6. 5x Division Property. CHAPTER 2 Linear Models, Equations, and Inequalities. Toolbox Exercises. 1. 3x = 6 Division Property

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1 CHAPTER Linear Models, Equations, and Inequalities CHAPTER Linear Models, Equations, and Inequalities Toolbox Exercises. x = 6 Division Property x 6 = x =. x 7= Addition Property x 7= x 7+ 7= + 7 x = 8. x + = 8 Subtraction Property x + = 8 x + = 8 x =. x = Addition Property x = x + = + x = 6. x = 0 Division Property x 0 = x = 7. x + 8= Subtraction Property and division Property x + 8= x + 8 8= 8 x = 0 x 0 = x = 0 x 8. = Addition Property and multiplication Property x = x + = + x = 8 x = (8) x = x. = 6 Multiplication Property x = 6 x = (6) x = 8 9. x = 6+ x x x = 6+ x x x = 6 x + = 6+ x = 9 x 9 = x = Copyright 0 Pearson Education. Inc.

2 CHAPTER Algebra Toolbox 0. x = 7x x+ 7x = 7x+ 7x 0x = 0x + = + 0x = 6 0x 6 = x = 0 x =. ( x ) = x 8 6x+ = x 8 6x x+ = x x 8 0x + = 8 0x + = 8 0x = 0 0x 0 = 0 0 x =... x = x = x = 8 x 8 = x = 6 x = 0 x = 0 x = 0 x 0 = x = ( x ) = x x = x x+ x = x+ x x = x + = + x = 0 x 0 = x =. 6. x 7= x x 7= 6x+ x+ 6x 7 = 6x+ 6x+ 8x 7 = 8x = + 7 8x = 9 8x 9 = 8 8 x = ( x 6) = x x+ = 9x x 9x+ = 9x 9x x + = x + = x = x = x = 7. y = x and x+ y = x+ ( x) = x = x = Copyright 0 Pearson Education. Inc.

3 CHAPTER Linear Models, Equations, and Inequalities 8. y= x and x+ y = x+ ( x) = x = x = 9. y= x and x+ y = x+ ( x) = x+ x= x = x = x x. = + x 0= x+ 8 x = 8 This is a conditional equation.. x + > x > 6 6 x > 0. y= 6 x and x+ y = 7 x+ (6 x) = 7 x+ x= 7 x = 7 x =. x x= x+ 7 x= x+ 7 x = 7 7 x = This is a conditional equation.. ( x ) + = x 7 x+ = x 7 = 7 This is a contradiction equation.. 9x x = x+ 0+ x 9x x+ 0= 7x+ 0 7x+ 0= 7x x 7 x 6 x x 7. > x > x 8. > 6 x > x 9. > x x > x + x> 0x+ 8 9x > 8 8 x < 9 This is an identity equation. Copyright 0 Pearson Education. Inc.

4 CHAPTER Algebra Toolbox x 0. + > 6x x+ 6> x x + 6 > 0 x > 6 6 x <. ( x ) < x + < x < 9 x > 9 + < 6 x < 6 x < 8 x > 6. ( x ) Copyright 0 Pearson Education. Inc.

5 6 CHAPTER Linear Models, Equations, and Inequalities Section. Skills Check. x = + 7x x 7x = + 7x 7x x = x + = + x = 7 x 7 = 7 x = x = 8. Applying the intersections of graphs method, graph y= x and y = + 7x. Determine the intersection point from the graph: [ 0, 0] by [ 00, 00]. x = 7x x 7x = 7x 7x x = x = x = x = x = =. x = 7x x 7x + = 0 x + = 0 Graph y = x+ and determine the x- intercept. The x-intercept is the solution to the equation. [ 0, 0] by [ 0, 0]. ( x 7) = 9 x x = 9 x x+ x = 9 x+ x x = 9 x + = 9+ x = 0 x 0 = x = 0 Applying the intersections of graphs method yields: [, ] by [ 0, 0] Applying the x-intercept method, rewrite the equation so that 0 appears on one side of the equal sign. Copyright 0 Pearson Education. Inc.

6 CHAPTER Section. 7 6 = 8 y. ( y ) y 0= 8 y 7y = 8 8 y = Applying the intersections of graphs method yields: Applying the intersections of graphs method yields: [ 0, 0] by [ 0, 0]. [ 0, 0] by [ 0, 0] Remember your calculator has solved for the x-value via the graph even though the original equation was written with the variable y. x = x+ 6 LCD : x = x+ 6 x 0 = 6x+ x 6x 0 = 6x 6x+ x 0 = x = + 0 x = x = x = x = x+ LCD : x = x+ 6x = 60x+ 9 x = 9 x = x = Applying the intersections of graphs method yields: [ 0, 0] by [ 0, 0] Copyright 0 Pearson Education. Inc.

7 8 CHAPTER Linear Models, Equations, and Inequalities 7. ( x ) x x = 6 9 LCD :8 ( x ) x 8 x = ( x ) 8x = 8 x x 8x= 8 x x = 8 x x = 8 x = 6 x = 6 Applying the intersections of graphs method yields: Applying the intersections of graphs method yields: [0, 70] by [ 0, ] Remember your calculator has solved for the x-value via the graph even though the original equation was written with the variable y. 8. [ 00, 0] by [ 0, 0] ( y ) y y = 6 LCD : ( y ) y y = 6 y y = 90 y y y = 90 y y y = 90 y y = 90 y y = y = t =.78t..9t.78t =..t =..t. =.. t = Applying the intersections of graphs method yields: [ 0, 0] by [ 0, 0] x+ 0.8 = 0.6x x = x = 0.7 x = 8 Copyright 0 Pearson Education. Inc.

8 CHAPTER Section. 9 Applying the intersections of graphs method yields: The solution to f( x ) = 0, the x-intercept of the function, and the zero of the function are all 0.. Answers a), b), and c) are the same. Let 0 f x = and solve for x. [0, 0] by [, ]. + x = x LCD = x = 60 x + x 0 = 8x 6x = x = = 6 6. x 6 = + x 6 LCD = x = 0 + x 6 0x 6 = + x x = x = =. Answers a), b), and c) are the same. Let 0 f x = and solve for x. +.6x = 0.6x = x =.6 x = 0 x 60 = 0 x = 60 x = The solution to f( x ) = 0, the x-intercept of the function, and the zero of the function are all.. Answers a), b), and c) are the same. Let 0 f x = and solve for x. x 6= 0 LCD : x 6 = 0 x = 0 x = x = The solution to f( x ) = 0, the x-intercept of the function, and the zero of the function are all. 6. Answers a), b), and c) are the same. Let 0 f x = and solve for x. x = 0 LCD : x = 0 x = 0 x = Copyright 0 Pearson Education. Inc.

9 60 CHAPTER Linear Models, Equations, and Inequalities The solution to f( x ) = 0, the x-intercept of the function, and the zero of the function are all. 7. a. The x-intercept is, since an input of creates an output of 0 in the function. b. The y-intercept is, since the output of corresponds with an input of 0. c. The solution to f( x ) = 0 is equal to the x-intercept for the function. Therefore, the solution to f( x ) = 0 is. 8. a. The x-intercept is 0., since an input of 0. creates an output of 0 in the function. b. The y-intercept is 7, since the output of 7 corresponds with an input of 0. c. The solution to f( x ) = 0 is equal to the x-intercept for the function. Therefore, the solution to f( x ) = 0 is The answers to a) and b) are the same. The graph crosses the x-axis at x = The answers to a) and b) are the same. The graph crosses the x-axis at x = Answers a), b), and c) are the same. Let 0 f x = and solve for x.. Answers a), b), and c) are the same. Let f( x ) = 0 and solve for x. 6x 0 = 0 6x = 0 x = 0 The zero of the function, the x-intercept of the graph of the function, and the solution to f( x ) = 0are all 0.. Answers a), b), and c) are the same. Let 0 f x = and solve for x x = 0 0x = 0 x = 8. The zero of the function, the x-intercept of the graph of the function, and the solution to f( x ) = 0are all 8... Answers a), b), and c) are the same. Let 0 f x = and solve for x. 0 + x = 0 x = 0 0 x = 9 The zero of the function, the x-intercept of the graph of the function, and the solution to f( x ) = 0are all 0/9. x 00= 0 x = 00 x = The zero of the function, the x-intercept of the graph of the function, and the solution to f( x ) = 0are all. Copyright 0 Pearson Education. Inc.

