CHAPTER FIVE. g(t) = t, h(n) = n, v(z) = z, w(c) = c, u(k) = ( 0.003)k,

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1 CHAPTER FIVE 5.1 SOLUTIONS 121 Solutions for Section 5.1 EXERCISES 1. Since the distance is decreasing, the rate of change is negative. The initial value of D is 1000 and it decreases by 50 each day, so D = d miles. 2. The initial height is 5 ft, which is 5 12 = 60 in, and the growth rate is 0.2 in/year, so the height after t years is h = t inches. 3. The initial value is 30 and the temperature decreases by 0.04 C per centimeter, so T = d. 4. This is already in the form f(x) = b + mx, so we have m = 3, b = Writing this as we have b = 5300, m = Writing this as we have b = 100, m = Writing this as we have b = 30, m = Writing this as we have b = 0, m = Writing this as we have b = 0.007, m = g(t) = t, h(n) = n, v(z) = z, w(c) = c, u(k) = ( 0.003)k, 10. (a) If the repairman works 0 hours, the cost will be $50. (b) The hourly rate is the coefficient of h. Thus, the repairman charges $25 per hour. 11. (a) The 300 is the initial cost of renting the limousine. If the renter spends up to four hours in the limousine, h = 0. (There are 0 hours above 4.) Therefore, the cost would be $300. (b) The hourly rate is the coefficient of h, in dollars per hour. Thus, the limousine company charges $100 per hour, for the number of additional hours the limousine is rented. 12. (a) When the town is founded we have t = 0, so the (b) After one year we have t = 1, so Population when founded = P(0) = (0) = Population after one year = P(1) = (1) = The population increased by = 350 during the first year. After two years we have t = 2, so Population after two yearsp(2) = (2) = The population increases by = 350 during the second year. In fact, the population increases by 350 every year.

2 122 Chapter Five /SOLUTIONS 13. The initial value is 600, which tells us that at the moment the probe is released, it is 600 km from Earth. The rate of change is 5, which tells us that the distance increases by 5 km every second, that is, that the probe is traveling away from the earth at 5 km/second. 14. The starting value is 50. This tells us that at time t = 0, or immediately after it rained, the trough held 50 gallons of water. The rate of change is 1.2. This tells us that 1.2 gallons of water evaporate each day. 15. If no minutes are used, we have n = 0. Thus, b = 25. This tells us that the fixed monthly charge is $25. Since the number of minutes is multiplied by 0.06, we know m = This represents the charge per minute to make a call on the cell phone. 16. In 2004, we have y = 0. Thus, b = 200. This is the number of people who enrolled in the course in The enrollment in the class is decreasing by 5 students per year. Thus, m = At noon, h = 0, and thus b = 50. This tells us that the temperature at noon is 50 degrees Fahrenheit. The temperature increases by 1.2 times the number of hours since noon. Thus, m = 1.2. This tells us that the temperature increases at a constant rate of 1.2 degrees Fahrenheit each hour after noon. 18. When the antique is purchased, n = 0. Thus, the initial value of the antique is $2500, and b = The value of the antique increases by $80 per year. Thus, m = The initial value, b = 100, represents the homework grade if no homework assignments are missing. For each missing assignment, 3 points are deducted. Thus, m = The initial value is 500 and represents the cost even if no students attend (for example, the cost of renting the hall and the band). When n = 1, the cost is 520, and the cost goes up by 20 every time n goes up by 1, so 20 is the slope and represents the cost per student (for example, food and drinks). 21. The initial value is 20,000 and represents the initial cost of the machine (for example, the price paid for it plus the cost of setting it up). When t = 1 (after one year) the cost is 20, = 21,500 dollars, and when t = 2 it is 20, = 23,000 dollars, and so on. Thus the amount spent on the machine goes up by $1500 each year. So the slope is 1500 and represents the annual cost in dollars, for example, the amount spent on maintenance and repairs. 22. The initial value, 9000, represents the initial population of the city, and the slope 500 is the rate of change of the population, in people per year. Since the change is positive, the population of the city is increasing. 23. The 120 is the distance from the shore when time is 0, when the surfer gets on top of the wave. The slope, 5, is the per second change in the distance to shore. 24. The y-intercept is 3 and the slope is 2, so the graph slopes upward from the point (0, 3), going up by 2 for each unit increase in x. See Figure 5.1. y x Figure 5.1

3 5.1 SOLUTIONS The y-intercept is 4 and the slope is 1, so the graph slopes downward from the point (0,4), going down by 1 for each unit increase in x. See Figure 5.2. y x Figure The y-intercept is 2 and the slope is 0.5, so the graph slopes upward from the point (0, 2), going up by 0.5 for each unit increase in x. See Figure 5.3. y 1 x Figure The y-intercept is 2 and the slope is 3, so the graph slopes upward from the point (0, 2), going up by 3 for each unit increase in x. See Figure 5.4. y x Figure 5.4

4 124 Chapter Five /SOLUTIONS 28. The y-intercept is 5 and the slope is 2, so the graph slopes downward from the point (0,5), going down by 2 for each unit increase in x. See Figure 5.5. y x Figure The y-intercept is 0.2 and the slope is 0.5, so the graph slopes downward from the point (0, 0.2), going down by 0.5 for each unit increase in x. See Figure 5.6. y x Figure 5.6 PROBLEMS 30. (iv). 31. (a) The coefficient 0.14 means that it costs $0.14 for each additional kilowatt hour of electricity that is used in excess of 250. (b) To find f(50), we substitute 50 in for h in our equation. f(h) = h f(50) = = 43.6 This tells us that it costs $43.60 when a customer uses = 300 kwh of electricity in a month. 32. (a) This collection begins with 200 baseball cards and grows at a rate of 100 baseball cards per year. (b) This collection begins with 100 baseball cards and increases at a rate of 200 baseball cards per year. (c) This collection begins with 2000 baseball cards and decreases at a rate of 100 baseball cards per year. In this case, cards are clearly being sold faster than they are being acquired. (d) This collection begins with 100 baseball cards and decreases at a rate of 200 baseball cards per year. In this case, the collection will be completely sold off in well under a year.

