4 4.52 4.71 E = + According to the models, the enrollments for men and women were approximately equal in 1971 (roughly 4.71 million). 5.Solve the system 2.45 72.823.08 12.58p tp t= += + Substitute2.45 72.82 t +for p in the second equation and solve for t. 2.45 72.82 3.08 12.585.53 60.2410.89t ttt + = + = Substitute this result into the first equation and solve for p. ( ) 2.45 10.89 72.82 46.14 p = + 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. 7.a.Solve the system 13.5 2295 365y ty t= += + Substitute13.5 229 t +for y in the second equation and solve for t. 13.5 229 5 36518.5 1367.35t ttt+ = += Substitute this result into the first equation and solve for y. ( ) 13.5 7.35 229328.23y = + According to the models, the two newspapers had equal circulations of roughly 328 thousand in 1997. b.Since the circulations were roughly equal in 1997, competition heated up as each newspaper tried to overtake the other. c.( ) ( ) 10 13.5 10 229135 229364D = += += ( ) ( ) 10 5 10 36550 365315R = += += ( ) 826 364 315 826 679 147 + = =According to the models, the combined increase due to bonus copies was roughly 147 thousand bonus copies. d.( ) ( ) 11 13.5 11 229377.5D = += ( ) ( ) 11 5 11 365310R = += 377.5 310 687.5 + =According to the models, the combined circulation from the two newspapers was roughly 688 thousand copies in 2001. e.The estimate in part d. was an overestimate. After joining revenue streams, the competition for subscribers ceased (orat leastreduced 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 3.3SSM: Intermediate Algebra 62 9.a.Start by plotting the data, then compute the regression line for each data set. Women: ( ) 1.22 338.47 Wt t = + Men: ( ) 0.36 240.44 Mt t = +b.Solve the system 1.22 338.470.36 240.44y ty t= += + Substitute1.22 338.47 t +for y in the second equation and solve for t. 1.22 338.47 0.36 240.440.86 98.03113.99t ttt + = + = Substitute this result into the first equation and solve for y. ( ) 1.22 113.99 338.47199.40y = + According to the models, the record times for men and women will both be approximately 199.40 seconds in 2014. c.Women: ( ) 1.01 323.57 Wt t = + Men: ( ) 0.36 240.73 Mt t = + Solve the system 1.01 323.570.36 240.73y ty t= += + Substitute1.01 323.57 t +for y in the second equation and solve for t. 1.01 323.57 0.36 240.730.65 82.84127.45t ttt + = + = Substitute this result into the first equation and solve for y. ( ) 1.01 127.45 323.57194.85y = + Now the models predict that the record times for men and women will be the same in 2027. 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 Taurusvalue decreases by a constant $1582 each year, the function( ) T tis linear and its slope is 1582. The T-intercept is( ) 0,12939 , since the car is worth $12,939 at year0 t =(2002). Doing similar work for( ) Et , we get the following equations: ( )( )1582 12, 9391024 9330T t tEt t= += + SSM: Intermediate AlgebraHomework 3.3 63 b.Solve the system 1582 12, 9391024 9330V tV t= += + Substitute1582 12, 939 t +in for V in the second equation and solve for t. 1582 12, 939 1024 9330558 36096.468t ttt + = + = Substitute this result into the first equation and solve for V. ( ) 1582 6.468 12, 9392707V = + The cars will both be worth roughly $2707 in 2008. c.Find the intersection point using a graphing utility. 