1. sports
2. skating
3. figure skating

5.1 SOLUTIONS
121
CHAPTER FIVE
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 = 1000 − 50d 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 = 60 + 0.2t inches.
3. The initial value is 30 and the temperature decreases by 0.04◦ C per centimeter, so T = 30 − 0.04d.
4. This is already in the form f (x) = b + mx, so we have m = 3, b = 12.
5. Writing this as
g(t) = −5300 + 250t,
we have b = −5300, m = 250.
6. Writing this as
h(n) = 100 + 0.01n,
we have b = 100, m = 0.01.
7. Writing this as
v(z) = 30 + 0 · z,
we have b = 30, m = 0.
8. Writing this as
w(c) = 0 + 0.5c,
we have b = 0, m = 0.5.
9. Writing this as
u(k) = 0.007 + (−0.003)k,
we have b = 0.007, m = −0.003.
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
Population when founded = P (0) = 5000 + 350(0) = 5000.
(b) After one year we have t = 1, so
Population after one year = P (1) = 5000 + 350(1) = 5350.
The population increased by 5350 − 5000 = 350 during the first year. After two years we have t = 2, so
Population after two yearsP (2) = 5000 + 350(2) = 5700.
The population increases by 5700 − 5350 = 350 during the second year. In fact, the population increases by 350
every year.
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 = 0.06. 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 2004. The enrollment
in the class is decreasing by 5 students per year. Thus, m = −5.
17. 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 = 2500. The value of the
antique increases by $80 per year. Thus, m = 80.
19. 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 = −3.
20. 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 20,000 + 1500 · 1 = 21,500 dollars, and when t = 2 it is
20,000 + 1500 · 2 = 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
15
10
5
3
x
−5
3
−3
−5
−10
Figure 5.1
5
5.1 SOLUTIONS
123
25. 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
10
5
4
x
−5
3
−3
5
−5
Figure 5.2
26. 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
−5
3
−3
5
−1
−2
−3
−4
−5
Figure 5.3
27. 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
15
10
5
x
−5
−3
−2
−5
−10
−15
−20
Figure 5.4
3
5
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
15
10
5
x
−5
3
−3
5
−5
Figure 5.5
29. 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
3
2
1
x
−5
−3
−0.2
−1
3
5
−2
−3
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) = 36.60 + 0.14h
f (50) = 36.60 + 0.14 · 50
= 43.6
This tells us that it costs $43.60 when a customer uses 250 + 50 = 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.
5.1 SOLUTIONS
125
33. (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)
0
0.5
1
1.5
2
2.5
3
v(t) (ft/sec)
48
32
16
0
−16
−32
−48
(b) Figure 5.7 shows the velocity over a three second interval.
velocity (ft/sec)
50
2
3
time (sec)
1
−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 = 220.
39. The slope is m = 1/3, and the y-intercept is b = −11.
40. Writing this as
f (x) =
we have slope m = 1/7 and y-intercept b = −12.
1
· x − 12,
7
126
Chapter Five /SOLUTIONS
41. Writing this as
20
2
− · x,
3
3
we have slope m = −2/3 and y-intercept b = 20/3.
42. Writing this as
f (x) = 15 − 2 · 3 − 2(−2x) = 9 + 4x,
we have slope m = 4 and y-intercept b = 9.
43. Writing this as y = πx + 0, we have slope m = π, and y-intercept b = 0.
44. In one hour n birds eating continuously consume V /T in3 of seed. In one hour one bird eating continuously consumes
V /(nT ) in3 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 5.
2. Not linear: it has 5 raised to the xth 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
3
1
1
3
3a + 1
= a + = + a,
4
4
4
4
4
so it is linear in a with constant term 1/4 and coefficient 3/4.
5. This expands out to
3a + 1
3a
1
1
=
+ =3+ ,
a
a
a
a
which is not linear because of the 1/a term.