5 5.1 SOLUTIONS (a) The function f(x) = s has slope m = 0 and y-intercept b = s. Its graph is a horizontal line that crosses the y-axis at y = s, or above the x-axis, so it matches graph (iii). (b) The function f(x) = kx has slope m = k and y-intercept b = 0. Its graph is a line with positive slope that crosses the y-axis at y = 0, or at the origin, so it matches graph (i). (c) The function f(x) = kx s has slope m = k and y-intercept b = s. Its graph is a line with positive slope that crosses the y-axis at y = s, or below the x-axis, so it matches graph (ii). (d) The function f(x) = 2s kx has slope m = k and y-intercept b = 2s. Its graph is a line with negative slope that crosses the y-axis at y = 2s, or above the x-axis, so it could match either graph (iv) or (v). The function f(x) = 2s 2kx from (e) has slope m = 2k and y-intercept b = 2s. Its graph is also a line with negative slope that crosses the y-axis at y = 2s. However, it has a steeper slope, 2k compared to k, so it matches graph (v), and function f(x) = 2s kx matches graph (iv). (e) The function f(x) = 2s 2kx has slope m = 2k and y-intercept b = 2s. Its graph is also a line with negative slope that crosses the y-axis at y = 2s, or above the x-axis, so it could match either graph (iv) or (v). The function f(x) = 2s kx from (d) has slope m = k and y-intercept b = 2s. Its graph is a line with negative slope that crosses the y-axis at y = 2s. However, it has a shallower slope, k compared to 2k, so it matches graph (iv), and this function matches graph (v). 34. (a) Table 5.1 gives values of the velocity over a 3 second interval. Table 5.1 t (sec) v(t) (ft/sec) (b) Figure 5.7 shows the velocity over a three second interval. velocity (ft/sec) time (sec) 50 Figure 5.7 (c) The 48 tells us that the initial velocity is 48 feet per second. That is, the object is tossed up in the air with an initial velocity of 48 feet per second. The 32 tells us that the velocity is decreasing by 32 feet per second for every second that the object is in the air. (d) A positive velocity indicates that the object is going up in the air. A negative velocity indicates that the object is falling. 35. A linear function of x can be written in the form y = b + mx, where m and b are the slope and vertical intercept, respectively. Since y = ax + 5a = 5a + ax, then y is a linear function with slope a and vertical intercept 5a. 36. Two graphs with the same slope are either parallel lines, or the same line. Since we also know that the lines have different x-intercepts, that means that they cannot be the same line, so the lines must be parallel. Therefore the lines cannot have the same y-intercept. 37. Yes. For example, the functions y = 2x and y = 3x have x- and y-intercepts at (0,0), but have different slopes. 38. The slope is m = 12, and the y-intercept is b = The slope is m = 1/3, and the y-intercept is b = Writing this as we have slope m = 1/7 and y-intercept b = 12. f(x) = 1 7 x 12,

6 126 Chapter Five /SOLUTIONS 41. Writing this as we have slope m = 2/3 and y-intercept b = 20/ Writing this as we have slope m = 4 and y-intercept b = x, f(x) = ( 2x) = 9 + 4x, 43. Writing this as y = πx + 0, we have slope m = π, and y-intercept b = In one hour n birds eating continuously consume V/T in 3 of seed. In one hour one bird eating continuously consumes V/(nT) in 3 of seed. 45. The units of W/(nT) are ounces per bird per hour. It represents the weight of seed one bird consumes in one hour. Solutions for Section 5.2 EXERCISES 1. Because 5t 3 = 3 + 5t this is linear in t, with constant term 3 and coefficient Not linear: it has 5 raised to the x th power, not 5 times x. 3. This can be written as 1 + 7r, so it is linear in r. 4. This can be written as 3a + 1 = a = a, so it is linear in a with constant term 1/4 and coefficient 3/4. 5. This expands out to which is not linear because of the 1/a term. 6. Not linear because of the r 2 term. 3a + 1 a 7. This is linear in x with constant term 4 2 and coefficient 1/3. 8. This simplifies to 15A 3 = A, which is linear in A. 9. The constant term is 4 and the coefficient is 3. = 3a a + 1 a = a, 10. Collecting like terms, we get 5x x + 5 = 4x + 5 = 5 + 4x. Thus the constant terms is 5 and the coefficient is We group the terms with x and the terms without x: The constant term is w + 1 and the coefficient is w. w + wx + 1 = wx + (w + 1) = (w + 1) + wx. 12. Since x + rx = (1 + r)x, the constant term is 0 and the coefficient is 1 + r. 13. We combine like terms by combining the terms with x and the terms without x: mx + mn + 5x + m + 7 = mx + 5x + mn + m + 7 The constant term is mn + m + 7 and the coefficient is m + 5. = (m + 5)x + (mn + m + 7) = (mn + m + 7) + (m + 5)x.

7 5.2 SOLUTIONS We rewrite the expression in the form of a constant term plus a coefficient times x, by distributing through and combining like terms. 5 2(x + 4) + 6(2x + 1) = 5 2x x + 6 = x. The constant term is 3 and the coefficient is We have 16. We have 17. We have 18. We have f(x) = (x 1) = x 3 = 9 + 3x. f(x) = (x + 3) = x = x. g(n) = 14 2/3(n 12) = 14 2/3n + 8 = 22 2/3n. j(t) = (t 5) = t 2 = t. 19. Since f(1) = = 5, the graph passes through the point (1, 5). The slope is 3. See Figure (1, 5) 3 3 x 5 10 Figure 5.8

8 128 Chapter Five /SOLUTIONS 20. Since f( 2) = = 4, the graph passes through the point ( 2,4). The slope is 2. See Figure 5.9. y 6 4 ( 2, 4) x 4 6 Figure Since g(1) = 0/2 + 3 = 3, the graph passes through the point (1, 3). The slope is 1/2. See Figure y 3 (1, 3) x Figure Since h(1) = 5 (1 1) = 5, the graph passes through the point (1, 5). The slope is 1. See Figure y 3 3 x (1, 5) 7 Figure 5.11