13.a.Since college As tuition increases by a constant $670 each year, the function( ) Atis linear and its slope is 670. The A-intercept is( ) 0, 6100 , since tuition is $6100 in year 0 t =(2000). Similar work for college B gives the following equations: ( )( )670 6100440 8500At tBt t= += + b.Solve the system 670 6100440 8500y ty t= += + Substitute670 6100 t +for y in the second equation and solve for t. 670 6100 440 8500230 240010.435t ttt+ = += Substitute this result in the first equation and solve for y. ( ) 670 10.435 610013091.45y = += The tuition at both colleges will be $13,091 in approximately 10 years (in 2010). c.Find the intersection point using a graphing utility. 15.a.Since Jenny Craigs 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 0 t = . So, an equation for( ) J tis: ( ) 72 19 J t t = + . Since Weight Watchersprogram 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, 0since there is no start-up fee at0 t = . An equation for( ) Wtis:( ) 77 Wt t = . b.Solve the system 72 1977y ty t= += Substitute72 19 t +for y into the second equation. 72 19 775 193.8t ttt+ === Substitute this result into the first equation and solve for y. ( ) 72 3.8 19292.6y = += The total cost at both Jenny Craig and Weight Watchers is approximately $293 in 4 weeks. c.Find the intersection point using a graphing utility. Homework 3.4SSM: Intermediate Algebra 64 17.Let( ) Ptrepresent the average price (in dollars) of a home in a community and( ) S trepresent 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, 214000since the price of a home is $214,000 in year0 t = . So, an equation for P is :( ) 8000 214000 Pt t = + . Similar work in finding the equation for the function S gives:( ) 3600 10000 S t t = + . (The slope of S is 3600 since the family plans to save $300 each month which is $3600 each year.) In 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 ( ) ( ) ( )( )0.1 0.1 8000 2140003600 10000y Pt ty S t t= = += = + Substitute( ) 0.1 8000 214000 t +for y in the second equation and solve for t. ( ) 0.1 8000 214000 3600 10000800 21400 3600 100002800 114004.07t tt ttt+ = ++ = + = The family will be able to pay a 10% down payment in 4 years (2004). Homework 3.4 1. Words Inequality Notation Graph Interval Notation Numbers greater than 33 x >0 3 ( ) 3, Numbers less than or equal to 4 4 x 4 0 ( ] , 4 Numbers less than 55 x + + > + + > + 2 11 11 4 1 112 4 102 4 4 10 46 106 106 653x xx xx x x xxxx + > + > > > < < Interval: 5,3 33 0 19.( ) 6.23 2.35 1.76 3 2.736.23 2.35 5.28 4.80486.23 2.35 2.35 5.28 4.8048 2.356.23 2.93 4.80486.23 4.8048 2.93 4.8048 4.80481.4252 2.931.4252 2.931.4252 1.42522.06x xx xx xx xx x x xxxx + < + < + < < + < + > Interval:( ) 2.06, 3 3 0 21.( ) ( ) 7 1 8 2 07 7 8 16 023 023231 123x xx xxxxx+ + + + Interval:[ ) 23, 23 26 Homework 3.4SSM: Intermediate Algebra 66 23. 2433 2 342 3 26xxx > < < Interval:( ) , 6 0 6 25. 12 3412 2 3 2415414 4 5420xxxxx < + < +< < < Interval:( ) , 20 15 20 0 27. 2 3 53 4 22 3 2 5 23 4 3 2 3xx