6. Not linear because of the r 2 term.
7. This is linear in x with constant term 42 and coefficient 1/3.
8. This simplifies to 15A − 3 = −3 + 15A, which is linear in A.
9. The constant term is 4 and the coefficient is 3.
10. Collecting like terms, we get 5x − x + 5 = 4x + 5 = 5 + 4x. Thus the constant terms is 5 and the coefficient is 4.
11. We group the terms with x and the terms without x:
w + wx + 1 = wx + (w + 1) = (w + 1) + wx.
The constant term is w + 1 and the coefficient is w.
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
= (m + 5)x + (mn + m + 7)
= (mn + m + 7) + (m + 5)x.
The constant term is mn + m + 7 and the coefficient is m + 5.
5.2 SOLUTIONS
127
14. 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 − 8 + 12x + 6 = 3 + 10x.
The constant term is 3 and the coefficient is 10.
15. We have
f (x) = 12 + 3(x − 1)
= 12 + 3x − 3
= 9 + 3x.
16. We have
f (x) = 1800 + 500(x + 3)
= 1800 + 500x + 1500
= 3300 + 500x.
17. We have
g(n) = 14 − 2/3(n − 12)
= 14 − 2/3n + 8
= 22 − 2/3n.
18. We have
j(t) = 1.2 + 0.4(t − 5)
= 1.2 + 0.4t − 2
= −0.8 + 0.4t.
19. Since f (1) = 3 · 0 + 5 = 5, the graph passes through the point (1, 5). The slope is 3. See Figure 5.8.
10
5
(1, 5)
x
3
−3
−5
−10
Figure 5.8
128
Chapter Five /SOLUTIONS
20. Since f (−2) = 4 − 2 · 0 = 4, the graph passes through the point (−2, 4). The slope is −2. See Figure 5.9.
y
6
4
(−2, 4)
2
x
3
−3
−2
−4
−6
Figure 5.9
21. Since g(1) = 0/2 + 3 = 3, the graph passes through the point (1, 3). The slope is 1/2. See Figure 5.10.
y
4
3
(1, 3)
2
1
x
3
−3
Figure 5.10
22. Since h(1) = −5 − (1 − 1) = −5, the graph passes through the point (1, −5). The slope is −1. See Figure 5.11.
y
3
−3
(1, −5)
−7
Figure 5.11
x
5.2 SOLUTIONS
129
23. (a) The y-value increases by 1 unit for each 2-unit increase in x, so
Slope =
Change in y
1
= .
Change in x
2
(b) Here the y-values decrease by 4 units for each 1-unit increase in x, so
Slope =
−4
Change in y
=
= −4.
Change in x
1
24. If the personal trainer works 40 hours, the payment will be P (40) = 500 + 18.75(40 − 40) = 500. Thus, the 500 is the
trainers’ weekly salary. The 18.75 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 40.
25. When two guests stay in the room, the cost will be C(2) = 79 + 10(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 2.
26. If the salesperson made $1000 in sales for a week, his income for the week would be T (1000) = 600 + 0.15(1000 −
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 $1000.
27. 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
h(−30) = b − 2(−30)
b = h(−30) + 2(−30) = 80 + 2(−30) = 20,
so h(x) = 20 − 2x.
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
f (20) = b − 1.2(20)
b = f (20) + 1.2(20) = 70 + 1.2(20) = 94,
so f (x) = 94 − 1.2x.
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
f (−12) = b − 0.5(−12)
b = f (−12) + 0.5(−12) = 60 + 0.5(−12) = 54,
so y = 54 − 0.5x.
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 45.
31. 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 20.
33. The area is x2 , 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 = 0.109 dollars/therm = 10.9 cents/therm.
(b) Using the distributive law to rewrite the expression for total cost as
Total cost = 8 + 0.82g + 0.109g = 8 + (0.82 + 0.109)g = 8 + 0.929g,
we see that the expression is linear in g.