9 5.2 SOLUTIONS (a) The y-value increases by 1 unit for each 2-unit increase in x, so Slope = Change in y Change in x = 1 2. (b) Here the y-values decrease by 4 units for each 1-unit increase in x, so Slope = Change in y Change in x = 4 1 = If the personal trainer works 40 hours, the payment will be P(40) = (40 40) = 500. Thus, the 500 is the trainers weekly salary. The is the rate of pay for hours worked in excess of 40. Thus, the trainer receives $18.75 per hour for each hour worked beyond When two guests stay in the room, the cost will be C(2) = (2 2) = 79. Thus, the 79 dollars is the daily charge for the room. The 10 dollars per guest is the daily rate the hotel charges for each additional guest above If the salesperson made $1000 in sales for a week, his income for the week would be T(1000) = ( ) = 600. Thus, the 600 dollars is the salesperson s income when he makes $1000 in sales for a week. The 0.15 is the commission rate paid to the salesperson for selling in excess of $1000. The salesperson would earn $0.15 for each $1 of sales above $ Since h(x) is linear, we have h(x) = b+mx for some constants b and m. We are given h( 30) = 80 and h(40) = 60, so m = (h(40) h( 30))/(40 ( 30)) = ( 60 80)/70 = 2. Solving for b, we have so h(x) = 20 2x. h( 30) = b 2( 30) b = h( 30) + 2( 30) = ( 30) = 20, 28. Since f(x) is linear, we have f(x) = b + mx for some constants b and m. We are given f(20) = 70 and f(70) = 10, so m = (f(70) f(20))/(70 20) = (10 70)/50 = 1.2. Solving for b, we have so f(x) = x. f(20) = b 1.2(20) b = f(20) + 1.2(20) = (20) = 94, 29. Since f(x) is linear, we have f(x) = b + mx for some constants b and m. We are given f( 12) = 60 and f(24) = 42, so m = (f(24) f( 12))/(24 ( 12)) = (42 60)/36 = 0.5. Solving for b, we have so y = x. f( 12) = b 0.5( 12) b = f( 12) + 0.5( 12) = ( 12) = 54, 30. You find the distance traveled by multiplying the number of hours, t, by the speed 45, giving 45t. This is a linear expression with constant term 0 and coefficient This is not linear, because the r is squared. 32. The area is 20w, which is linear in w with constant term 0 and coefficient The area is x 2, which is not linear. PROBLEMS 34. (a) The term 0.109g represents the rate adjustment. Since the 8 is in dollars, the rate adjustment is 0.109g dollars, so Rate adjustment per therm = dollars/therm = 10.9 cents/therm. (b) Using the distributive law to rewrite the expression for total cost as we see that the expression is linear in g. Total cost = g g = 8 + ( )g = g,

10 130 Chapter Five /SOLUTIONS 35. (a) The term 15(d/15) represents the cost of gasoline, the 0.3d represents the other expenses, and the 20 represents the cost of renting the car. (b) Cost is in dollars and distance d is in miles. (c) Using the distributive law to rewrite the expression for cost as ( d Cost = d ) we see that the expression is linear in d. = 1.5 d + 0.3d = ( )d + 20 = 0.4d + 20, 36. Linear, because it can be written as (1/2)b + (1/2)a = Constant + Constant a. 37. Not linear, because of the term in r Linear, because regarding r as constant, 2πr 2 + πrh = Constant + Constant h. 39. Linear, because ax 2 + bx + c 3 = (bx + c 3 ) + (x 2 )a = Constant + Constant a. 40. Not linear because of the term in x Linear, because 2ax + bx + c = (bx + c) + (2x)a = Constant + Constant a. 42. Linear, because 2ax + bx + c = c + (2a + b)x = Constant + Constant x. 43. Linear, because 3xy + 5x y = (2 10y) + (3y + 5)x = Constant + Constant x. 44. Linear, because 3xy + 5x y = (5x + 2) + (3x 10)y = Constant + Constant y. 45. Not linear, because P(P c) = P 2 cp contains P Linear, because P(P c) = P 2 + ( P)c = Constant + Constant c. 47. The slope is R S = 1 = 0.1 inches of rain/inches of snow. 10 The slope tells us that, for every 10 inches the amount of snow increases, the amount of rain increases by 1 inch. Thus 10 inches of snow is equivalent to 1 inch of rain. The vertical intercept is 0. This tells us that 0 inches of snow is equivalent to 0 inches of rain. The function is R = 0.1S. 48. We have Slope = m = = 100 h = 50. The units of the slope are miles per hour, and it represents the speed. Thus the speed would be 50 mph. Using point-slope form with the point (1, 70), we have m = (h 1) = h 50 = h, so the vertical intercept is 20, which represents the distance from home when the journey starts. 49. The slope is T = h 3 1 = 6 = 3 degrees/hour. 2 This represents the rate at which temperature is decreasing each hour. The vertical intercept is 40, representing the temperature at midnight. The function is T = 40 3h. 50. When the boy is 20 we have t = 10. The point-slope form in (ii) best shows the value of h when t = 10, and the height is 6ft. 51. From (iii) we clearly see that b = 7 when c = 6, so this is the best form. 52. (a) We know that P(4) = (4 4) = This tells us that, after 4 months of operation, the company s profit is $1000.

11 5.2 SOLUTIONS 131 (b) We have P(t) = (t 4) = t 2000 = t. The 1000 tells us that, when the company began operating its business, it was $1000 in debt. The 500 tells us that the company s profit increases at a rate of $500 per month. 53. (a) Since D(t) = (t 3), we know that D(3) = (3 3) = 138. This tells us that, after 3 hours, Liza is 138 miles from home. (b) We have D(t) = (t 3) = t 120 = t. The 18 tells us that, when Liza began her trip, t = 0, she was 18 miles from home. The 40 tells us that Liza travels at a rate of 40 miles per hour. 54. (a) Since f(12) = (12 12) = 45, we see that 12 minutes after reaching the flat the cyclist is 45 km from the finish line. The coefficient 0.5 is the speed of the cyclist in km/min. (b) We transform into slope-intercept form, because the vertical intercept gives us the distance of the cyclist from the finish line at the beginning of the flat: f(t) = (t 12) = t + 6 = t. In this form, we clearly see that the start of the flat is 51 km from the finish line. 55. (a) Since B(20) = (20 20) = 50, we see that after 20 years there are 50 butterflies in the collection. The coefficient 2 is the number of butterflies added per year. (b) We put the function in slope-intercept form B(x) = (x 20) = x 40 distributing the 2 = x. This shows clearly that the collection started with 10 butterflies. 56. (a) Since C(w) = (w 1), we know that C(1) = (1 1) = This tells us that it costs $0.88 to mail a letter weighing one ounce or less. (b) The 0.17 tells us that it costs an additional $0.17 per ounce to mail letters weighing more than one ounce. (c) If the letter weighs 9.1 ounces, we have to find C(10). Thus, C(10) = (10 1) = (9) = Thus, the cost would be $ The profit is the revenue from sales minus the costs of production. Since the widgets sold for $27 each, the amount made on sales is 27q. The cost of production is the set-up cost of $1000 plus the manufacturing cost of q widgets at $15 each, or 15q. Thus the profit is a sum (or difference) of terms, each of which is either a constant or a constant times q, so we expect it to be linear. Thus, Profit = Sales Cost = 27q ( q) = 12q 1000 = q dollars. 58. The four toll booths cost $400,000 and the mileage cost is $500,000d. The total cost is their sum, which is linear in d because the expression for total cost is calculated using only addition and multiplication of d by a constant. The total cost is given by the linear expression 400, ,000d dollars.