10 CHAPTER Section. 6. Applying the intersections of graphs method yields: y = x 8. Applying the intersections of graphs method yields: y = (x+ ) + y = ( x ) y = 7 x [ 0, 0] by [ 0. 0] The solution is the x-coordinate of the intersection point or x =. [ 0, ] by [ 70, 0] The solution is the x-coordinate of the intersection point or x =. 6. Applying the intersections of graphs method yields: y = x+ y = x 8 9. Applying the intersections of graphs method yields: x y = + x y = [ 0, 0] by [ 0. 0] The solution is the x-coordinate of the intersection point or x =. [ 0, ] by [ 0, 0] The solution is the x-coordinate of the intersection point. t =. 7. Applying the intersections of graphs method yields: y = ( x ) + 6 y = ( x 8) 0. Applying the intersections of graphs method yields: x y = ( x ) y = 6 [ 0, ] by [ 70, 0] The solution is the x-coordinate of the intersection point or s =. [ 0, 0] by [ 0, 0] The solution is the x-coordinate of the intersection point or x = 6. Copyright 0 Pearson Education. Inc.

11 6 CHAPTER Linear Models, Equations, and Inequalities. Applying the intersections of graphs method yields: [ 0, 0] by [, ] The solution is the x-coordinate of the 7 intersection point, which is t =. =.. Applying the intersections of graphs method yields: x y = + [ 0, 0] by [, ] x + y = 9 The solution is the x-coordinate of the 9 intersection point, which is x =. =. 9. A= P( + rt) A= P+ Prt A P= P P+ Prt A P= Prt A P Prt = Pt Pt A P A P = r or r = Pt Pt x y = x y = x +. V = π r h LCD : ( V) π V = π = r h r h V π r h = πr πr V V = h or h= πr πr. F 9C = 60 F 9C+ 9C = C F = C F 60+ 9C = 9 60 F = C+ 9 F = C+ 6. c ( a x) = x+ LCD : c ( ( a x) ) = x+ a x = x+ c a x = x+ c a x x = x x+ c a 9x = c a a 9x = c a 9x = c a c a x = or 9 c a a c x = = 9 9 Copyright 0 Pearson Education. Inc.

12 CHAPTER Section P + A = m n LCD : P + A = m n P+ A= 0m n P+ A 0m= 0m n 0m P+ A 0m n = P+ A 0m = n P A 0m n = + m P A n = 8. y y = m( x x ) y y = mx mx y y + mx = mx mx + mx y y + mx mx = m m y y+ mx x = m 0. x+ y = 6 y = x+ 6 x + 6 y = y = x+. x + y = 6 y = 6 x 6 x y = y = x y x or = + 9. x y= y= x+ x + y = y= x. + = x y 8 = + y x 8 + y = y = x + x 8 Copyright 0 Pearson Education. Inc.

13 6 CHAPTER Linear Models, Equations, and Inequalities Section. Exercises. Let y = 690,000 and solve for x. 690,000 = 88,000 00x 8,000 = 00x 8,000 x = 00 x = 60 After 60 months or years the value of the building will be $690,000.. Let C = 0 and solve for F. F 9C = 60 F 9( 0) = 60 F 80 = 60 F = 0 0 F = = Fahrenheit equals 0 Celsius.. S = P( + rt) 9000 = P( + (0.0)()) 9000 = P( + 0.0) 9000 =.P 9000 P = = $6000 must be invested as the principal. 6. I = t 0.( h)( t 8) 79 = 80 0.( h)( 80 8) 79 = 80 0.( h)( ) 79 = 80.( h) 79 = 80.+.h 79 = h.h =.. h = =. h = A relative humidity of 9% gives an index value of M = 0.89W.. = 0.89W. 0.89W =.7.7 W = = The median annual salary for whites is approximately $, Recall that F 9C = 60. Let F = C, and solve for C. C 9C = 60 C = C = C = 0 Therefore, F = C when the temperature is Let B(x) = 0.9, and calculate x. 0.9 = x.87x = x = = 0.87 Copyright 0 Pearson Education. Inc.

14 CHAPTER Section. 6 In the year 00, 0 years after 980, the average monthly bill will be $ Let y = 9., and solve for x. 9. = 0.x = 0.x.0 = 0.x.0 x = 0. x = 6 An x-value of 6 corresponds to the year 996 ( ). The average reading score was 9. in Let y = 706, and solve for x. 706 = 80.9x + 8, ,8 = 80.9x,7 = 80.9x 7 x = 80.9 x = 0 An x-value of 0 corresponds to the year 00 ( ). The number of banks was 706 in 00. b. 0x x x = 00 0 Based on the model, producing and selling 00 specialty golf hats will avoid a loss.. Let F(x) = 9., and calculate x. 9. = 0.06x x =.9.9 x = = In the year 00 ( ), the percent of world forest area land will be 9.%.. Let y =.8, and calculate x..8 = 0.6x x =.7.7 x = = In the year 06 ( ), the marriage rate per 000 population will be.8.. a. Let P(x) = $8000, and calculate x = 0x 000 0x = x = = Based on the model, producing and selling 600 specialty golf hats will result in a profit of $ Note that p is in millions. A population of 0,000,000 corresponds to a p-value of 0. Let p = 0 and solve for x. 0 =.6x =.6x =.6x x =.6 x = An x-value of corresponds to the year 0 (960 + ). Based on the model, in Copyright 0 Pearson Education. Inc.

15 66 CHAPTER Linear Models, Equations, and Inequalities 0, the population is estimated to be 0,000, When the number of inmates is,787,000, then y =,787. Let y =,787, and calculate x.,787 = 76x +,787 = 76x 67 = 76x 67 x = = 76 An x-value of corresponds to the year 0 (990 + ). The number of inmates is estimated to be,787,000 in Let T(x) =,, and calculate x., = 6,66x 6,60 6, 66x = 9,97 9,97 x = = 6,66 In the year 0 (000 + ), the number of tweets will be, thousand. 8. Let p =.7, and solve for x x = 8 9x = 9x = x = 9 x = 0 In the year 00 ( ), the percent using marijuana daily was.7%. 9. Let y =.8, and solve for x..8 = 0.876x = 0.876x 8.76 = 0.876x 8.76 x = x = An x-value of 0 corresponds to the year 990 ( ). The model shows that it was in 990 that the number of Hispanics in the U.S. civilian population was.8 million. 60. Let I(x) = 0,9, and solve for x. 0,9 = 86.7x +. 0,9. = 86.7x 0,6.68 = 86.7x 0,6.68 x = 86.7 x 7 An x-value of 7 corresponds to the year 007 ( ). Based on the model, U.S. personal income reached $0,9 billion in a. Since the given function is described as linear, the increase of $.68 billion dollars between the years 00 and 00 represents the rate of change per year, or the slope of the function. Using the point-slope formula to determine the equation gives the following, where x =, the number of years since 000 (00-000), and y = $., the expenditure in that year. y y = m( x x) y. =.68( x ) y. =.68x 8.0 y =.68x+ 7. Copyright 0 Pearson Education. Inc.

16 CHAPTER Section. 67 b. Let y = 60, and solve for x. 60 =.68x =.68x.89 x = = Since x = 0 and x is the number of years after 000, the year in which expenditures are expected to exceed 60 billion dollars is If the number of customers is,700,000, then the value of S(x) is.70. Let S(x) =.70, and solve for x..70 = x = x 0 = x 0 x = = 0 An x-value of 0 corresponds to the year 00 (99 + 0). The model indicates that there were.7 million subscribers in Let x represent the score on the fifth exam x 90 = LCD: x 90 = = + x x = 97 The student must score 97 on the fifth exam to earn a 90 in the course. 6. Since the final exam score must be higher than 79 to earn a 90 average, the 79 will be dropped from computation. Therefore, if the final exam scores is x, the student s average x is. To determine the final exam score that produces a 90 average, let x = 90. LCD : x = 90 x = 60 x + 8 = 60 x = x = = 89 The student must score at least an 89 on the final exam. 6. Let x = the company s 999 revenue in billions of dollars. 0.9x = 7 7 x = 0.9 x 78.7 The company s 999 revenue was approximately $78.7 billion. 66. Let x = the company s 999 revenue in billions of dollars..79x = 6 6 x =.79 x 7. The company s 999 revenue was approximately $7. billion. Copyright 0 Pearson Education. Inc.

17 68 CHAPTER Linear Models, Equations, and Inequalities 67. Commission Reduction = ( 0% )( 0,000) = 0,000 New Commission = 0,000 0,000 = 0,000 To return to a $0,000 commission, the commission must be increased $0,000. The percentage increase is now based on the $0,000 commission. Let x represent the percent increase from the second year. 0,000x = 0,000 x = 0. = % 69. Total cost = Original price + Sales tax Let x = original price. 9,998 = x+ 6% x 9,998 = x+ 0.06x 9,998 =.06x 9,998 x = = 8,00.06 Sales tax = 9,998 8,00 = $ Let x = total in population. x 0 = 0 0 x = x = Salary Reduction = ( % )( 00,000) = 000 New Salary = 00, = 9,000 To return increase to a $0,00 salary, the new $9,000 must be increased $9,00. The percentage increase is now based on the $9,000 salary. Let x represent the percent raise from the reduced salary x = = 0 The estimated population is sharks. A= P+ Prt A P= P P+ Prt A P= Prt A P Prt = Pr Pr A P A P = t or t = Pr Pr 9,000x = x = = 0.0 = 0% 9, A= P( + rt) A = P + rt Copyright 0 Pearson Education. Inc.