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
Cost = 1.5
d
15
+ 0.3d + 20
1.5
d + 0.3d + 20
15
= (0.1 + 0.3)d + 20
=
= 0.4d + 20,
we see that the expression is linear in d.
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 2 .
38. Linear, because regarding r as constant, 2πr 2 + πrh = Constant + Constant · h.
39. Linear, because ax2 + bx + c3 = (bx + c3 ) + (x2 )a = Constant + Constant · a.
40. Not linear because of the term in x2 .
41. 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 + 2 − 10y = (2 − 10y) + (3y + 5)x = Constant + Constant · x.
44. Linear, because 3xy + 5x + 2 − 10y = (5x + 2) + (3x − 10)y = Constant + Constant · y.
45. Not linear, because P (P − c) = P 2 − cP contains P 2 .
46. Linear, because P (P − c) = P 2 + (−P )c = Constant + Constant · c.
47. The slope is
1
∆R
=
= 0.1 inches of rain/inches of snow.
∆S
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
∆m
170 − 70
100
=
=
= 50.
∆h
3−1
2
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
Slope =
m = 70 + 50(h − 1) = 70 + 50h − 50 = 20 + 50h,
so the vertical intercept is 20, which represents the distance from home when the journey starts.
49. The slope is
∆T
31 − 37
−6
=
=
= −3 degrees/hour.
∆h
3−1
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) = 1000 + 500(4 − 4) = 1000. This tells us that, after 4 months of operation, the company’s
profit is $1000.
5.2 SOLUTIONS
131
(b) We have
P (t) = 1000 + 500(t − 4)
= 1000 + 500t − 2000
= −1000 + 500t.
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) = 138 + 40(t − 3), we know that D(3) = 138 + 40(3 − 3) = 138. This tells us that, after 3 hours, Liza
is 138 miles from home.
(b) We have
D(t) = 138 + 40(t − 3)
= 138 + 40t − 120
= 18 + 40t.
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) = 45 − 0.5(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) = 45 − 0.5(t − 12)
= 45 − 0.5t + 6
= 51 − 0.5t.
In this form, we clearly see that the start of the flat is 51 km from the finish line.
55. (a) Since B(20) = 50 + 2(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) = 50 + 2(x − 20)
= 50 + 2x − 40
distributing the 2
= 10 + 2x.
This shows clearly that the collection started with 10 butterflies.
56. (a) Since C(w) = 0.88 + 0.17(w − 1), we know that C(1) = 0.88 + 0.17(1 − 1) = 0.88. 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) = 0.88+0.17(10−1) = 0.88+0.17(9) = 2.41.
Thus, the cost would be $2.41.
57. 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 − (1000 + 15q) = 12q − 1000 = −1000 + 12q 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 400,000 + 500,000d dollars.
132
Chapter Five /SOLUTIONS
59. The two gates cost 2 · 300 = 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 600 + 40x − 80 = 520 + 40x 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 x2 + 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 = −2 + 10x. 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))/(−50 − 100) = (15 − 30)/(−150) = 0.1. Solving for b, we have
g(100) = b + 0.1(100)
b = g(100) − 0.1(100) = 30 − 0.1(100) = 20,
so g(x) = 20 + 0.1x.
Solutions for Section 5.3
EXERCISES
1. (a) Since C is 8, we have Tuition cost = 300 + 200C = 300 + 200(8) = 1900. 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:
1700 = 300 + 200C
7 = C.
Thus, $1700 is the cost of taking 7 credits.
2. We want to know the value of t that makes V (t) equal to $2000, so we solve
18,000 − 1700t = 2000
−1700t = −16,000
−16,000
t=
= 9.412
−1700
subtract 18,000 from both sides
divide both sides by −1700.
Thus the car is worth $2000 about halfway through its 10th year—after about nine and a half years.
3. (a) We have
f (11) =
2 · 11 + 3
25
=
= 5.