3 114 64 3 4 113 4 3 6229xxx Interval: 22,9 02 29 29. 1 1 12 6 31 1 1 1 12 6 6 3 61 12 21 12 22 21xxxxx + + Interval:( ] , 1 1 0 31. 1582 12, 939 1024 93301582 12, 939 12, 939 1024 9330 12, 9391582 1024 36091582 1024 1024 3609 1024558 3609558 3609558 5586.47T Et tt tt tt t t tttt> + > + + > + > + > + > < < The value of the Taurus is more than the value of the Escort for years up until 2008 ( 6 t < ). 33.a.Since U-Hauls charge increases at a constant rate of $0.69 per mile, the equation is linear with slope0.69 . The U-intercept is 19.95 since U-Haul charges a flat fee of $19.95. An equation for U-Hauls charge is: ( ) 0.69 19.95 U x x = + . After similar work for Penske, an equation for Penskes charge is: ( ) 0.39 29.95 Px x = + . b. 0.69 19.95 19.95 0.39 29.95 19.950.69 19.95 0.39 29.95x xU Px x+ < + She must marry a man who is born in 1986 or later. 37.The student made a mistake. It is not necessary to switch the direction of the inequality when dividing by a positive number. 39.a.Solve the inequality for x. ( ) 3 2 1 7 43 6 1 7 43 5 7 43 5 5 7 4 53 12 43 4 12 4 47 127 127 7127x xx xx xx xx xx x x xxxx + + + + + + Any three numbers that are greater than or equal to 127 are possible solutions. b.From part a., any number less than 127 is not a solution. 41.We havemx c . This implies thatm must be negative so that dividing both sides by m produces: mx cmx cm mcxm> Since we want2cm = , this means c must also be negative. Thus, c and m can be any negative numbers such that2cm = . 43.A solution of the form2 5 x < when2.8 x < . 51.Written response. Answers may vary. Chapter 3 Review Exercises 1. 312164y xy x= += 4yx4 4(4, 5) The solution is( ) 4, 5 . 2.3 5 15 3 13 15 5x yy xy x = = = + ( ) 2 42 8y xx= = + 48yx4 8(3, 2) The solution is( ) 3, 2 . 3.Solve using substitution. Substitute2 5 x +for y in the second equation. 3 102 5 3 102 5 5 3 10 52 3 52 3 3 5 35 51y xx xx xx xx x x xxx= ++ = ++ = + = ++ = + +== Substitute this result into the first equation and solve for y. ( ) 2 1 52 57y = += += The solution is( ) 1, 7 . 4.Solve using elimination. Multiply the first equation by 3 and the second equation by4 , then add the equations. 12 15 6612 8 20 23 46 2x yx yyy = = = = Substitute2 y =into3 2 5 x y + = and solvefor x. ( ) 3 2 2 53 4 53 93xxxx+ = + = = = The solution is( ) 3, 2 . 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. 2 3 74 6 14x yx y + = = Multiply the first equation by 2 and add the equations. 4 6 144 6 140 0 truex yx y + = = = This is an identity. The system is dependent. The solution set contains all ordered pairs( ) , xysuch that 2 73 3y x = + . 6.Solve using elimination. Multiply the first equation by 2 and add the equations. 6 14 106 14 10 11 falsex yx y + = = = This is a contradiction. The system is inconsistent. The solution set is the empty set. SSM: Intermediate AlgebraChapter 3 Review Exercises 69 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. 4 5 38 10 6x yx y =+ = Multiply the first equation by 2 and add the equations. 8 10 6 8 10 60 0 truex yx y =+ = = This is an identity. The system is dependent. The solution set contains all ordered pairs( ) , xysuch that 4 35 5y x = . 8.Solve using substitution. Substitute4.2 7.9 x for y in the second equation. 2.8 1.14.2 7.9 2.8 1.14.2 7.9 7.9 2.8 1.1 7.94.2 2.8 94.2 2.8 2.8 9 2.87 91.29y xx xx xx xx x x xxx= + = + + = + += ++ = + += Substitute this result into the first equation and solve for y. ( ) 4.2 1.29 7.92.5y = The solution is approximately( ) 1.29, 2.5 . 9.Solve using substitution. Substitute4.9xfor y in the second equation and solve for x. ( )3.23.2 4.915.6816.68 00y xx xx xxx = = = == Substitute this result into the first equation and solve for y. ( ) 4.9 00y == The solution is( ) 0, 0 . 10.Solve using elimination. Multiply the first equation by 2 and add the equations. 