12 132 Chapter Five /SOLUTIONS 59. The two gates cost = 600 dollars. The perimeter of the field is 4x meters, but only 4x 8 meters of fence are needed because there is no fence where the gates are. The cost of the fence is (4x 8) 10 = 40x 80 dollars. The project costs, gates and fence together, are x 80 = x dollars. This expression is linear in x because the total cost is calculated using only addition and multiplication of x by a constant. 60. Starting with x, we add 5 to get x + 5, then multiply by 2 to get 2(x + 5), then subtract x to get 2(x + 5) x, which simplifies to x + 10 = 10 + x. This expression is linear. 61. Starting with x, we add 5 to get x + 5, then multiply by x to get x(x + 5), then we subtract 2 to get x(x + 5) 2, which simplifies to x 2 + 5x 2. This expression is not linear. 62. Starting with x, we add x to get 2x, then multiply by 5 to get 10x, then subtract 2 to get 10x 2 = x. This expression is linear. 63. We know two points on the graph of f: The first is (100, 30), because the value of y = g(x) at x = 100 is y = 30. The second is ( 50,15), because the solution g(x) = 15 is x = 50. We have g(100) = 30 and g( 50) = 15. From these two points, we see that m = (g( 50) g(100))/( ) = (15 30)/( 150) = 0.1. Solving for b, we have so g(x) = x. g(100) = b + 0.1(100) b = g(100) 0.1(100) = (100) = 20, Solutions for Section 5.3 EXERCISES 1. (a) Since C is 8, we have Tuition cost = C = (8) = Thus, taking 8 credits costs $1900. (b) Here the tuition cost is 1700 dollars, so we set the expression for the cost equal to 1700 and solve for C: Thus, $1700 is the cost of taking 7 credits = C 7 = C. 2. We want to know the value of t that makes V (t) equal to $2000, so we solve 18, t = t = 16,000 subtract 18,000 from both sides t = 16,000 = divide both sides by Thus the car is worth $2000 about halfway through its 10 th year after about nine and a half years. 3. (a) We have (b) We have f(11) = t + 3 = 2 5 2t + 3 = 10 2t = 7 t = 7 2. = 25 5 = 5.

13 5.3 SOLUTIONS We have 5x 2x 3 = 7 5x = 7(2x 3) 5x = 14x 21 9x = 21 x = We want to find the value of s that makes s = 200. So we solve this equation for s: s = s = 100 s = = So 1000 sales must be made to make $200 under option (a). 6. It is not possible to make $200 dollars under option (c), since the salary is $175 no matter what the sales. 7. We want to find the value of s that makes s = s, so we solve this equation for s: s = s 0.10s 0.05s = s = 50 s = = We want to find the value of s that makes s = 175, so we solve this equation for s: s = s = 25 s = = We want to find the value of s that makes s = 175, so we solve this equation for s: s = s = 75 s = = We have 7 3y = 17 3y = 24 3y 3 = 24 3 y = We have 13t + 2 = 49 13t = 47 13t 13 = t = =

14 134 Chapter Five /SOLUTIONS 12. The common denominator for this fractional equation is 3. If we multiply both sides of the equation by 3, we obtain: ( ) 2(t 1) 3 3t + = 3(4) 3 9t + 2(t 1) = 12 9t + 2t 2 = 12 11t 2 = 12 11t = 14 t = = We have 2(r + 5) 3 = 3(r 8) r = 3r r + 7 = 3r 4 2r = 3r 11 r = 11 r = The equation 2x + x = 27 becomes 3x = 27 and x = The equation 4t + 2(t + 1) 5t = 13 becomes t + 2 = 13 and t = x x = 0 9 x 3 = 5 1 x 9(1 x) = 5(x 3) 9 9x = 5x 15 14x = 24 x = Positive, because 3x must be positive. 18. Negative, because 3a must be negative. 19. Negative, because 5z must be negative. 20. Positive, because 3u must be positive. 21. Positive, because 5w must be negative. 22. Zero, because, collecting like terms, we see that 5y must equal zero. 23. Negative, because, collecting like terms, we see that 5b must be negative. 24. Positive, because, collecting like terms, we see that 3p must be negative. 25. Positive, because, collecting like terms, we see that 6r must be positive. 26. Positive, because, collecting like terms, 8t must be negative. 27. Negative, because, collecting like terms, we see that 8c must be positive. 28. Zero, because, collecting like terms, we see that 3d must equal zero.

15 29. This reduces to 4x = 4, so x = 1 is the only solution. 30. This reduces to 4x = 10, so x = 5/2 is the only solution. 5.3 SOLUTIONS The right-hand side reduces to 4x + 3, so the equation is 4x + 3 = 4x + 3, which is true for all x, so we have an infinite number of solutions. 32. The right-hand side reduces to 4x + 5, so the equation is 4x + 3 = 4x + 5, or 3 = 5, which has no solutions. 33. This reduces to 4x = 0, which has the solution x = 0. Thus, there is one solution. (Note: There is a difference between a solution being zero, and there being zero solutions.) 34. The right-hand side reduces to 4x + 1, so the equation is 4x + 3 = 4x + 1, or 3 = 1, which has no solutions. 35. The right-hand side reduces to 4x + 3, so the equation is 4x + 3 = 4x + 3, which is true for all x, so we have an infinite number of solutions. 36. We have t(t + 3) t(t 5) = 4(t 5) 7(t 3) t 2 + 3t (t 2 5t) = 4t 20 (7t 21) t 2 + 3t t 2 + 5t = 4t 20 7t t = 3t t = 1 t = PROBLEMS 37. If p is the tag price in dollars then p 20 is the price using the coupon while 0.80p is the price using the 20% discount. Then when the discounts are the same, Thus, the tag price is $100. p 20 = 0.80p p 0.80p = p = 20 p = If a is the number of pounds of apples purchased and p is the number of pounds of pears, then the total cost of the purchase is 0.99a p dollars. If I spend $4 then 4 = 0.99a p. If the weight of the apples I buy is twice the weight of the pears then a = 2p. Thus, 4 = p p, so p = 4/( ) = pounds. 39. (a) To find the cost, we add $37 to the mileage charge. The mileage charge is the cost per mile multiplied by the number of miles traveled. Therefore, Cost = $37 + $ = $37 + $25 = $62. (b) The cost of renting the car for three days is 3 $37. To find the total cost, we add 3 37 to the mileage charge. The mileage charge is the cost per mile multiplied by 400 miles. Therefore, Cost = 3 $ $0.25 = $211. (c) If we let m be the number of miles driven, then the cost would be This must be $385, so we need to solve Cost = 5 $37 + m $ $37 + m $0.25 = $385