5
5
(b) We have
2t + 3
=2
5
2t + 3 = 10
2t = 7
7
t= .
2
5.3 SOLUTIONS
4. We have
5x
=7
2x − 3
5x = 7(2x − 3)
5x = 14x − 21
−9x = −21
7
x= .
3
5. We want to find the value of s that makes 100 + 0.10s = 200. So we solve this equation for s:
100 + 0.10s = 200
0.10s = 100
100
s=
= 1000.
0.10
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 100 + 0.10s = 150 + 0.05s, so we solve this equation for s:
100 + 0.10s = 150 + 0.05s
0.10s − 0.05s = 150 − 100
0.05s = 50
50
s=
= 1000.
0.05
8. We want to find the value of s that makes 150 + 0.05s = 175, so we solve this equation for s:
150 + 0.05s = 175
0.05s = 25
25
= 500.
s=
0.05
9. We want to find the value of s that makes 100 + 0.10s = 175, so we solve this equation for s:
100 + 0.10s = 175
0.10s = 75
75
= 750.
s=
0.10
10. We have
7 − 3y = −17
−3y = −24
−3y
−24
=
−3
−3
y = 8.
11. We have
13t + 2 = 49
13t = 47
47
13t
=
13
13
47
t=
= 3.615.
13
133
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
= 3(4)
9t + 2(t − 1) = 12
9t + 2t − 2 = 12
11t − 2 = 12
11t = 14
14
t=
= 1.273.
11
13. We have
2(r + 5) − 3 = 3(r − 8) + 20
2r + 10 − 3 = 3r − 24 + 20
2r + 7 = 3r − 4
2r = 3r − 11
−r = −11
r = 11.
14. The equation 2x + x = 27 becomes 3x = 27 and x = 9.
15. The equation 4t + 2(t + 1) − 5t = 13 becomes t + 2 = 13 and t = 11.
16.
5
9
−
=0
x−3
1−x
9
5
=
x−3
1−x
9(1 − x) = 5(x − 3)
9 − 9x = 5x − 15
14x = 24
12
x=
.
7
17. 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.
5.3 SOLUTIONS
135
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.
31. 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)
t2 + 3t − (t2 − 5t) = 4t − 20 − (7t − 21)
t2 + 3t − t2 + 5t = 4t − 20 − 7t + 21
8t = −3t + 1
11t = 1
1
.
t=
11
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,
p − 20 = 0.80p
p − 0.80p = 20
0.20p = 20
p = 100.
Thus, the tag price is $100.
38. 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 + 1.25p dollars. If I spend $4 then 4 = 0.99a + 1.25p. If the weight of the apples I buy is twice the weight of
the pears then a = 2p. Thus, 4 = 0.99 × 2p + 1.25p, so p = 4/(1.98 + 1.25) = 1.238 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 + $0.25 · 100 = $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 × $37 + 400 × $0.25 = $211.
(c) If we let m be the number of miles driven, then the cost would be
Cost = 5 × $37 + m × $0.25.
This must be $385, so we need to solve
5 × $37 + m × $0.25 = $385
136
Chapter Five /SOLUTIONS
for m. Isolating m, we find
0.25m = 385 − 5 × 37
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
= 800.
$0.25
40. 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 = 10.
41. Since the total area is the area of the triangle ((1/2) · 4 · x) plus the area of the rectangle (6x) we have:
4x
+ 6x = 144,
2
so 8x = 144 and x = 18 feet.
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,
1 + 50/V hours. Thus, the average speed for the entire journey is
100
100V
mph.
=
V
+ 50
1 + 50
V
(c) If you want to average 75 mph for the entire journey, then
100V
= 75.
V + 50
We must solve this for V . Thus,
100V = 75(V + 50)
25V = 75 · 50
75 · 50
= 150 mph.
V =
25
(d) If you want to average 100 mph for the entire journey, then
100V
= 100.