18 CHAPTER Section A= P( + rt) A= P+ Prt A P= P P+ Prt A P= Prt A P Prt = Pt Pt A P = r Pt = r 000(6) 00 = = r r =. = 0% 7. A= P( + rt) A = P + rt 888 = P +.07() P = $00 I I 7. = r r 90 x = (.08) x = = $6.. I I 76. = t t = x 78.8() x = = 8 years Yes, the circumference of a circle varies directly with the radius of a circle since the relationship fits the direct variation format of y = kx, or in the case of a circle, C = πr, where the constant of variation is π. 78. Since y = kx is the direct variation format, let y = H (heat) and x = A (amount of protein). H = ka = k(), so k = Then H = A 80 = A A =.6 Thus the amount of proteins to be burned to produce 80 calories is.6 grams. 79. a. Since y = kx is the direct variation format, let y = B (BMI) and x = w (weight of person in kilograms). B= kw 0 = k(), so k = 9 Thus the constant of variation is /9. b. Then = A 9 9 () = A A = 7 The woman would weigh 7 kilograms. 80. Since y = kx is the direct variation format, let y = C (cost of land) and x = S (size of land). C = ks 7,800 = k(00), so k = 69. Then C = 69.(00) C = 80,60 Thus the cost of a piece of land that is 00 square feet would be $80,60. Copyright 0 Pearson Education. Inc.

19 70 CHAPTER Linear Models, Equations, and Inequalities Section. Skills Check. No. The data points do not lie close to a straight line.. Yes. The data points lie approximately in a straight line. 9. Using a spreadsheet program yields y 0 y =.x x. Can be modeled approximately since not all points lie exactly on a line.. Can be modeled exactly since all points line up with a straight edge.. 0. Using a spreadsheet program yields y y = 0.8x x. 6.. Yes. The points appear to lie approximately along a line.. Using a spreadsheet program yields 7. Exactly. The first differences of the outputs are constantly three. 8. No. The first differences are not constant. Also, a line will not connect perfectly the points on the scatter plot. y y =.9x x Copyright 0 Pearson Education. Inc.

20 CHAPTER Section. 7. f () =.9().7 = = f () =.9().7 =.09.7 = The second equation, y=.x+ 8, is a better fit to the data points. 6. Yes. The points appear to lie approximately along a line y y =.77x x 8. f () =.77()+.89 = = The first equation, y =.x+, is a better fit to the data points. f () =.77()+.89 = = Copyright 0 Pearson Education. Inc.

21 7 CHAPTER Linear Models, Equations, and Inequalities. a. Exactly linear. The first differences are constant. b. Nonlinear. The first differences vary, but don t grow continuously. c. Approximately linear. The first differences increase continuously.. The difference between inputs is not constant. The inputs are not equally spaced. Section. Exercises. a. Discrete. There are gaps between the years. b. Continuous. Gaps between the years no longer exist. c. No. A line would not fit the points on the scatter plot. A non-linear function is better.. a. Discrete. There are gaps between the years. b. No. A line would not fit the points on the scatter plot. c. Yes. Beginning in 00, a line would fit the points on the scatter plot well.. This data cannot be modeled exactly by a linear function since the increases from one bar top to the next are not constant. 6. a. Yes. There is a one unit gap between the years and a constant 60 unit gap in future values. b. S = f( t) = 60t+ 000, where 60 is the constant rate of change and (0, 000) is the y-intercept. c. For t = 7, S(7) = 60(7) = 0. Thus the future value of this investment at the end of the 7 th year is $0. This is an extrapolation since the result is beyond the original table values. d. Discretely, since the interest payments occur only at the end of the years. 7. a. Yes, the first differences are constant for uniform inputs. b. T = 0.x 8 Copyright 0 Pearson Education. Inc.

22 CHAPTER Section. 7 c. For x = 0,00, T = $680.00, and for x = 0,00, T = $ These results agree with the values found in the table. d. For x = 0,, T = $68.7. Since 0, is within the given tables values, the result is an interpolation from the data. The model and the scatterplot appear to be a fairly good fit.. a. Using a spreadsheet program yields e. No, just tax due for income in the % tax bracket. 8. No, the first differences are not equal. 9. a. Using a spreadsheet program yields y = 0.6x +.09 b. With x = 8, y = 0.9 or approximately 0.% of U.S. adults. c. With y =.0, x = 7.979, or approximately 8 years after 000, in the year 08. [0, 0] by [00, 600] b. Let x = = 0, and solve for y. y =.68(0) y = The model estimates the number of persons who took trips in 00 to 8.9 million, or.89 billion. 0. a. Using a spreadsheet program yields b. y = 7.070x +, c. c. Let y = 77., and solve for x. 77. =.68x =.68x.68x = x = = The model estimates that the year in which 77. million trips were taken was 6 years after the year 000, in 06. Copyright 0 Pearson Education. Inc.

23 7 CHAPTER Linear Models, Equations, and Inequalities. a. Using a spreadsheet program yields. a. Using a spreadsheet program yields b. y =.0x 8,69.8 c. The slope is.0, and for each increase of $.00 of median salary for whites, the black median salary will increase by $.0.. a. y = 0.x +.79 b. b. Yes the fit appears to be a reasonable one for this data. c. y =.9x +.99 d. The U.S. personal consumption for 0 (x = ) is estimated to be $,98. billion. 6. a. Using a spreadsheet program yields U.S. Population Projections 600 y =.90x [0, 0] by [0, ] Not close to an exact fit.. a. y =.7x b. When the CPI = 6.9, x = 0, which represents the year 00, 0 years after 970. Population, Millions Years past 000 b. f ( 6).7. The projected population in the year 06 is.7 million. c. In 080, x = 80. The population is predicted to be 99. million people, fairly close to the value in the table. Copyright 0 Pearson Education. Inc.

24 CHAPTER Section a. Using a spreadsheet program yields b. y = 6.90x.6 c. c. To determine the percent increase in the amount of spending from 009 to 00, subtract the y-values for each year and divide by the y-value of the year 009. Thus, from the table, =.69 = 6.9% 60 d. To determine the average rate of change of spending from 006 to 0, find the slope between the years = = Thus, the average rate of change of spending from 006 to 0 is $9.6 million dollars per year. 0. a. Using a spreadsheet program yields: S = 9.80x + 0,66.60 It appears to be a good fit. 8. a. y = 0.7x b. Since 0 is years after 990, x =. Solve for y = 8.7%. c. Let y= 6, and solve for x: 6 = 0.7x = 0.7x x = 0 Thus, when x = 0 years after 990, in the year 00, the percent will be a. Using a spreadsheet program yields: S =.9x 77.8 b. The annual increase in the amount of online spending in this linear function is the slope of the line, $.9 million dollars. b.. a. Since =, Let x=, and solve for y. y = 9.80( ) + 0,66.60 y =,0 Since 00 is within the table of values for the years 980 to 007, this is interpolation. c. Let y= 78,70, and solve for x: 78,70 = 9.80x + 0, ,8.7 = 9.80x x = Based on the model, the anticipated number of those imprisoned will be 78,70 in the year 0, years from 980. D= W b. Fits exactly Copyright 0 Pearson Education. Inc.

25 76 CHAPTER Linear Models, Equations, and Inequalities c. D = (0) = 68.8 mg d. Discretely. e. No, the U.S./Canada box office will not ever reach the International box office since the slope is less and thus rises slower.. a. P= 0.8W a. b. It is not a perfect fit.. a. b. y=.67x c. b. The model fits the data well. c. d. Copyright 0 Pearson Education. Inc.

26 CHAPTER Section. 77. a. b. y = 0.9x+ 9.6 c. The model fits the data well. d. 0.9x = x = =.. x = =. 0.9 Thus, in the year 0 (990 + ), the marriage rate is expected to be 6. per 000 population. Copyright 0 Pearson Education. Inc.

27 78 CHAPTER Linear Models, Equations, and Inequalities Section. Skills Check. To determine if an ordered pair is a solution of the system of equations, substitute each ordered pair into both equations. a. For the first equation, and the point (,) x+ y = () + () = + = 7 For the second equation, and the point (,) x y = 6 () () = = The point (, ) works in neither equation, thus is not a solution of the system. b. For the first equation, and the point (,) x+ y = ( ) + () = + = For the second equation, and the point (,) x y = 6 ( ) () = = 6 The point (,) works in both of the equations, thus is a solution of the system.. To determine if an ordered pair is a solution of the system of equations, substitute each ordered pair into both equations. a. b. For the first equaton, and the point (, ) x y = 7 ( ) = 6 + = 7 For the second equaton, and the point (, ) x+ y = + ( ) = = The point (, ) works in both of the equations, thus is a solution of the system. For the first equaton, and the point (, ) x y = 7 ( ) = + = For the second equaton, and the point (, ) x+ y = + ( ) = = The point (, ) works in only one of the equations, thus is not a solution of the system. Copyright 0 Pearson Education. Inc.