16 136 Chapter Five /SOLUTIONS for m. Isolating m, we find 0.25m = so m = 800 miles. Alternatively, the five days cost is 5 $37 = $185. So, the mileage cost came to $385 $185 = $200. At $0.25 per mile, the number of miles driven is $200 $0.25 = Since the sum of the two distances represented by 2s and 15 equals the distance represented by 35, we have the equation 2s + 15 = 35. Then 2s must be 20. Therefore s = Since the total area is the area of the triangle ((1/2) 4 x) plus the area of the rectangle (6x) we have: so 8x = 144 and x = 18 feet. 4x 2 + 6x = 144, 42. (a) The time taken on each leg is the distance traveled (50 miles) divided by the speed for that leg. The first 50 miles took (50 miles)/(50 mph) = 1 hour, whereas the second took (50 miles)/(v mph) = 50/V hours. (b) The average speed over the entire journey is the total distance traveled, 100 miles, divided by the total time taken, /V hours. Thus, the average speed for the entire journey is V (c) If you want to average 75 mph for the entire journey, then We must solve this for V. Thus, = 100V V + 50 mph. 100V V + 50 = V = 75(V + 50) 25V = V = = 150 mph. 25 (d) If you want to average 100 mph for the entire journey, then We must solve this for V. Thus, 100V V + 50 = V = 100(V + 50) 0 = This is not possible. Thus, there is no speed that will allow you to average 100 mph on the entire journey. 43. (a) A = 0, because if x = 0 then 3x = 0. (b) A > 0 because 3x has the same sign as x. (c) There is a solution for all values of A. 44. (a) No value of A satisfies A 0 = 3. (b) If x > 0 then A must be positive because Ax > 0. (c) There is no solution for A = 0 because 0 x = 3 has no solution. 45. (a) At x = 0, we have 3x + 5 = 5 and so A = 5. (b) We have 3x = A 5 and so A 5 must be positive for x > 0, which means A > 5. (c) There is a solution for all values of A.

17 46. (a) If x = 0 then A = 5. (b) We see that 3x = 5 A so if x > 0 then A must be less than 5. (c) There is a solution for all values of A. 47. The only solution to this equation occurs at x = 0. (a) Any value of A. (b) No value of A. (c) None. 48. This equation has no solutions. (a) No value of A. (b) No value of A. (c) Any value of A. 49. (a) No value of A, because 7/0 is undefined. (b) If x > 0 then 7/x > 0 and so A > 0. (c) A = 0, because for all values of x, 7/x (a) None, because A/0 is undefined. (b) Since A/x is positive, if x > 0, then A > 0. (c) A = 0, because 0/x = 0 for all x SOLUTIONS The equation is 2t + 7 = 7, giving t = 0, one solution. (Note: There is a difference between a solution being zero, and there being zero solutions.) 52. The equation is 2(2t + 7) = 2(2t) + 7, giving 14 = 7, which has no solutions. 53. The equation is 2t+7 = (2(t+1)+7) 2, giving 2t+7 = 2t+7, which is true for all t, so we have an infinite number of solutions. 54. The equation is 2t + 7 = 2( t) + 7, giving 2t = 2t, so t = 0 is the only solution. (Note: There is a difference between a solution being zero, and there being zero solutions.) 55. The equation is 2t + 7 = (2t + 7), giving 4t = 14, so t = 7/2 is the only solution. 56. The equation is (2t + 7) + 1 = 2(t + 1) + 7, giving 8 = 9, which has no solutions. 57. The equation is (2t + 7) + (2(3t) + 7) 2(2(2t) + 7) = 0, giving 0 = 0, which is true for all t, so we have an infinite number of solutions. 58. (a) One solution means that the graphs of the lines y = b 1 + m 1x and y = b 2 + m 2x must cross, so they must not be parallel. Thus, their slopes m 1 and m 2 must be different, m 1 m 2. (b) No solution means that the graphs of the lines y = b 1 + m 1x and y = b 2 + m 2x must not cross, so they must be parallel, but must not be the same line. Thus, their slopes m 1 and m 2 must be the same, m 1 = m 2, but their y-intercepts b 1 and b 2 must be different, b 1 b 2. (c) An infinite number of solutions means that the lines y = b 1+m 1x and y = b 2+m 2x must be identical, so m 1 = m 2 and b 1 = b 2. Solutions for Section 5.4 EXERCISES 1. We have, y = 100 3(x 20) = 100 3x 3( 20) = 100 3x + 60 = 160 3x,

18 138 Chapter Five /SOLUTIONS so m = 3, b = We have, 80x + 90y = y = x y = x = x, so b = 10/9 and m = 8/9. 3. We have, x y 300 = 1 y 300 = 1 x ( 100 y = x = = 300 3x, 100 ) x 100 so b = 300 and m = We have, so b = 45, m = 3/2. x = y 2 3 y = 30 x y = 3 (30 x) 2 = x, 5. The equation can be put in slope-intercept form, y = 6 + 3x. The y-intercept is 6, so we put a point on the y-axis at y = 6. From there, for every 1 unit you move to the right, also move up 3 units. See Figure y 3 3 x 15 Figure 5.12

19 5.4 SOLUTIONS The line y = 5 passes through the y-axis at 5 and is a horizontal line that remains parallel to the x-axis. See Figure y x Figure We put the equation in slope-intercept form, y = 8 +(2/3)x. This tells us that the y-intercept is y = 8, so we plot the point (0, 8). Next we draw a line of slope 2/3 through this point, so we go up 2/3 for every increase of 1 in x, which is the same as going up 2 for every increase of 3 in x. See Figure y x 6 10 Figure The line x = 7 passes through the x-axis at 7 and is a vertical line that remains parallel to the y-axis. See Figure y 7 x 10 Figure The equation can be put into slope-intercept form y = 4 2 x. Put a point on the y-axis of 4. From there, for every 1 3 unit you move to the right, also move down 2/3 units. See Figure y x 6 Figure 5.16