V + 50
We must solve this for V . Thus,
100V = 100(V + 50)
0 = 100 · 50.
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.
5.4 SOLUTIONS
137
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 6= 0.
50. (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 6= 0.
51. 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 = b1 + m1 x and y = b2 + m2 x must cross, so they must not be
parallel. Thus, their slopes m1 and m2 must be different, m1 6= m2 .
(b) No solution means that the graphs of the lines y = b1 + m1 x and y = b2 + m2 x must not cross, so they must
be parallel, but must not be the same line. Thus, their slopes m1 and m2 must be the same, m1 = m2 , but their
y-intercepts b1 and b2 must be different, b1 6= b2 .
(c) An infinite number of solutions means that the lines y = b1 +m1 x and y = b2 +m2 x must be identical, so m1 = m2
and b1 = b2 .
Solutions for Section 5.4
EXERCISES
1. We have,
y = 100 − 3(x − 20)
= 100 − 3x − 3(−20)
= 100 − 3x + 60
= 160 − 3x,
138
Chapter Five /SOLUTIONS
so m = −3, b = 160.
2. We have,
80x + 90y = 100
90y = 100 − 80x
80
100
−
x
y=
90
90
8
10
− x,
=
9
9
so b = 10/9 and m = −8/9.
3. We have,
x
y
+
=1
100
300
y
x
= 1−
300
100
x
100
x
= 300 − 300 ·
100
= 300 − 3x,
y = 300 1 −
so b = 300 and m = −3.
4. We have,
x = 30 −
2
y
3
2
y = 30 − x
3
3
y = (30 − x)
2
3
= 45 − x,
2
so b = 45, m = −3/2.
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 5.12.
y
3
x
3
−3
−15
Figure 5.12
5.4 SOLUTIONS
139
6. 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 5.13.
y
10
x
3
−3
Figure 5.13
7. 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 5.14.
y
x
−3
3
−2
−6
−10
Figure 5.14
8. 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 5.15.
y
10
x
7
−10
Figure 5.15
2
9. The equation can be put into slope-intercept form y = −4 − 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 5.16.
y
x
3
−3
−2
−6
Figure 5.16
140
Chapter Five /SOLUTIONS
10. This is a linear equation. The y-intercept is b = 200 and the slope is m = −4. See Figure 5.17.
y
200
160
120
80
40
0
x
10
20
30
40
Figure 5.17
11. (a)
(b)
(c)
(d)
(e)
(f)
(g)
is (V), because slope is positive, vertical intercept is negative.
is (IV), because slope is negative, vertical intercept is positive.
is (I), because slope is 0, vertical intercept is positive.
is (VI), because slope and vertical intercept are both negative.
is (II), because slope and vertical intercept are both positive.
is (III), because slope is positive, vertical intercept is 0.
is (VII), because it is a vertical line with positive x-intercept.
12. Using the point-slope form, we have m = 5 and (x0 , y0 ) = (2, 3), so
y = y0 + m(x − x0 )
y = 3 + 5(x − 2).
13. Using the point-slope form, we have m = 6 and (x0 , y0 ) = (−1, 7), so
y = y0 + m(x − x0 )
y = 7 + 6(x − (−1))
y = 7 + 6(x + 1).
14. Using the point-slope form, we have m = −3 and (x0 , y0 ) = (8, 10), so
y = y0 + m(x − x0 )
y = 10 − 3(x − 8).
15. Using the point-slope form, we have m = −2/3 and (x0 , y0 ) = (2, −9), so
y = y0 + m(x − x0 )
2
y = −9 − (x − 2).
3
16. We first find the slope:
∆y
7−1
6
=
= = 2.
∆x
4−1
3
Using the point-slope form, we have m = 2 and (x0 , y0 ) = (4, 7), so
m=
y = y0 + m(x − x0 )
y = 7 + 2(x − 4).