28 CHAPTER Section. 79. Applying the substitution method y = x and y = x x= x x x= x x x = x = x = x = Substituting to find y y = () =, is the intersection point.. Applying the elimination method x+ y = ( Eq) x y= ( Eq) 9x+ 6y = Eq 0x 6y= Eq 9x = 7 7 x = = 9 Substituting to find y + y = 9+ y = y = y =, is the intersection point. The solution for # is(, ). 6. Solving the equations for y x y= 6 y = x+ 6 x + 6 y = y = x and x+ y = 0 y = x+ 0 x + 0 y = y = x+ Applying the intersection of graphs method [ 0, 0] by [ 0, 0] The solution is (, ).. Applying the intersection of graphs method, for y = x and y = x + [ 0, 0] by [ 00, 0] Copyright 0 Pearson Education. Inc.

29 80 CHAPTER Linear Models, Equations, and Inequalities 7. Solving the equations for y x y = y = x x y = y = x+ and x y = y = x x y = y = x+ Applying the intersection of graphs method 8. Solving the equations for y x 6y = 6y = x+ x + y = 6 y = x 6 and x y = 6 y = x+ 6 x + 6 y = y = x Applying the intersection of graphs method [ 0, 0] by [ 0, 0] The solution is ( 0.7,0.7) or, [ 0, 0] by [ 0, 0] The solution is (, ). x+ y = 6 Eq x +.y = Eq x+ y = 6 ( Eq ) x y = 6 ( Eq ) 0= 0 There are infinitely many solutions to the system. The graphs of both equations represent the same line. Copyright 0 Pearson Education. Inc.

30 CHAPTER Section x+ y = Eq x+ y = Eq 6x+ y = Eq 6x y = 6 ( Eq ) 0= There is no solution to the system. The graphs of the equations represent parallel lines. x = y+ x+ y = Substituting the first equation into the second equation ( y+ ) + y = y+ 6 + y = 9y + 6 = 9y = 8 y = Substituting to find x x = ( ) + x = 0 + x = The solution is,. ( ). x y= y = x 8 Substituting the second equation into the first equation ( x ) x 8 = x x+ = x + = x = x = Substituting to find y y = 8 y = y = 6 y = The solution is,. Copyright 0 Pearson Education. Inc.

31 8 CHAPTER Linear Models, Equations, and Inequalities. x y = x+ y = Solving the first equation for x x y = x= y+ y + x = Substituting into the second equation y + + y = y + + y [] = y+ + 8y = y + = y = y = Substituting to find x x ( ) = x + = x = x = The solution is,.. x y = 7 x+ y = 7 Solving the first equation for x x y = 7 x= y 7 y 7 x = Substituting into the second equation 7 y + y = 7 y 7 + y [ 7] = y + 8y = 8 y = 8 y = y = Substituting to find x x () = 7 x = 7 x = x = The solution is,.. () x+ y = Eq x+ y = 8 Eq x 6y = 0 Eq x+ y = 8 Eq y = y = Substituting to find x x + = x + = x = The solution is,. Copyright 0 Pearson Education. Inc.

32 CHAPTER Section ( ) x y = Eq x+ 6y = Eq 8x 6y = 6 Eq x+ 6y = Eq x = x = Substituting to find x ( ) y = y = y = 9 y = The solution is,. x+ y = 8 Eq x+ y = 8 Eq 0x 6y = 6 Eq 0x+ 0y = 0 Eq y = y = = 7 Substituting to find x x + = x + 7[ 8] 7 = x + 8 = 6 x = 8 8 x = = 7 The solution is, x+ y = Eq x+ y = 8 Eq x y = 0 Eq 6x+ y = Eq 6x = x = = 6 Substituting to find y + y = + y = y = 7 7 y = 7 The solution is, x+ 0.y =. Eq x y= Eq 9x+ y = 7 0 Eq 0x y = Eq 9x = 6 6 x = = 9 Substituting to find y y = 0 y = y = 9 y = The solution is,. Copyright 0 Pearson Education. Inc.

33 8 CHAPTER Linear Models, Equations, and Inequalities x y = 0 Eq 0.x+ 0.y =. Eq x y = 0 Eq 0x+ y = 88 0 Eq x = 88 x = Substituting to find y 8 y = 0 6 y = 0 y = 6 y = The solution is,. ( Eq ) x+ 6y = Eq y 8= x Eq x+ 6y = Eq x+ y = 8 Eq 6x y = 6x+ y = Eq 0= 0 Infinitely many solutions. The lines are the same. This is a dependent system. 6y = x Eq 0x y= 0 Eq x+ 6y = Eq 0x y= 0 Eq 0x+ 0y= 60 Eq 0x 0y = 60 Eq 0= 0 Infinitely many solutions. The lines are the same. This is a dependent system.... 6x 9y= Eq x.y = 6 Eq 6x 9y = Eq 6x + 9y = Eq 0= No solution. Lines are parallel. This is an inconsistent system. x 8y = Eq 6x y= 0 Eq x y = Eq x + y = 0 Eq 0= No solution. Lines are parallel. This is an inconsistent system. y= x y = x 6 Substituting the first equation into the second equation x = x 6 x = 6 x = x = Substituting to find y y = = 6 = The solution is,. Copyright 0 Pearson Education. Inc.

34 CHAPTER Section y = 8x 6 y = x Substituting the first equation into the second equation 8x 6 = x 6x = 6 x = Substituting to find y y = () y = The solution is,. 7. x+ 6y = x= y+ 8 Substituting the second equation into the first equation ( y+ 8) + 6y = 6y+ + 6y = y + = y = 8 8 y = = Substituting to find x x = x = + = The solution is, y= x x y = 7 Substituting the first equation into the second equation ( x ) x = 7 x 6x+ 0= 7 x = x = Substituting to find y () y = y = ( ) The solution is,. ( ) x y = 6 Eq 6x 8y = Eq 6x+ y = 8 Eq 6x 8y = Eq 7y = y = Substituting to find x x = 6 x + 0= 6 x = 6 x = The solution is,. Copyright 0 Pearson Education. Inc.

35 86 CHAPTER Linear Models, Equations, and Inequalities 0.. x y = Eq 6x+ y = Eq x y = Eq 6x+ y = Eq 8x = 7 7 x = = 8 Substituting to find y y = 6 y = y = y = The solution is,. () x= 7y Eq x+ y = Eq x 7y = Eq x+ y = Eq x+ 8y = Eq x+ 9y = Eq 7 y = 7 7 y = = 7 Substituting to find x x 7 = x 7= x = 6 x = The solution is,.... x= + y Eq x y= 8 Eq x y = Eq x y= 8 Eq x+ 9y = 6 Eq x y= 0 Eq 6y = y = = 6 Substituting to find x x = x + = x + = 0x + = 8 0x = 9 x = = 0 9 The solution is,. x y = 9 Eq 8x 6y = 6 Eq 8x+ 6y = 8 Eq 8x 6y = 6 Eq 0= No solution. Lines are parallel. x y = 8 Eq x + y = Eq x y= Eq x + y = Eq 0= No solution. Lines are parallel. Copyright 0 Pearson Education. Inc.

36 CHAPTER Section. 87 Section. Exercises. 6. R= C 76.0x= x 9.0x = x = 9.0 x = 60 Applying the intersections of graphs method yields 60 x =. [ 0, 00] by [ 0, 0,000] Break-even occurs when the number of units produced and sold is 60. R= C 89.7x=.0x x = x = 66. x = 8 Applying the intersections of graphs method yields 8 x =. [ 0, 00] by [ 000, 000] R= C.80x= x.60x = x =.60 x = 68 Break-even occurs when the number of units produced and sold is 68. R= C 6.0x= x 9.90x = x = 9.90 x = 0 Break-even occurs when the number of units produced and sold is a. p+ q= 0 ( Eq) p 8q = 0 ( Eq) p+ 8q= 80 Eq p 8q = 0 ( Eq ) p = p = = 60 The equilibrium price is $60. b. Substituting to find q 60 8q = 0 8q = 0 0 q = = 0 8 The equilibrium quantity occurs when 0 units are demanded and supplied. Break-even occurs when the number of units produced and sold is 8. Copyright 0 Pearson Education. Inc.