20 140 Chapter Five /SOLUTIONS 10. This is a linear equation. The y-intercept is b = 200 and the slope is m = 4. See Figure y x Figure (a) is (V), because slope is positive, vertical intercept is negative. (b) is (IV), because slope is negative, vertical intercept is positive. (c) is (I), because slope is 0, vertical intercept is positive. (d) is (VI), because slope and vertical intercept are both negative. (e) is (II), because slope and vertical intercept are both positive. (f) is (III), because slope is positive, vertical intercept is 0. (g) is (VII), because it is a vertical line with positive x-intercept. 12. Using the point-slope form, we have m = 5 and (x 0, y 0) = (2, 3), so y = y 0 + m(x x 0) y = 3 + 5(x 2). 13. Using the point-slope form, we have m = 6 and (x 0, y 0) = ( 1, 7), so y = y 0 + m(x x 0) y = 7 + 6(x ( 1)) y = 7 + 6(x + 1). 14. Using the point-slope form, we have m = 3 and (x 0, y 0) = (8, 10), so y = y 0 + m(x x 0) y = 10 3(x 8). 15. Using the point-slope form, we have m = 2/3 and (x 0, y 0) = (2, 9), so y = y 0 + m(x x 0) y = 9 2 (x 2) We first find the slope: m = y x = = 6 3 = 2. Using the point-slope form, we have m = 2 and (x 0, y 0) = (4, 7), so y = y 0 + m(x x 0) y = 7 + 2(x 4).

21 5.4 SOLUTIONS 141 Note that we could have used the point (1,1) instead. y = y 0 + m(x x 0) y = 1 + 2(x 1). To show that the two equations are equivalent, we can rewrite them each in slope-intercept form. y = 7 + 2(x 4) y = 7 + 2x 8 y = 1 + 2x. y = 1 + 2(x 1) y = 1 + 2x 2 y = 1 + 2x. 17. We first find the slope: m = y x = = 4 1 = 4. Using the point-slope form, we have m = 4 and (x 0, y 0) = (6, 5), so y = y 0 + m(x x 0) y = 5 4(x 6). Note that we could have used the point (7,1) instead. y = y 0 + m(x x 0) y = 1 4(x 7). To show that the two equations are equivalent, we can rewrite them each in slope-intercept form. y = 5 4(x 6) y = 5 4x + 24 y = 29 4x. y = 1 4(x 7) y = 1 4x + 28 y = 29 4x. 18. We first find the slope: m = y x = = 12 4 = 3. Using the point-slope form, we have m = 3 and (x 0, y 0) = ( 2, 8), so y = y 0 + m(x x 0) y = 8 + 3(x ( 2)) y = 8 + 3(x + 2). Note that we could have used the point (2,4) instead. y = y 0 + m(x x 0) y = 4 + 3(x 2).

22 142 Chapter Five /SOLUTIONS To show that the two equations are equivalent, we can rewrite them each in slope-intercept form. 19. We first find the slope: y = 8 + 3(x + 2) y = 8 + 3x + 6 y = 2 + 3x. y = 4 + 3(x 2) y = 4 + 3x 6 y = 2 + 3x. m = y 7 ( 1) = = 6 x 6 ( 6) 12 = 1 2. Using the point-slope form, we have m = 1/2 and (x 0, y 0) = (6, 7), so Note that we could have used the point ( 6, 1) instead. y = y 0 + m(x x 0) y = 7 1 (x 6). 2 y = y 0 + m(x x 0) y = 1 1 (x ( 6)) 2 y = 1 1 (x + 6). 2 To show that the two equations are equivalent, we can rewrite them each in slope-intercept form. y = 7 1 (x 6) 2 y = x + 3 y = x. y = 1 1 (x + 6) 2 y = x 3 y = x. 20. If two lines are parallel, their slopes are equal. Thus, the slope of our line is m = 5. Using the point-slope form, we have m = 5 and (x 0, y 0) = ( 1, 8), so y = y 0 + m(x x 0) y = 8 + 5(x ( 1)) y = 8 + 5(x + 1). 21. If two lines are parallel, their slopes are equal. Thus, the slope of our line is m = 5/4. Using the point-slope form, we have m = 5/4 and (x 0, y 0) = (3, 6), so y = y 0 + m(x x 0) y = (x 3). 4

23 5.4 SOLUTIONS If two lines are perpendicular, their slopes are negative reciprocals. Thus, the slope of our line is m = 1/( 4) = 1/4. Using the point-slope form, we have m = 1/4 and (x 0, y 0) = (12, 20), so y = (x 12) To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: x = 3y 2 x 3y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: y = 2 + 4(x 3) y = 2 + 4x 12 y 4x = x + y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: 5x = 7 2y 5x + 2y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: y 6 = 5(x + 2) y 6 = 5x x + y = x + y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: x + 4 = 3(y 1) x + 4 = 3y 3 x 3y = 3 4 x 3y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: 6(x + 4) = 3(y x) 6x + 24 = 3y 3x 6x + 3x 3y = 24 9x 3y = 24.

24 144 Chapter Five /SOLUTIONS 29. To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: 9(y + x) = 5 9y + 9x = 5 9x + 9y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: 3(2y + 4x 7) = 5(3y + x 4) 6y + 12x 21 = 15y + 5x 20 6y + 12x 15y 5x = x 9y = To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: y = 5x + 2a 5x + y = 2a. 32. To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the other: 5b(y + bx + 2) = 4b(4 x + 2b) 5by + 5b 2 x + 10b = 16b 4bx + 8b 2 5b 2 x + 4bx + 5by = 16b 10b + 8b 2 (5b 2 + 4b)x + 5by = 6b + 8b (a) If a line is parallel to y = 3 + 5x, its slope is 5. And since the y-intercept is 10, we have y = x. (b) We transform the line into slope-intercept form to find the slope: 4x + 2y = 6 2y = 6 4x y = 3 2x. If a line is parallel to y = 3 2x, its slope is 2. And since the y-intercept is 12, we have y = 12 2x. (c) If a line is parallel to y = 7x + 2, its slope is 7, so its equation is y = b + 7x.