50
5.4 SOLUTIONS
Note that we could have used the point (1, 1) instead.
y = y0 + m(x − x0 )
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:
5−1
4
∆y
=
=
= −4.
∆x
6−7
−1
Using the point-slope form, we have m = −4 and (x0 , y0 ) = (6, 5), so
m=
y = y0 + m(x − x0 )
y = 5 − 4(x − 6).
Note that we could have used the point (7, 1) instead.
y = y0 + m(x − x0 )
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:
∆y
−8 − 4
−12
=
=
= 3.
∆x
−2 − 2
−4
Using the point-slope form, we have m = 3 and (x0 , y0 ) = (−2, −8), so
m=
y = y0 + m(x − x0 )
y = −8 + 3(x − (−2))
y = −8 + 3(x + 2).
Note that we could have used the point (2, 4) instead.
y = y0 + m(x − x0 )
y = 4 + 3(x − 2).
141
142
Chapter Five /SOLUTIONS
To show that the two equations are equivalent, we can rewrite them each in slope-intercept form.
y = −8 + 3(x + 2)
y = −8 + 3x + 6
y = −2 + 3x.
y = 4 + 3(x − 2)
y = 4 + 3x − 6
y = −2 + 3x.
19. We first find the slope:
−7 − (−1)
−6
1
∆y
=
=
=− .
∆x
6 − (−6)
12
2
Using the point-slope form, we have m = −1/2 and (x0 , y0 ) = (6, −7), so
m=
y = y0 + m(x − x0 )
1
y = −7 − (x − 6).
2
Note that we could have used the point (−6, −1) instead.
y = y0 + m(x − x0 )
1
y = −1 − (x − (−6))
2
1
y = −1 − (x + 6).
2
To show that the two equations are equivalent, we can rewrite them each in slope-intercept form.
1
(x − 6)
2
1
y = −7 − x + 3
2
1
y = −4 − x.
2
y = −7 −
1
(x + 6)
2
1
y = −1 − x − 3
2
1
y = −4 − x.
2
y = −1 −
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 (x0 , y0 ) = (−1, −8), so
y = y0 + m(x − x0 )
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 (x0 , y0 ) = (3, −6), so
y = y0 + m(x − x0 )
5
y = −6 + (x − 3).
4
5.4 SOLUTIONS
143
22. 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 (x0 , y0 ) = (12, 20), so
y = 20 +
1
(x − 12).
4
23. 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 = −2.
24. 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 = 2 − 12
−4x + y = −10.
25. 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 = 7.
26. 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 + 10
−5x + y = 10 + 6
−5x + y = 16.
27. 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 = −7.
28. 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.
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 = 5.
30. 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 = −20 + 21
7x − 9y = 1.
31. 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 + 5b2 x + 10b = 16b − 4bx + 8b2
5b2 x + 4bx + 5by = 16b − 10b + 8b2
(5b2 + 4b)x + 5by = 6b + 8b2 .
33. (a) If a line is parallel to y = 3 + 5x, its slope is 5. And since the y-intercept is 10, we have
y = 10 + 5x.
(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.
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 + 7 · 3
22 = b + 21
1 = 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:
y = b − 9x
15 = b − 9 · 5
15 = b − 45
60 = b.
Thus, we have y = 60 − 9x.
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 = 1 + 1 · 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:
2x + 3y = 5
3y = 5 − 2x
5
2
y = − x,
3
3
and
4x + 6y = 7
6y = 7 − 4x
7
4
y= − x
6
6
7
2
y = − x.
6
3
The slope of each is −2/3. Since they have the same slope, they are parallel.
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
q
3
y = − x,
r
r
and
qx + ry = 4
ry = 4 − qx
4
q
y = − x.
r
r
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)
6
y = 5 + 18 x −
18
1
.
y = 5 + 18 x −
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 = y0 + m(x − x0 ).