37 88 CHAPTER Linear Models, Equations, and Inequalities 0. a.. a. Let p= 60 and solve for q. Supply function 60 = q + 0 q = 0 q = 8 Demand function 60 = 8 q q = q = = 7 When the price is $60, the quantity supplied is 8, while the quantity demanded is 7. b. Equilibrium occurs when the demand equals the supply, q+ 0 = 8 q 9q + 0= 8 9q = q = = 9 Substituting to calculate p p = + 0= 80 When the price is $80, units are produced and sold. This level of production and price represents equilibrium. For the drug Concerta, let x = 0, y =., and x =, y = 0. y y 0. Then, = = 0.69 x x 0 Since x = 0, y is the y-intercept. Thus the equation is y = 0.69x+., representing the market share for Concerta as a linear function of time. b. For the drug Ritalin, let x = 0, y = 7.7, and x =, y = 6.9. y y = = Then, 0.07 x x 0 Since x = 0, y is the y-intercept. Thus the equation is y = 0.07x+ 7.7, representing the market share for Ritalin as a linear function of time. c. To determine the number of weeks past the release date when the two market shares are equal, set the two linear equations equal, and solve for x. Thus, 0.69x+. = 0.07x x =. x = weeks. Equilibrium occurs when the demand equals the supply, q= 00.70q.0q = 9.0 q = Substituting to calculate p p = = $6.90 When the price is $6.90, units are produced and sold. This level of production and price represents equilibrium.. a. Applying the intersection of graphs method [0, 0] by [ 00, 000] The solution is 7.,.787. The number of active duty Navy personnel Copyright 0 Pearson Education. Inc.

38 CHAPTER Section. 89 equals the number of active duty Air Force personnel in 987. b. Considering the solution in part a, approximately,787 people were on active duty in each service branch.. Applying the intersection of graphs method 6. B = 0.6W B =.0W 8.69 Setting the first equation equal to the second equation solve for x: 0.6W =.0W W = W = = $, Thus, the annual earning amount by whites that will give the same median annual earnings for blacks in both models is $,70. [0, 0] by [ 0, 00] The solution is ( 8.7,.6 ). In 979, the percent of male students enrolled in college within months of high school graduation equaled the percent of female students enrolled in college within months of high school graduation.. yh = 0.x+ 9.0 (the % of Hispanics) yb = 0.07x+. (the % of blacks) Setting the first equation equal to the second equation, solve for x: 0.x+ 9.0 = 0.07x+..67x =.. x = = Let h = the 0 revenue, and let l = the 008 revenue. h+ l =.9 Eq h l =. Eq h+ l =.9 Eq h l = 7 Eq h =.9.9 h = = 80. Substituting to calculate l 80. l =. l = l = The 008 revenue is $669.8 million, while the 0 revenue is $80. million. Thus, 0 years after 980, in the year 000, the percent of Hispanics in the U.S. civilian non-institutional population equaled the percent of blacks. Copyright 0 Pearson Education. Inc.

39 90 CHAPTER Linear Models, Equations, and Inequalities 8. Let l = the low stock price, and let h= the high stock price. h+ l = 8. ( Eq ) h l =.88 ( Eq ) h = h = =.69 Substituting to calculate l.69 + l = 8. l = 0.8 The high stock price is $.69, while the low stock price is $ a. x+ y = 00 b. 0x c. y 0. a. x+ y = 00,000 b. 0% x or 0.0x c. % y or 0.y d. 0.0x+ 0.y =,000 e. x+ y= 00,000 Eq 0.0x+ 0.y=,000 Eq 0.0x 0.0y = 0, Eq 0.0x+ 0.y=,000 Eq 0.0y = y = = 0, Substituting to calculate x x + 0, 000 = 00, 000 x = 0, 000 d. 0x+ y = 8,000 e. x+ y = 00 Eq 0x+ y = 8,000 Eq 0x 0y = 7,000 Eq 0x+ y = 8,000 0 Eq y =,000,000 y = = 800 Substituting to calculate x x = 00 x = 600 The promoter needs to sell 600 tickets at $0 per ticket and 800 tickets at $ per ticket.. a. $0,000 is invested in the 0% property, and $0,000 is invested in the % property. Let x = the amount in the safer account, and let y = the amount in the riskier account. x+ y = 00,000 Eq 0.08x+ 0.y = 9,000 Eq 0.08x 0.08y = 8, Eq 0.08x+ 0.y = 9,000 Eq 0.0y = y = =, Substituting to calculate x x +,000 = 00,000 x = 7,000 b. $7,000 is invested in the 8% account, and $,000 is invested in the % account. Using two accounts minimizes investment risk. Copyright 0 Pearson Education. Inc.

40 CHAPTER Section. 9. a. Let x = the amount in the safer fund, and let y = the amount in the riskier fund. x+ y =, 000 ( Eq ) 0.0x+ 0.y =,70 ( Eq ) 0.0x 0.0 y =, Eq 0.0x+ 0.y =,70 ( Eq ) 0.0y = 0 0 y = =, Substituting to calculate x x +,000 =,000 x = 9, 000 Thus, $9,000 is invested in the 0% fund, and $,000 is invested in the % fund.. Let x = the amount in the money market fund, and let y = the amount in the mutual fund. x+ y = 00,000 Eq 0.06x y =.076 x+ y Eq 6x+ 9y = 76x+ 76y 000 Eq 6y= x 7 y= x= x Then substituting x for y in Eq, gives 8 7 x+ x= 00, x+ 7x=,00,000 x =,00,000 x = 60,000 Substituting to calculate y. Let x = the amount in the money market fund, and let y = the amount in the mutual fund. x+ y = 0,000 Eq 0.066x y =.07 x+ y Eq 66x+ 86y = 70x+ 70y 000 Eq 6y= x y = x Then substituting y for x in Eq, gives y+ y = 0,000 y = 0,000 y = 0,000 Substituting to calculate x x + 0,000 = 0,000 x = 00,000 Thus, $00,000 is invested in the money market fund, and $0,000 is invested in the mutual fund.. 7 y = (60,000) = 0,000 8 Thus, $60,000 is invested in the money market fund, and $0,000 is invested in the mutual fund. Let x = the amount (cc's) of the 0% solution, and let y = the amount (cc's) of the % solution. x+ y = Eq x 80 x y = 00 Eq Then using the elimination method x = 00 x = 60 cc's of the 0% solution Substituting to calculate y y = 0 cc's of the % solution 00 (cc's of the final solution) ( ) x+ y = ( Eq ) + y = 0 00 ( Eq ) Thus, 60 cc s of the 0% solution should be mixed with 0 cc s of the % solution to get 00 cc s of the 8% solution. Copyright 0 Pearson Education. Inc.

41 9 CHAPTER Linear Models, Equations, and Inequalities 6. Let x = the amount (cc's) of the 6% solution, and let y = the amount (cc's) of the 6% solution. 00 (cc's of the final solution) ( ) x+ y = ( Eq ) + = ( q ) x+ y = Eq x 6y E 6x 6y = 00 Eq Then using the elimination method 0x = 00 x = 0 cc's of the 6% solution Substituting to calculate y y = 80 cc's of the 6% solution Thus, 0 cc s of the 6% solution should be mixed with 80 cc s of the 6% solution to get 00 cc s of the % solution. 7. Let x = the number of glasses of milk, and let y = the number of /-pound servings of meat. Then the protein equation is: 8.x+ y = 69. Eq and the iron equation is: + = x+ y = ( Eq ) + y = 7 0 ( Eq ) x y ( Eq ) 0.x.y 7. Eq x = y = 0 0 y = = 670 Substituting to calculate x 8.x + = x + = x =.. x = = 8. Thus, glasses of milk and servings of meat provide the desired quantities for the diet. Copyright 0 Pearson Education. Inc.

42 CHAPTER Section Let x = the amount of substance A, and let y = the amount of substance B. Nutrient equation: 6% x+ 0% y = 00% x+ y = ( Eq) 0.06x+ 0.0y= ( Eq) 0.06x 0.06y = ( Eq) 0.06x+ 0.0y = ( Eq ) 0.0y = y = = 0.0 Substituting to calculate x x + = x = 0 The patient needs to consume 0 ounces of substance A and ounces of substance B to reach the required nutrient level of 00%. 9. To find when the percent of those enrolled equals that of those not enrolled, set the equations equal to each other and solve for x. Then find the percent by substituting into either one of the original equations. 0.8x+ 9. = 0.086x x =.9 x = 0., the number of years after the year 000, during the year 0. Substituting to calculate y: y = 0.8(0.) + 9. y = % It is estimated that in the year 0, the percent of young adults aged 8 to who use alcohol will be the same for those enrolled in college as for those not enrolled in college. That percent will be % Let x = the amount of the 0% solution, and let y = the amount of the % solution. x+ y = Medicine concentration: 0% x+ % y = 0% ( ) x+ y = ( Eq) 0.0x+ 0.y = 9 ( Eq) 0.0x 0.0y=. 0.0 ( Eq) 0.0x+ 0.y = 9 ( Eq ) 0.y =.. y = = 0 0. Substituting to calculate x x + 0 = x = The nurse should mix cc of the 0% solution with 0 cc of the % solution to obtain cc of a 0% solution. a. Demand function: Finding a linear model using L as input and L as output yields p = q+. Copyright 0 Pearson Education. Inc.