25 5.4 SOLUTIONS 145 To find b, we use the fact that the line passes through (3,22) by substituting these x- and y-values into the equation: y = b + 7x 22 = b = b = b. Thus, we have y = 1 + 7x. (d) We put 9x + y = 5 into slope-intercept form to find its slope: 9x + y = 5 y = 5 9x. If a line is parallel to 9x + y = 5, its slope is 9, so its equation is y = b 9x. To find b, we use the fact that the line passes through (5,15) by substituting these x- and y-values into the equation: Thus, we have y = 60 9x. y = b 9x 15 = b = b = b. 34. Both lines are in slope-intercept form, and the slope of each is a. Since they have the same slope, they are parallel. 35. Both lines are in slope-intercept form. Since 1 + x = x, the slope of the first line is 1. The slope of the second is 2. Since they have different slopes, they are not parallel. 36. The first line is in point-slope form, and the second is in slope-intercept form. The slope of each is 4. Since they have the same slope, they are parallel. 37. Both lines are in point-slope form. However, the slope of the first is 3, and the slope of the second is 4. Since they have different slopes, they are not parallel. 38. Both lines are in standard form. To find the slope, we transform them into slope-intercept form: and 2x + 3y = 5 3y = 5 2x 4x + 6y = 7 y = x, 6y = 7 4x y = x y = x. The slope of each is 2/3. Since they have the same slope, they are parallel.

26 146 Chapter Five /SOLUTIONS 39. Both lines are in standard form. To find the slope, we transform them into slope-intercept form: qx + ry = 3 ry = 3 qx y = 3 r q r x, and qx + ry = 4 ry = 4 qx y = 4 r q r x. The slope of each is q/r. Since they have the same slope, they are parallel. 40. The first line is in point-slope form, and the slope is 4. The second is not quite in point-slope form. We put it into pointslope form (we could use slope-intercept as well): y = 8 + 2(2x + 3) y = 8 + (4x + 6) y = 8 + 4(x + 6/4) y = 8 + 4(x + 1.5). The slope of the second is also 4. Since they have the same slope, they are parallel. 41. The first line is in point-slope form, and the slope is 6. The second is not quite in point-slope form. We put it into pointslope form (we could use slope-intercept as well): y = 5 + 6(3x 1) y = 5 + (18x 6) ( y = x 6 ) 18 ( y = x 1 ). 3 The second line has slope 18. Since they have different slopes, they are not parallel. PROBLEMS 42. Notice that several of these equations can easily be written in (or are already in) point-slope form, y = y 0 + m(x x 0). We see that Equation I: y = 9 + 2(x 4) contains point (x 0, y 0) = (4, 9) Equation II: y 9 = 3(x 4) Equation V: y = 9 2(4 x) y = 9 + ( 3)(x 4) contains point (x 0, y 0) = (4, 9) = 9 2( 1)(x 4) = 9 + 2(x 4) contains point (x 0, y 0) = (4, 9). We can quickly verify that the other three equations do not contain this point by letting x = 4 to show that y 9: Equation III: y + 9 = 2(4 4) y = = 9 Equation IV: y = 4(4) + 9 = 25 Equation VI: y = = 9 ( 7) = 16. 4

27 5.4 SOLUTIONS Four of the equations are in (or almost in) point-slope form, y = y 0 + m(x x 0): Equation I: y = 9 + 2(x 4) Slope is m = 2 Equation II: y 9 = 3(x 4) y = 9 + ( 3)(x 4) Equation III: y + 9 = 2(x 4) Equation V: y = 9 2(4 x) Slope is m = 3 y = 9 + ( 2)(x 4) Slope is m = 2 = 9 2( 1)(x 4) = 9 + 2(x 4) Slope is m = 2. Equation IV is in slope-intercept form with m = 4. Finally, rewriting equation VI gives y = 9 4 8x 4 = 9 1 (4 8x) 4 = x) = 8 + 2x, so m = 2. This means that Equations I, V, and VI have the same slope. 44. Equations I, II, and VI are all in point-slope form, y = y 0 + m(x x 0), where (x 0, y 0) = (8,20). Thus, the graphs of these three equations all contain the point (8,20). 45. Placing equation V into slope-intercept form, we have y = 2 3 x Thus, the graphs of equations III and V have the same y-intercept, b = Placing equation IV into slope-intercept form, we have y = x. We see that both equations III and IV have the same slope, m = The line intersects the graph of y = x 3 x + 3 at x = 2 and x = 2. This means that: At x = 2: y = ( 2) 3 ( 2) + 3 = 3. At x = 2: y = = 9. Thus, the line contains the points ( 2, 3) and (2, 9). We have m = y x = 9 ( 3) 2 ( 2) = 12 4 = 3. Using the point-slope form and the point (x 0, y 0) = (2,9), we can write y = 9 + 3(x 2) = 3 + 3x.

28 148 Chapter Five /SOLUTIONS 48. First we place 5x 3y = 6 into slope-intercept form: The slope of this line is 5/3, and at x = 15 we have 5x 3y = 6 3y = 5x 6 y = 5 3 x 2. y = = 25 2 = 23, 3 so the graph contains the point (15, 23). Since the line we want is perpendicular to this line so its slope is m = 1/(5/3) = 3/5, and it has an equation of the form Putting x = 15 and y = 23 we get so y = 32 (3/5)x. y = b (3/5)x. 23 = b + m = b 3 5 (15) 23 = b 9 b = 32, 49. We have y = b + mx where m is the negative reciprocal of the slope of y = x, and so m = 1/( 0.2) = 5. At x = 1.5 we have y = (1.5) = 0.4, so Thus, y = 5x 7.1. b + 5(1.5) = 0.4 b = 0.4 5(1.5) = At x = 12, we know that y = (12) = 700. The slope of our line is m = 1/25 = The point-slope form gives y = (x 12) = x. 51. Since we are given the slope and the y-intercept, slope-intercept form would be easier. 52. When we are given two points, we first have to find the slope. Once we calculate the slope, we now have a point and a slope, so point-slope form would be easier. 53. We have a point and a slope. Point-slope form seems to be the obvious choice. 54. We know the slope and the y-intercept, so slope-intercept form would be easier. 55. We have a point and a slope. Point-slope form would be easier. 56. We have y y 0 = r(x x 0) y = y 0 + r(x x 0) so b = y 0 rx 0 and m = r. = y 0 + rx rx 0 = y 0 rx }{{} 0 +rx, b