0.8 0.6 0.80.8 1.2 6.41.8 7.24x yx yyy+ = + === Substitute4 y =into0.4 0.3 0.4 x y + =and solve for x. ( ) 0.4 0.3 4 0.40.4 1.2 0.40.4 0.82xxxx+ =+ == = The solution is( ) 2, 4 . 11.Solve using substitution. Substitute 142 x for y in the first equation. 13 5 4 21253 20 212120 2121122x xx xxxx = + =+ === Substitute2 x =into the second equation and solve for y. ( )12 421 43y = = = The solution is( ) 2, 3 . 12.Solve using elimination. 6 4 85 36 845 34433x yx yyy = + = == Substitute3 y =into 3 245 3x y =and solvefor x. Chapter 3 Review ExercisesSSM: Intermediate Algebra 70 ( )3 23 45 332 4536510xxxx = === The solution is( ) 10, 3 . 13.First use the distributive property to simplify each equation. ( ) ( ) 2 3 4 3 2 1 56 8 6 3 56 6 11 56 6 6x yx yx yx y + = + = + = + = ( ) ( ) 3 2 1 4 3 76 3 4 12 76 4 9 76 4 16x yx yx yx y + + + = + + = + + = + = The system is: 6 6 66 4 16x yx y+ = + = Solve using elimination.Add the two equations together. 6 6 66 4 1610 10 1x yx yyy+ = + = = = Substitute1 y = into6 6 6 x y + =and solvefor x. ( )6 6 66 6 1 66 6 66 122x yxxxx+ =+ = === The solution is( ) 2, 1 . 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 5 152 9x yx y =+ = Multiply the second equation by1 and add the equations. 2 5 152 9 6 6 1x yx yyy = = == Substitute1 y = into2 5 15 x y =and solve for x. ( ) 2 5 1 152 5 152 105xxxx =+ === The solution is( ) 5, 1 . Substitution: Substitute2 9 x +for y in the first equation and solve for x. ( )2 5 152 5 2 9 152 10 45 1512 605x yx xx xxx = + =+ === Substitute this result into the second equation and solve for y. ( )2 92 5 910 91y x = += += += The solution is( ) 5, 1 . Graphically: 2 5 155 2 15235x yy xy x = = += 2 9 y x = +4yx842 9 y x = +235y x = (5, 1) The solution is( ) 5, 1 . SSM: Intermediate AlgebraChapter 3 Review Exercises 71 15.Answers may vary. Possible answers: a.2 43 6 12x yx y+ =+ = Solving this system, you will encounter an identity such as0 0 = . The solution set is the set of ordered pairs( ) , xysuch that 2 4 x y + = . b.2 42 6 10x yx y+ =+ = Solving this system, you will encounter a contradiction such as0 2 = . The solution set is the empty set,. c.103 3 6x yx y+ = = The point( ) 4, 6satisfies both equations. 16.The approximate solutions for this system are ( ) 2.6, 21.9and( ) 0.4,15.1 . Although answers may vary, your answers should be close to these points. 17.Begin by substituting 5 for x and 3 for y in the equations and solve for a and b. ( ) ( ) 2 5 3 310 919aaa+ =+ == ( ) ( ) 6 5 4 330 1218bbb = == Substitute 19 for a and 18 for b in the original system. This gives: 2 3 196 4 18x yx y+ = = Verify: ( ) ( ) 2 5 3 3 1910 9 1919 19true+ =+ ==( ) ( ) 6 5 4 3 1830 12 1818 18 = == To verify graphically, put each equation in slope-intercept form and then find the intersection point using a graphing utility. 2 3 193 2 192 193 3x yy xy x+ == += + 6 4 184 6 183 92 2x yy xy x = = +=