43 9 CHAPTER Linear Models, Equations, and Inequalities b. Supply function: Finding a linear model using L as input and L as output yields p = q+ 0. Supply function: Finding a linear model using L as input and L as output yields p = q+ 0 or p = q. c. Applying the intersection of graphs method Applying the intersection of graphs method [0, 00] by [ 0, 00] When the price is $8, 0 units are both supplied and demanded. Therefore, equilibrium occurs when the price is $8 per unit. [0, 800] by [ 00, 800] When the price is $00, 600 units are both supplied and demanded. Therefore equilibrium occurs when the price is $00 per unit Applying the intersection of graphs method Demand function: Finding a linear model using L as input and L as output yields p= q [, ] by [-, ] Although the lines do intersect, they intersect at a negative x-value which indicates they were equal prior to 990. Since the problem calls for a time after the year 000, there is no solution to the system. Copyright 0 Pearson Education. Inc.

44 CHAPTER Section % x+ % y =.% ( x+ y) 0.0x+ 0.0y = 0.x+ 0.y 0.0x= 0.0y 0.0x 0.0y = y = x 7 Therefore, the amount of y must equal of the amount of x. 7 If x= 7, y = ( 7) =. 7 The % concentration must be increased by cc. 6. a. 00x+ 00y= 00,000 b. x= y 00( y) + 00y = 00, y = 00,000 00,000 y = = 800 Substituting to calculate x x = = 0 There are 0 clients in the first group and clients in the second group. 66. The slope of the demand function is y y m = = = =. x x Calculating the demand equation: y y = m( x x) y 0 = ( x 800) 8 y 0 = x y = x+ 0 or 8 p= q+ 0 8 Likewise, the slope of the supply function is y y m = = = =. x x Calculating the supply equation: y y = m( x x) y 80 = ( x 700) 0 y 80 = x 0 0 y = x+ 7 or 0 p= q+ 7 0 To find the quantity, q, that produces market equilibrium, set the equations equal. q+ 0 = q LCD : 0 0 q+ 0 = 0 q q+ 8,000 = 6q q =,000,000 q = = 000 To find the price, p, at market equilibrium, solve for p. p = (000) + 0 = 8 Thus, the market equilibrium point is (000, ). If the price per unit is $, both supply and demand will be 000 units. Copyright 0 Pearson Education. Inc.

45 96 CHAPTER Linear Models, Equations, and Inequalities 67. The slope of the demand function is y y m = = = =. x x Calculating the demand equation: y y = m x x y 0 = y 0 = x y = x+ 00 or 0 p= q ( x ) To find the price, p, at market equilibrium, solve for p. p = ( 700 ) p = p = units priced at $0 represents the market equilibrium. Likewise, the slope of the supply function is y y m = = = =. x x Calculating the supply equation: y y = m( x x) y 0 = ( x 700) 70 y 0 = x 0 70 y = x+ 0 or 70 p= q To find the quantity, q, that produces market equilibrium, set the equations equal. q+ 00 = q q+ 00 = 70 q q = q q = q = = Copyright 0 Pearson Education. Inc.

46 CHAPTER Section. 97 Section. Skills Check. Algebraically: x 7 x x 7 x x x Graphically: y = x y = x 7 Graphically: y = x+ 6 y = x+ [, ] by [, ] x+ 6 < x+ implies that the solution region is x >. The interval notation is,. The graph of the solution is: [ 0, 0] by [ 0, 0] x 7 x implies that the solution region is x. The interval notation is (,]. The graph of the solution is:. Algebraically: x+ 6< x+ x + 6< x < x > (Note the inequality sign switch) x >. Algebraically: ( x ) x 9 x 8 x 9 7x 7x 7 7 x 7 Graphically: y = (x ) [, ] by [, ] y = x 9 Copyright 0 Pearson Education. Inc.

47 98 CHAPTER Linear Models, Equations, and Inequalities ( x ) x 9 implies that the solution region is x. 7 The interval notation is,. 7 The graph of the solution is:. Algebraically:. Algebraically: x+ < x+ ( x+ ) < x+ 0x+ < x+ x + < x < 0 0 x < ( x ) > x+ 6 0x > x+ 6 6x > 6 6x > x > x > Graphically: 6 7 y = x+ 6 y = (x ) Graphically: y = x+ y = x+ [ 0, 0] by [ 0, 0] x+ < x+ implies that the solution 0 region is x <. 0 The interval notation is,. [ 0, 0] by [, ] The graph of the solution is: ( x ) >x+ 6 implies that the solution 7 region is x >. 7 The interval notation is,. The graph of the solution is: Copyright 0 Pearson Education. Inc.

48 CHAPTER Section Algebraically: x x + x 6 x 6 + x + x x x 9 9 x 7. Algebraically: x 8 < x 8 0 < 0 ( x ) < ( 8) x < 6 x < 6 6 x < Graphically: y = x Graphically: 8 y = x y = + y = x [ 0, 0] by [ 0, 0] x x + implies that the solution 9 region is x. 9 The interval notation is,. The graph of the solution is: [ 0, 0] by [ 0, 0] x 8 < implies that the solution 6 region is x <. 6 The interval notation is,. The graph of the solution is Copyright 0 Pearson Education. Inc.

49 00 CHAPTER Linear Models, Equations, and Inequalities 8. Algebraically: x 6 < x 6 < ( x ) < 6 x 9< 6 x < 7 7 x < Graphically: 6 y = 9. Algebraically: ( x ) 6 x ( x 6) x 0 0 ( ( x )) ( x) ( x ) x x 90 x 0 x 90 0 x 0 0 x y = x Graphically: ( x ) 6 y = [ 0, 0] by [ 0, 0] x 6 < implies that the solution 7 region is x <. 7 The interval notation is,. The graph of the solution is: x y = [ 0, 0] by [, ] ( x ) 6 x implies that the solution 0 region is x. 0 The interval notation is,. The graph of the solution is: Copyright 0 Pearson Education. Inc.

50 CHAPTER Section Algebraically: ( x ) x 8 ( x ) x 8 0x 0 9x 0 x 80 Graphically: y = x 8 y ( x ) = Graphically: y =.x.6 y = 6 0.8x [ 0, 0] by [ 0, 0].x x implies that the solution region is x.86. The interval notation is.86,. The graph of the solution is: ) [ 90, 70] by [ 60, 0] ( x ) x 8 implies that the solution region is x 80. The interval notation is [ 80, ). The graph of the solution is:. Algebraically:.x x x.. x x. Graphically:. Algebraically:.x x.0x.6 6.0x x.0 x.86 y = 8 0.x [ 0, 0] by [ 0, 0] y =.x 6..x x implies that the solution region is x.. The interval notation is (,.]. Copyright 0 Pearson Education. Inc.

51 0 CHAPTER Linear Models, Equations, and Inequalities The graph of the solution is: Let f( x) = x+, and determine graphically where f( x) 0. f ( x) = x+. Applying the intersection of graphs method yields: y = 7x+ y = x 7 [ 0, 0] by [ 0, 0] [ 0, 0] by [ 0, 0] 7x+ < x 7 implies that the solution region is x <. The interval notation is (, ).. Applying the intersection of graphs method yields: y = x+ y = 6x f( x) 0 implies that the solution region is x 8. The interval notation is [ 8, ). 6. To apply the x-intercept method, first rewrite the inequality so that zero is on one side of the inequality. ( x ) < ( x ) x+ < 6x 9x + < 0 Let f( x) = 9x+, and determine graphically where f( x) < 0. f ( x) = 9x+ [ 0, 0] by [ 0, 0] x+ 6x implies that the solution region is x. The interval notation is [, ).. To apply the x-intercept method, first rewrite the inequality so that zero is on one side of the inequality. ( x+ ) 6( x ) 0x+ 0 6x x + 0 [ 0, 0] by [ 0, 0] f( x) < 0 implies that the solution region is x >.. The interval notation is,. 9 Copyright 0 Pearson Education. Inc.