29 5.4 SOLUTIONS We have y = β x α = β 1 ( α x = β + 1 ) x, α }{{} m so b = β, m = 1/α. 58. We have Ax + By = C By = C Ax so b = C/B and m = A/B. 59. We have y = C B A B x = C ( + A ) x, }{{} B B }{{} b m so b = b 1 + b 2, m = m 1 + m 2. y = b 1 + m 1x + b 2 + m 2x = b 1 + b }{{} 2 + (m 1 + m 2) x, }{{} b m 60. Since the lines are parallel, they have the same slope, m = 3. The new line passes through the point (x 0, y 0) = ( 3, 8). By the point-slope formula, we have y = y 0 + m(x x 0) = 8 3(x 3). This can also be written in slope-intercept form as y = x. 61. (a) Since the slopes are 6 and 3, we see that y = 3 + 6x has the greater slope. (b) Since the y-intercepts are 3 and 5, we see that y = 5 3x has the greater y-intercept. 62. (a) Since the slopes are 1 and 6, we see that y = 1 x has the greater slope. 5 5 (b) Since the y-intercepts are 0 and 1, we see that y = 1 6x has the greater y-intercept. 63. We begin by putting 2x = 4y + 3 into slope-intercept form by solving for y: 2x = 4y + 3 2x 3 = 4y 4y = 2x 3 y = 2x y = x. (a) The slope of the first line is 1/2, which is greater than 1, so 2x = 4y + 3 has the greater slope. (b) The y-intercept of the first line is 3/4, which is greater than 2, so 2x = 4y + 3 has the greater y-intercept.

30 150 Chapter Five /SOLUTIONS 64. We begin by putting 3y = 5x 2 into slope-intercept form by solving for y: 3y = 5x 2 y = x. (a) The slope of the first line is 5/3, which is less than 2, so y = 2x + 1 has the greater slope. (b) The y-intercept of the first line is 2/3, which is less than 1, so y = 2x + 1 has the greater y-intercept. 65. We begin by putting y + 2 = 3(x 1) into slope-intercept form by solving for y: y + 2 = 3(x 1) y = 3x 3 2 y = 5 + 3x. (a) The slope of the first line is 3, which is greater than 50, so y + 2 = 3(x 1) has the greater slope. (b) The y-intercept of the first line is 5, which is less than 6, so y = 6 50x has the greater y-intercept. 66. We begin by putting each equation into slope-intercept form by solving for y: and y 3 = 4(x + 2) y 3 = 4x 8 y = 4x y = 5 4x, 2x + 5y = 3 5y = 3 + 2x y = x. (a) The slope of the first line is 4, which is less than 2/5, so 2x + 5y = 3 has the greater slope. (b) The y-intercept of the first line is 5, which is less than 3/5, so 2x + 5y = 3 has the greater y-intercept. 67. The y-intercepts are 2, 2/3, 1, and 1/2, of which 2 is the largest, so (a). 68. The slopes are 4,5, 2, and 3, of which the smallest is 4, so (a). 69. The graphs of the equation y = 14x 18 has a positive slope and a negative y-intercept, while the graph of the equation y = 14x + 18 has a negative slope and positive y-intercept. The graph of the equation y = 14x 18 rises from left to right, while the graph of the equation y = 14x + 18 falls from left to right. 70. Writing this as 3xt + 2xt = x(3t + 2t 2 ) + 5 = (2t 2 + 3t) x + 5, }{{} m factor out x so m = 2t 2 + 3t and b = 5. Notice that the values of m and b involve t. 71. The points (0, 12) and (3, 0) fall on a line whose slope is m 1 = y x = = 4. Since this line crosses the y-axis at (0,12), its equation is y = 12 4x. The points (3,0) and (17/3, 2/3) fall on a line whose slope is m 2 = y x

31 5.5 SOLUTIONS 151 = = = = = = 1 4. Since this line contains the point (x 0, y 0) = (3, 0), its equation is y = (x 3) = 1 4 x 3 4. Both these lines contain the point (3,0), one corner of the triangle. These lines form a right angle, since their slopes are negative reciprocals: 1 4 = 1 4. Thus, this triangle having sides formed by these lines and (0,3) as a corner is a right triangle. Its third side is the line containing the points (0,12) and (17/3, 2/3), but we do not need to find the equation of this line. 72. (a) (i) The slope is given by We know that b = 5, so the equation is m = y2 y1 = 0 5 x 2 x = 5 2. y = x. (ii) The new intercepts are (4,0) and (0,10), so the slope is Now b = 10, so the equation is m = y2 y1 = 0 10 x 2 x = 5 2. y = x. (b) Since the slope is 5/2 in both cases, the lines are parallel. (c) One way of generalizing is to say that doubling the intercepts always gives a line which is parallel to the original line. Here we have generalized to other intercepts besides (2,0) and (0,5). Another way of generalizing is to say that tripling or multiplying the intercepts by any integer n gives a parallel line. (This can be either (2, 0) and (0, 5) or more general intercepts.) Here we have generalized from doubling to multiplying by other factors. Solutions for Section 5.5 EXERCISES 1. The x-values go up in steps of 3, while the y-values go down in steps of 4. Since the y-values change by the same amount each time, the table satisfies a linear equation. 2. The x-values go up in steps of 2, while the y-values go up in steps of 3. Since the y-values change by the same amount each time, the table satisfies a linear equation.

32 152 Chapter Five /SOLUTIONS 3. The x-values first increase by 2, then 4, then 8. The y-values go up in steps of 5. Although the y-values change by the same amount each time, the x-values do not change by the same amount. Therefore, this table does not satisfy a linear equation. 4. The x-values first increase by 2, then 4, then 8, and then 16. The y-values also increase by 2, then 4, then 8, and then 16. The increase in the y-value is always the same as the corresponding increase in the x-value, so the slopes between successive points are the same, namely 1. Therefore, this table satisfies a linear equation. 5. To decide whether the data are linear we calculate the slope between successive data points. We find ( 2) = 1 2, = 1 3 0, and = 1 2. Since the slopes are the same, the data satisfies a linear equation. 6. The value of Q is very nearly constant: Q goes down by almost 5 each time t goes up by 1. However, the change in Q does vary slightly, and so the slope is not constant and the table does not represent the values of a linear function. See Table 5.2. Table 5.2 t Q m The value of Q is different each time, jumping from 0.88 to 2.64 to 0.44 and so on. However, this is because the value of t changes each time, from 4 to 12 to 2 and so on. As we can see from Table 5.3, the slope itself is a constant m = 0.22, and so this data can be described using a linear equation. We can use an entry from the table to find the value of b: and so Q = t. Q = b t 5.79 = b (3) b = = 5.13, Table 5.3 t Q m We compute the slope between each successive pair of points: First pair of points: m = y = x 2 0 = 8 2 = 4. Second pair of points: m = y = x 10 2 = 32 8 = 4. Third pair of points: m = y = x = = 4. Since the slope is constant, the data satisfies a linear equation. The line has slope 4 and we can see from the table that the y-intercept is 50, so the equation is y = x.