18.The coordinate for A is( ) 0, 0since it lies at the origin. The coordinate for B is the y-intercept of 1l , which is( ) 0, 4 . The coordinate for C is the point of intersection of 1land 2l . Solve the following system: 12: 3 4: 3 2 34y xy x= ++ =ll Substitute3 4 x +for y in the second equation. ( ) 3 3 4 2 349 12 2 3411 222x xx xxx+ + =+ + === Substitute2 x =into the first equation and solve for y. ( ) 3 2 46 410y = += += The solution is( ) 2,10so the coordinate of C is ( ) 2,10 . The coordinate for D is the point of intersection of 2land 3l . Solve the following system: 23: 3 2 34: 4 28y xy x+ =+ =ll Multiply the second equation by3 and add the equations. 3 2 343 12 84 10 50 5y xy xxx+ = = = = Substitute5 x =into3 2 34 y x + =and solve for y. ( ) 3 2 5 343 248yyy+ === Chapter 3 Review ExercisesSSM: Intermediate Algebra 72 The solution is( ) 5,8so the coordinate of D is ( ) 5,8 . The coordinate for E is the point of intersection of 3land 4l . Solve the following system. 34: 4 28: 3 14y xy x+ == ll Substitute3 14 x in for y in4 28 y x + = . ( ) 3 14 4 287 426x xxx + === Substitute6 x =into3 14 y x = and solve for x. ( ) 3 6 1428 144y = = = The solution is( ) 6, 4so the coordinate of E is ( ) 6, 4 . The coordinate of F is the x-intercept of 4l . Let0 y =in 4land solve for x. 3 140 3 143 14143y xxxx= = == The coordinate of F is 14, 03 . 8yx8B(0,4)C(2,10)D(5,8)FE(6,4)A(0,0)14, 03 16 19.Graph each of the equations. 22 73 6y xy xy x= += += 4yx842 y x = +2 7 y x = +3 6 y x = The three graphs do not share a common point. Thus, the solution set is the empty set. 20.2 182 182 29xxx Interval:[ ) 9, 4 9 21.3 8 133 8 8 13 83 213 213 37xxxxx + + Interval:( ] , 7 10 7 4 22.( ) 8 3 2 2 98 3 6 2 98 5 158 5 5 15 53 153 153 35x x xx x xx xx x x xxxx > + > + + > + + > + + >< < Interval:( ) , 5 10 5 0 SSM: Intermediate AlgebraChapter 3 Review Exercises 73 23.( ) 4.2 3.6 3.9 2.14.2 3.6 3.9 8.194.2 3.6 4.2 3.9 8.19 4.23.6 3.9 4.013.6 3.9 3.9 4.01 3.97.5 3.997.5 3.997.5 7.50.532x xx xx xx xx x x xxxx + + + + + Interval:( ] , 0.532 400.532 24.( ) ( ) 5 2 3 2 3 410 15 6 8x xx x + 10 15 15 6 8 1510 6 710 6 6 7 616 716 716 16716x xx xx x x xxxx + + + + Interval: 7,16 1

1 ] 4 0716 25. 2 5 73 2 32 5 5 7 53 2 2 3 2xx + < + < 2 14 153 6 62 13 63 2 3 12 3 2 614xxxx < < > > Interval: 1,4 _ , 1 014 26. 3 54 24 3 4 53 4 3 2103xxx Interval: 10,3 _

, 2 2 0103 27. 2 335 22 33 3 35 2x xx x > > 2 95 22 95 27 95 25 7 5 97 5 7 24514x xx x x xxxx > > > < < Interval: 45,14 _ , 04514 28.a.Solve the inequality for x. ( ) 7 2 3 5 4 17 6 10 4 16 3 4 16 3 3 4 1 36 4 46 4 4 4 410 410 410 1025x xx xx xx xx xx x x xxxx + < + < + < + + < + + < + < + > Any three numbers that are greater than 25are possible solutions. Chapter 3 Review ExercisesSSM: Intermediate Algebra 74 b.From part a., any number less than or equal to 25is not a solution. 29.The students work is incorrect. When dividing both sides of an inequality by a negative number, you must switch the direction of the inequality. 30.( ) 4 1 f =31.( ) 4 3 g = 32.( ) 0 f x =when1 x = . 33.( ) 0 g x =when5 x = . 34.( ) ( ) f x g x =when2 x = . 35.The graph of( ) f xis above the graph of( ) g xfor values of x that are greater than2 . Thus, ( ) ( ) f x g x >when2 x > . 36.Any values for a, b, and c so that5c ba= , 0 a < , andb c > . For example,1 a = ,6 b = , and1 c =will work. 37.a.Since U-Hauls charge increases at a constant rate of $0.69 per mile, the equation is linear with slope0.69 . The U-intercept is 29.95 since U-Haul charges a flat fee of $29.95. An equation for U-Hauls charge is: ( ) 0.69 29.95 U x x = + . After similar work for Rent A Wreck, an equation for Rent A Wrecks charge is: ( ) 0.22 75.00 Rx x = + . b.