52 CHAPTER Section a. The x-coordinate of the intersection point is the solution. x =. b. (, ) 8. a. x = 0 b. (,0] c. No solution. f( x ) is never less than h( x ).. 6x 8 > and 6x 8 < < 6x 8 < 0 < 6x < 0 0 6x 0 < < < x < x> and x< The interval notation is, x < 7+ x + < + x < 6 x 6 < x < The interval notation is, < 0x < 0x < 0x 00 8< x 0 ( ] The interval notation is 8,0.. x+ 6 and x+ 6 x + x 0 x 0 x and x 0 The interval notation is,0.. x+ < 7 and x > 6 Inequality x + < 7 x < 8 8 x < Inequality x > 6 x > x > 8 x< and x> Since and implies that these two inequalities must both be true at the same time, and there is no set of numbers where this occurs, there is no solution to this system of inequalities. Copyright 0 Pearson Education. Inc.

53 0 CHAPTER Linear Models, Equations, and Inequalities. 6x or x+ > 9 Inequality 6x 6x x Inequality x + > 9 x > x > x or x> Since or implies that one or the other of these inequalities is true, the solution for this system, in interval notation, is,,. x 6 x or x x Inequality x ( 6 x) x 8 8x x x Inequality x ( x ) x 6x 6 x x x or x system, in interval notation, is,, 6. x < x or x > 6x Inequality x < ( x) x 6< 0x 9x < 6 x > Inequality x > ( 6x) x > 0x 8x > x < 8 x> or x< 8,, 8 Since or implies that one or the other of these inequalities is true, the solution for this Copyright 0 Pearson Education. Inc.

54 CHAPTER Section x x x x x 6 [ ] The interval notation is 7, x < x 70 < x x x x < The interval notation is 6,. [ ) Copyright 0 Pearson Education. Inc.

55 06 CHAPTER Linear Models, Equations, and Inequalities Section. Exercises 9. a. V =, t b., t < 8000 c., t a. p 0. b. Considering x as a discrete variable representing the number of drinks, then if x 6, the 0-lb male is intoxicated.. F 9 C + 9 C 0 C 0. A Celsius temperature at or below zero degrees is freezing. C 00 ( F ) F F 060 F ( F ) [ ]. Position income = 00 Position income = x,where x represents the sales within a given month. The income from the second position will exceed the income from the first position when x > x > x > 0.0 x >, 000 When monthly sales exceed $,000, the second position is more profitable than the first position.. Original value = ( 000)( ) = $,000 Adjusted value =,000 (,000)( 0% ) =, = 7,600 Let x = percentage increase 7, ,600x >,000 7,600x > x > 7,600 x > 0. x > % The percent increase must be greater than % in order to ensure a profit. A Fahrenheit temperature at or above degrees is boiling. Copyright 0 Pearson Education. Inc.

56 CHAPTER Section Let x = Stan's final exam grade x x x x 0 60 x 60 x 80 x 07 If the final exam does not contain any bonus points, Stan needs to score between 80 and 00 to earn a grade of B for the course. Let x = John's final exam grade x x x 9 8 x 7 8 x 7 6 x 86. b. Solve for y between the years 000 ( x = 0) and 009 ( x= 9) : 8. y y y. 6. y., or. y 6. Thus the range of cigarette use for the years 000 to 009 is.% to 6.%. 8. a. y x x x 0.97 x or approximately x 899 b. Old scores greater than or equal to 899 are equivalent to new scores. John needs to score between 6 and 86. to earn a grade of C for the course. 7. a..x+ y = 8. Solve for x :.x= 8. y 8. y x =. 9. a. Copyright 0 Pearson Education. Inc.

57 08 CHAPTER Linear Models, Equations, and Inequalities y = 0 8.x + 0. = 0 8.x = x = 8. x = 7 Thus, the number of doctorates employed was 0 seven years after the year 000, in the year 007. c. Prior to 007, the number of doctorates employed was below 0. b. 0. If x is the number of years from 99, and if y =.8x +7., then the percent of households in the U.S. with Internet access greater than 88 is given by:.8x x x.9 years after 99 Therefore, years from 99 is 009 and after.. Let x represent the actual life of the HID headlights. x 00 0% % x x 60. Remember to convert years into months. < y < 6 8 < 0.x.886 < < 0.x < < 0.x < x < < < x <.78 or approximately, 9 < x < Copyright 0 Pearson Education. Inc.

58 CHAPTER Section. 09 Thus, the prison sentence should be between 9 months and months.. Let y > 9. y y m = 0.6x +.07 > 9 x x 0.6x > , , 000 = 0.6x > x < =. 80, = = 0, 000 Thus when x < years after 980, before 99, the marriage rate per 000 is greater than 9. Solving for the equation: y y = m( x x) Let y < 8. y 90,000 = 0,000( x 0) 0.6x +.07 < 8 y 90,000 = 0,000x 0.6x < 8.07 y = 0,000x+ 90, x < b. x > =. 0.6 y > 00,000 0,000x + 90,000 > 00,000 Thus when x > years after 980, after 0,000x > 0,000 00, the marriage rate per 000 is less than 8. 0,000 x > 0,000 x > 0.. Note that W and B are in thousands. 00 corresponds to B 00 x = =..0W Therefore, x..0w W.0 W.09 Or, y > 00,000 between 007 and 00, inclusive. The value of the home will be greater than $00,000 between 007 and 00. The median salary for whites would be at least $,09.. a. Since the rate of increase is constant, the equation modeling the value of the home is linear. c. Since the housing market took such a down turn in the last year or so, it does not seem reasonable that this model remained accurate until the end of 00. Copyright 0 Pearson Education. Inc.

59 0 CHAPTER Linear Models, Equations, and Inequalities 6. Cost of cars =,00 = 90,000 Cost of cars =,00 = 7,00 th Let x = profit on the sale of the car. (.% )( 7,00) + x ( 6% )( 90,000) 9, x,00 x 77.0 th The price of the car needs to be at least, = 6,7.0 or $6,8. 7. Px > 0,900 6.x 000 > 0,900 6.x >,900,900 x > 6. x > 000 A production level above 000 units will yield a profit greater than $0, Px > 8, 0, x > 8, 9.80x >, 60,60 x > 9.80 x >,7.06 Rounding since the data is discrete: x >,7 The number of units produced and sold should exceed,7. 9. Px 0 6.x x x 6. x Generating a loss implies that Px < 0 0, x < x < 0, 0, x < 9.80 x < Rounding since the data is discrete: x < 08 Producing and selling fewer than 08 units results in a loss.. Recall that Profit = Revenue Cost. Let x = the number of boards manufactured and sold. Px = Rx Cx Rx = 89x Cx = x+,000 Px = 89x ( x+,000) Px = 89x x,000 Px = 6x,000 To make a profit, Px > 0. 6x,000 > 0. T 8 0.m m m 0. m or approximately, m 9 The temperature will be at most 8 F for the first 9 minutes. Sales of 00 feet or more of PVC pipe will avoid a loss for the hardware store. Copyright 0 Pearson Education. Inc.

60 CHAPTER Section.. Since at least implies greater than or equal to, H( x).6. Thus, 0.x x.6.6 x 0. x Thus, the Hispanic population is at least.6%, years after 990, on or after the year 0.. a. < y < 8 < 0.x +.7 < 8.7 < 0.x < < 0.x < < x < < x <. Considering x as a discrete variable yields < x<. From 97 (970 + ) until 99 (970 + ), the reading scores were between and 8.. a. x = = 8 p(8) = (8) = =.9.% In 008, the percent of high school seniors who used cigarettes is estimated to be.%. b. 6. a. Since "at least" implies greater than or equal to, Bx.. Thus, 0.07x x.. x 0.07 x 0 Thus, the percent of blacks is at least.%, 0 years after 990, on or after the year 00. p 7.08 when x c. Considering part b) above, the model suggests that the percent is at least 7.08% before the year 00. Copyright 0 Pearson Education. Inc.

61 CHAPTER Linear Models, Equations, and Inequalities Chapter Skills Check. a.. a. x+ = 8x x 8x+ = 8x 8x x + = x + = x = x = x = = 6.8 b. Applying the intersection of graphs method yields x = 6.8. [, ] by [ 0, 60] ( x 7) = ( x+ ) x x = x+ x x = x+ x x = x x+ x = x + = + x = 9 x 9 = 9 x = =.. a. b. Applying the intersection of graphs method yields x =.. [ 0,0] by [ 0, 0] ( x ) x x = 8 LCD: ( x ) x x = 8 ( ( x ) ) x = 0 x ( x 6) x = 0 x 9x 8 x = 0 x 6x 8 = 0 x x 8 = 0 x = 8 x = 8 b. Applying the intersection of graphs method yields 8 x =. [ 0, 0] by [ 0, 00] Copyright 0 Pearson Education. Inc.