( ) ( )0.69 0.22 0.22 45.05 0.220.69 29.95 0.22 750.69 29.95 29.95 0.22 75 29.950.69 0.22 45.050.47 45.0595.9x x x xU x Rxx xx xx xxx = + =+ = ++ = + = += The two charges will be the same when the number of miles driven is roughly 95.9 miles. c. 0.22 75 75 0.69 29.95 750.22 75 0.69 29.950.22 0.69 45.050.22 0.69 0.69 45.05 0.690.47 45.050.47 45.050.47 0.4795.9x xR Ux xx xx x xxxx+ < + > Rent A Wreck will be cheaper for miles driven more than 95.9 miles. 38.Let( ) Ptrepresent the average price (in dollars) of a home in a community and( ) S trepresent 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 $9000 per year, the function P is linear and its slope is 9000. the P-intercept is ( ) 0, 250000since the price of a home is $250,000 in year0 t = . So, an equation for P is : ( ) 9000 250000 Pt t = + . Similar work in finding the equation for the function S gives:( ) 4800 12000 S t t = + . (The 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 average-priced house, solve the following system for t when ( ) ( ) ( )( )0.1 0.1 9000 2500004800 12000y Pt ty S t t= = += = + Substitute( ) 0.1 9000 250000 t +for y in the second equation and solve for t. ( ) 0.1 9000 250000 4800 12000900 25000 4800 120003900 130003.33t tt ttt+ = ++ = + = The family will be able to pay a 10% down payment in 3 years (2003). SSM: Intermediate AlgebraChapter 3 Test 75 39.a. Start by plotting the data sets, then find the regression line for each region. North America: ( ) 0.52 35.04 nt t = + Far East: ( ) 1.50 13.89 f t t = +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.( ) ( )0.52 35.04 1.50 13.890.52 35.04 35.04 1.50 13.89 35.040.52 1.50 21.150.52 1.50 1.50 21.15 1.500.98 21.1521.6nt f tt tt tt tt t t ttt=+ = ++ = + = = = The consumption of petroleum was the same for the Far East and North America in 2002. d.( ) ( )0.52 35.04 1.50 13.890.52 35.04 35.04 1.50 13.89 35.040.52 1.50 21.150.52 1.50 1.50 21.15 1.500.98 21.150.98 21.150.98 0.9821.6nt f tt tt tt tt t t tttt > The consumption of petroleum in North America will be less than the consumption of petroleum in the Far East for years after 2002. 40.a.Solve the system ( )( )21.14 100.0015.63 220.68y Ht ty At t= = += = + Solve using substitution. Substitute21.14 100.00 t +for y in the second equation and solve for t. 21.14 15.63 15.63 120.68 15.6315.63 220.6821.14 100.00 15.63 220.6821.14 100 100 15.63 220.68 10021.14 15.63 120.6836.77 120.683.3t t t ty tt tt tt ttt+ = + += ++ = ++ = + = += Substitute3.3 t =into21.14 100.00 y t = +and solve for y. ( )21.14 100.0021.14 3.3 100.00169.8y t = += + According to the models, there were the same number of jobs in Hollywood as in the aerospace industry in the year 1993 ( 3 t = ). b.Solve for t when( ) ( ) Ht At > . This inequality is true when3.3 t > . Therefore, there are more jobs in Hollywood than in the aerospace industry in years after 1993 ( 3 t > ). c.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 1.Solve using substitution. Substitute3 1 x for y in the second equation and solve for x. ( )3 2 13 2 3 1 13 6 2 13 2 13 31x yx xx xxxx = = + = + = = = (cont.) Chapter 3 TestSSM: Intermediate Algebra 76 Substitute1 x =into the first equation and solve for y. ( ) 3 1 13 12y = = = 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. 2 5 36 15 9x yx y = = Multiply the first equation by3 and add the equations. 6 15 96 15 9 0 0 truex yx y + = == This is an identity. The system is dependent. The solution set is the set of ordered pairs( ) , xysuch that2 5 3 x y = . 3.Solve using elimination. Multiply the first equation by 3 and the second equation by2 , then add the equations. 12 18 1512 18 4 0 19 falsex yx y = + == 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. 2 385 49 335 411115 5x yx yxx =+ === Substitute5 x =into 9 335 4x y + =and solvefor y. ( )3 15 15 413 1411 341248yyyyy+ =+ == = = The solution is( ) 5, 8 . 5.First use the distributive property to simplify each equation. ( ) ( ) 4 2 3 2 1 214 8 6 3 214 6 11 214 6 322 3 16x yx yx yx yx y + + = + = + = + = = ( ) ( ) 5 3 2 4 3 5915 10 4 3 5915 4 13 5915 4 46x yx yx yx y + = = = = The system can be rewritten as: 2 3 1615 4 46x yx y = = Solve using elimination. Multiply the first equation by 4 and the second equation by3 , then add the equations. 8 12 6445 12 13837 742x yx yxx = + = == Substitute2 x = into2 3 16 x y = and solve for y. ( ) 2 2 3 164 3 163 124yyyy = = = = The solution is( ) 2, 4 . 6.Answers may vary. One possible answer: 2 13 17x yx y =+ = SSM: Intermediate AlgebraChapter 3 Test 77 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. 4 3 92 5x yx y + Multiply the second equation by 2 and add the equations. 4 3 94 2 10 1 1x yx yyy + Substitute1 y into4 3 9 x y and solvefor x. ( ) 4 3 1 94 3 94 123xxxx The solution is( ) 3,1 . Substitution: Substitute2 5 x for y in the first equation. ( )4 3 94 3 2 5 94 6 15 92 15 92 63x yx xx xxxx + + Substitute3 x into2 5 y x and solve for y. ( ) 2 3 51y The solution is( ) 3,1 . Graphically: 4 3 9433x yy x 2 5 y x 44yx422 5 y x 433y x (3, 1) The solution is( ) 3,1 . 8.If the solution set is the empty set, the system is two parallel lines. Therefore, m iny mx b + is 5 since this is the slope in5 13 y x and parallel lines have the same slope.The y-intercept iny mx b +is any number other than13 since5 13 y x and y mx b +intersect the y-axis at different points ( 13 b ). 9.2 10 3 142 10 2 3 14 210 3 1210 3 3 12 313 1213 1213 131213x xx xx xx x x xxxx + + + + Interval: 12,13 1

1 ] 2 20 10.( ) ( ) 3 4 1 5 23 12 1 5 103 13 5 10x xx xx x+ + < + + < + < 3 13 13 5 10 133 5 233 5 5 23 52 232 232 2232x xx xx x x xxxx+ < < < < > > Interval: 23,2 _ , 12 6 0 Chapter 3 TestSSM: Intermediate Algebra 78 11.( ) 2.6 3.1 4.7 5.92.6 8.06 4.7 5.92.6 8.06 8.06 4.7 5.9 8.062.6 4.7 2.162.6 4.7 4.7 2.16 4.72.1 2.162.1 2.162.1 2.11.03x xx xx xx xx x x xxxx > > + > +> + > + >< < Interval:( ) , 1.03 1 20 12. 5 1 73 6 45 1 5 7 53 6 3 4 31 416 12x xx x x xx + + + +

12 1 12 4141 6 41 12241241xxx Interval: 2,41 2 1 0 13.( ) 5 3 f =14.( ) 3 g x =when4 x = . 15.( ) ( ) f x g x =when2 x = . 16.The graph of f is below the graph of g for2 x + > > > < > Interval:( ) 2, 2c. 4yx422 d.Solve using substitution. Substitute 3 6 x +for y in the second equation and solve for x. 4 13 6 4 13 4 77 71y xx xx xxx= + = = = = Substitute this result into the first equation and solve for y. ( )3 63 1 63y xy= += += The solution is( ) 1, 3 . 39.Answers may vary. Possible answers: a.3 2 82 63xxx+ === 3 b. 223y x = +444yx4 c.2 12y xy x= = + 444yx4(1, 1)2 y x =+2 1 y x = d.4 2 104 82xxx+ 0 2SSM: Intermediate AlgebraCumulative Review Chapters 1-3 85 40.a.The grass height is declining at a constant rate. Thus,( ) htis linear and has the form ( ) ht mt b = + . Selecting two ordered pairs allows us to find the slope. We will use 155,4 and 195,8 . 1 11/ 8 18 495 55 40 320m= = = Using the slope and the first point, we can find the y-intercept. ( )1 1554 3201 114 642764y mx bbbb= += += += Therefore,( )1 27320 64ht t = + . b.