62 CHAPTER Skills Check. a. 6x+ ( x) = 6x+ 0 x = LCD: 6 6x+ 0 x 6 = 6 6 ( x+ ) = 0 ( x) 8x+ = 0 0x 8x+ + 0x= 0 0x+ 0x 8x + = 0 8x + = 0 8x = x = b. Applying the intersection of graphs method yields x = / b. Applying the intersection of graphs method yields x = /.096. [ 0,0] by [ 0, 0] 6. a..9x =.8(8.6x+.97).9x =.77x x = 80.7 x = b. Applying the intersection of graphs method yields x =. [ 0,0] by [ 0, 0] [ 0, 0] by [ 00, 0]. a. x = x 6 x = x + 9 LCD: 6 x 6 = 6 x + 9 7x = 6 x+ x = x = a. f( x) = 7x 0 7x 0 = 0 7x = 0 x = b. The x-intercepts of the graph are the same as the zeros of the function. c. Solving the equation f( x ) = 0is the same as finding the zeros of the function and the x-intercepts of the graph. Copyright 0 Pearson Education. Inc.

63 CHAPTER Linear Models, Equations, and Inequalities 8. Solve for y : m Pa ( y) = + m Pa Py = + m Py = + Pa + m Pa Py = + m Pa y = P.. No. The data points in the table do not fit the linear model exactly. 9. Solve for y : x y = 6 y = x+ 6 x+ 6 x 6 y = = y = x. ( ) x+ y = 0 Eq x y = 7 Eq x+ y = 0 Eq x y= Eq 7x = x = Substituting to find y + y = 0 6+ y = 0 y = 6 y = The solution is,. 0.. y =.897x. Copyright 0 Pearson Education. Inc.

64 CHAPTER Skills Check x+ y = Eq x y = Eq 9x+ 6y = 9 Eq x 6y = 6 Eq x = x = Substituting to find y + y = y = 0 y = y = The solution is,. x+ y= Eq x y = 7 Eq x+ y = Eq x y= Eq 0= 0 Dependent system. Infinitely many solutions. 6x+ y= 0 Eq x y = Eq 6x+ y = 0 Eq 6x y = 0 Eq 0= 0 No solution. Lines are parallel ( ) x+ y = 9 Eq x y = Eq x+ y = 9 Eq x y = Eq y = Substituting to find x x + ( ) = 9 x + = 9 x = 6 x = The solution is,. x+ y = Eq x y = 0 Eq x+ y = 6 Eq x y = 0 Eq 8x = x = Substituting to find y + y = + y = y = The solution is,. 0. Algebraically: x+ 8< x x + 8< x < x < = 0.8 Copyright 0 Pearson Education. Inc.

65 6 CHAPTER Linear Models, Equations, and Inequalities Graphically: y = x+ 8 y = x x x + implies that the solution is x. 8 The interval notation is,. 8 [ 0, 0] by [ 0, 0] x+ 8 < x implies that the solution is x <. The interval notation is,.. Algebraically: x x + x 0 x 0 + 0x x+ 0 8x 0 8x x Graphically:. Algebraically: 8 x + 6 < 8 6 x < 6 x < 6 x 6 < 6 x < 8 Graphically: y = y = x+ 6 y = 8 [, ] by [ 0, 0] 8 x + 6 < implies that the solution is 6 x < 8. The interval notation is 6,8. [ ) x y = + y = x [ 0, 0] by [ 0, 0] Copyright 0 Pearson Education. Inc.

66 CHAPTER Review 7 Chapter Review Exercises. a. Yes. As each amount borrowed increases by $000, the monthly payment increases by $89.6. b. Yes. Since the first differences are constant, a linear model will fit the data exactly. c. Car Loans Monthly Payment y = 0.08x Amount Borrowed Rounded model: P= f( A) = 0.08A Unrounded model: f( A) = 0.079A t +,90.77 = 0,760. t = 8, and the year is 998 ( ). 6. f( x ) = a. Let x = the number of years after 990, and f( x) = the asparagus production in Switzerland. Then f( x) = 00. b. This is a constant function. 8. a. Let x = monthly sales. 00 = % x 00 = x 00 = 0.0x 00 x = =, If monthly sales are $,000, both positions will yield the same monthly income.. a. Using the linear model from part c) in problem : f ( 8,000) = 0.08( 8,000) = 0.0 The predicted monthly payment on a car loan of $8,000 is $0.0. b. Considering the solution from part a), if sales exceed $,000 per month, the second position will yield a greater salary. b. Yes. Any input could be used for A. c. f( A) A A A A 7,89.00 The loan amount should be less than or equal to $7, Copyright 0 Pearson Education. Inc.

67 8 CHAPTER Linear Models, Equations, and Inequalities 9. Original Cost =,000 = 88,000 Desired Profit = 0% ( 88,000) = 8,800 Revenue from 8 sold cars, with % avg profit = 8(,000 + %,000) = 8(,000 +,880) =,00 Solve for x where x is the selling price of the remaining four cars. The total Revenue will be,00 + x. The desired Profit = Revenue Cost 8,800 = (,00 + x) 88,000 8,800 = x 7,960 x = 0,760 0, 760 x = =,0 The remaining four cars should be sold for $,0 each. 0. Let x = amount invested in the safe account, and let 0,000 x = amount invested in the risky account. 6% x+ 0% ( 0,000 x) = 0, x+, x= 0, x =,000,000 x = 0.0 x = 00,000 The couple should invest $00,000 in the safe account and $0,000 in the risky account.. Let y= 8, and solve for x. 8 = 0.69x = 0.69x = 0.69x 0.69x 8.87 = x. Therefore, the writing score is 8 in 99 (980 + ).. Profit occurs when R( x) > C( x). 00x> 8, x 00x > 8,000 x > 0 The company will make a profit when producing and selling more than 0 units.. a. Px = 6x ( 0, x) = 6x 0,000 6x = 00x 0,000 b. 00x 0,000 > 0 00x > 0,000 x > 80 c. There is a profit when more than 80 units are produced and sold.. a. y+,000x= 00,000 y = 00,000,000x 00,000,000x < 0,000,000x < 0,000 x > 0 b. After 0 years, the property value will be below $0,000. Copyright 0 Pearson Education. Inc.

68 CHAPTER Review 9 y y a. m = = = =. x x The average rate of change is $.0 per unit. ( x ) y y = m x x y =. 0 y =.x 660 y =.x 0 Px =.x 0 b. Profit occurs when Px > 0..x 0 > 0.x > 0 0 x = = d. y > 8 6 = x +.70 > x > x > x >.69 In 00 (90 + ) and beyond, the average woman is expected to live more than 8 years. 7. a. y = 0.06x+. b. The company will make a profit when producing and selling at least 7 units. 6. a. y = 0.068x+.70 b. c. g (0) = 0.06(0) +. = = c. f (99) = =.08 In 09 ( ), the average woman is expected to live years beyond age 6. Her life expectancy is 87 years. d. In 080 (00 + 0), a 6-year old male is expected to live 0.8 more years. The overall life expectancy is 86 years. A life expectancy of 90 years translates into 90-6 = years beyond age 6. Therefore, let gx =. 0.06x +. = 0.06x =. 0.06x = x = = In approximately the year (90 +9), male life expectancy will be 90 years. Copyright 0 Pearson Education. Inc.

69 0 CHAPTER Linear Models, Equations, and Inequalities e. A life expectancy of 8 years translates into 8 6 = 6 years beyond age 6. gx x x x 0.06 x 6. Since = 006, the average male could expect to live less than 8 years prior to 007. Using the unrounded model: The unrounded model predicts that population in 00 will be approximately,7.66 thousand, or,76, a. 9. a. b. [0, 0] by [0000, 000] Yes. A linear model would fit the data well. The data points lie approximately along a line. b. [0, 0] by [0, 0] Yes. A linear model would fit the data well. The data points lie approximately along a line. [0, 0] by [0000, 000] c. In 00, x = = 0. y = 7.9(0) + 6,8.76 =.9 + 6,8.76 =,7.66 In 00, the predicted population is,76,000. [0, 0] by [0, 0] Copyright 0 Pearson Education. Inc.

70 CHAPTER Review c. In 0, x = =. y =.8() = = In 0, the predicted personal expenditures for higher education is 0.9 billion dollars. Using the unrounded model: The unrounded model predicts that the personal expenditures for higher education in 0 will be approximately $0.9 billion, or $0,900,000,000.. y > 6 0.6x +.07 > 6 0.6x > x < 0.6 (Note the inequality sign switch.) x <.7 y < 7 0.6x +.07 < 7 0.6x < x > 0.6 (Note the inequality sign switch.) x > The marriage rate with be between 6 and 7 during the years between 008 ( ) and 0 (980 + ). 0. a. [, ] by [0,.] b. y= 0.09x 0.9 c.. 0 p 9x= ( + 9 x) Then p = 0 Since. p 7.0 ( + 9 x) x 0 9 9x 09 x Thus, from 99 through 00, marijuana use was in the range of.% 7.0%. The line seems to fit the data points very well. Copyright 0 Pearson Education. Inc.