( ) ( )1 27105 105 0.094320 64h = + The grass on the green will have a height of about 0.094 inches in 2005. c. 1 1 2716 320 6423 164 320115ttt= + = = The grass on the green will have a height that is 116 of an inch in 2015. d. 1 11/ 8 18 495 55 40 320m= = = The slope of h is 1320 . The height of the grass on the green gets shorter each year by 1320 of an inch. Note: 10.0031320 e. 1 270320 641 27320 64135ttt= +== The t-intercept is( ) 135, 0 . 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 tis linear. The slope is12 since the number hit decreases by 12 each year. The n-intercept is ( ) 0, 446since 446 bicyclists younger than 13 were hit by motorists in 1975. Thus, an equation for( ) f tis: ( ) 12 446 f t t = +b.(see a.) The slope of f is12 . 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. d.Find the t-intercept by solving( ) 0 f t = . 0 12 44612 44637.17ttt= += 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.Since Company As sales increase by a constant 1.3 million dollars each year, the slope for( ) Atis 1.3. The A-intercept is ( ) 0, 9.5since the total sales are 9.5 million dollars in year0 t = . Therefore, the equation for ( ) Atis( ) 1.3 9.5 At t = + . Similar work for( ) Btgives the equation ( ) 1.8 5.2 Bt t = + . Cumulative Review Chapters 1-3SSM: Intermediate Algebra 86 b. yt366 9 121215182430Millions of dollarsYe ars The intersection of A and B is( ) 8.6, 20.68 . According to the sketch, sales at the companies will both equal $20.68 million dollars in 2006 ( 9 t = ). c.Solve for t when( ) ( ) At Bt = . ( ) ( )1.3 9.5 1.8 5.21.3 9.5 9.5 1.8 5.2 9.51.3 1.8 4.31.3 1.8 1.8 4.3 1.80.5 4.38.6At Btt tt tt tt t t ttt=+ = ++ = + = = = = Substitute 8.6 for t in( ) 1.3 9.5 At t = + . ( ) ( ) 8.6 1.3 8.6 9.5 20.68 A = + =In 2009 ( 9 t = ), the sales at both companies will equal $20.68 million dollars.This is the same result as in part b. d.Solve the inequality( ) ( ) At Bt < . ( ) ( )1.3 9.5 1.8 5.21.3 9.5 9.5 1.8 5.2 9.51.3 1.8 4.31.3 1.8 1.8 4.3 1.80.5 4.30.5 4.30.5 0.58.6At Btt tt tt tt t t tttt > According to the models, Company As sales will be less than Company Bs sales after 2009 ( 9 t = ). 43.a.Start by plotting each set of data and then find the regression lines. Women: ( ) 0.064 27 Wt t = + Men: ( ) 0.029 22.10 Mt t = +b.( ) ( ) 107 0.064 107 2720.15W = += ( ) ( ) 107 0.029 107 22.1019M = += The models predict that 200-Meter Run record time in 2007 will be 20.15 seconds for women and 19 seconds for men. 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( ) ( ) Wt Mt = . ( ) ( )0.064 27 0.029 22.10.064 0.029 4.90.035 4.9140Wt Mtt tt ttt= + = + = = = The models predict that the record times will be equal in 2040. SSM: Intermediate AlgebraCumulative Review Chapters 1-3 87 f.( ) ( )0.064 27 0.029 22.10.064 0.029 4.90.035 4.90.035 4.90.035 0.035140Wt Mtt tt tttt> + > + > > < < Womens record times are greater than mens record times for years before 2040.g.Women: 0 0.064 270.064 27421.88ttt= += Men: 0 0.029 22.10.029 22.1762.07ttt= += The t-intercept for women is( ) 421.88, 0and the t-intercept for men is( ) 762.07, 0 . According to the models, the womens record time will be 0 seconds in 2322 and the mens record time will be 0 seconds in 2662. This is not feasible so model breakdown has occurred. h.Answers may vary. One possible answer: rtMW