We see that
Equation I:
Equation II:
Equation V:
y = 9 + 2(x − 4)
contains point (x0 , y0 ) = (4, 9)
y = 9 + (−3)(x − 4)
contains point (x0 , y0 ) = (4, 9)
y − 9 = −3(x − 4)
y = 9 − 2(4 − x)
= 9 − 2(−1)(x − 4)
= 9 + 2(x − 4)
contains point (x0 , y0 ) = (4, 9).
We can quickly verify that the other three equations do not contain this point by letting x = 4 to show that y 6= 9:
Equation III:
Equation IV:
Equation VI:
y + 9 = −2(4 − 4)
y = −9 + 0 = −9
y = 4(4) + 9 = 25
4−8·4
y = 9−
= 9 − (−7) = 16.
4
5.4 SOLUTIONS
147
43. Four of the equations are in (or almost in) point-slope form, y = y0 + m(x − x0 ):
Equation I:
y = 9 + 2(x − 4)
Slope is m = 2
y = 9 + (−3)(x − 4)
Slope is m = −3
y = −9 + (−2)(x − 4)
Slope is m = −2
Equation II:
y − 9 = −3(x − 4)
Equation III:
y + 9 = −2(x − 4)
Equation V:
y = 9 − 2(4 − x)
= 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
4 − 8x
4
1
= 9 − (4 − 8x)
4
= 9 − 1 + 2x)
y = 9−
= 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 = y0 + m(x − x0 ), where (x0 , y0 ) = (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
x + 30.
3
Thus, the graphs of equations III and V have the same y-intercept, b = 30.
46. Placing equation IV into slope-intercept form, we have
y = −30 + 5x.
We see that both equations III and IV have the same slope, m = 5.
47. The line intersects the graph of y = x3 − x + 3 at x = −2 and x = 2. This means that:
At x = −2:
At x = 2:
y = (−2)3 − (−2) + 3 = −3.
y = 23 − 2 + 3
= 9.
Thus, the line contains the points (−2, −3) and (2, 9). We have
m=
9 − (−3)
∆y
12
=
=
= 3.
∆x
2 − (−2)
4
Using the point-slope form and the point (x0 , y0 ) = (2, 9), we can write
y = 9 + 3(x − 2) = 3 + 3x.
148
Chapter Five /SOLUTIONS
48. First we place 5x − 3y = 6 into slope-intercept form:
5x − 3y = 6
3y = 5x − 6
5
y = x − 2.
3
The slope of this line is 5/3, and at x = 15 we have
y=
5
· 15 − 2 = 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
y = b − (3/5)x.
Putting x = 15 and y = 23 we get
23 = b + m · 15
3
23 = b − (15)
5
23 = b − 9
b = 32,
so y = 32 − (3/5)x.
49. We have y = b + mx where m is the negative reciprocal of the slope of y = 0.7 − 0.2x, and so m = −1/(−0.2) = 5.
At x = 1.5 we have y = 0.7 − 0.2(1.5) = 0.4, so
b + 5(1.5) = 0.4
b = 0.4 − 5(1.5) = −7.1.
Thus, y = 5x − 7.1.
50. At x = 12, we know that y = 400 + 25(12) = 700. The slope of our line is m = −1/25 = −0.04. The point-slope
form gives y = 700 − 0.04(x − 12) = 700.48 − 0.04x.
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 − y0 = r(x − x0 )
y = y0 + r(x − x0 )
= y0 + rx − rx0
= y0 − rx0 +rx,
| {z }
b
so b = y0 − rx0 and m = r.
5.4 SOLUTIONS
149
57. We have
x
α
1
= β− ·x
α 1 =β+ −
x,
α
y=β−
| {z }
m
so b = β, m = −1/α.
58. We have
Ax + By = C
By = C − Ax
A
C
− x
y=
B
B
A
C
+ −
=
x,
B
B
|{z}
b
so b = C/B and m = −A/B.
| {z }
m
59. We have
y = b1 + m 1 x + b2 + m 2 x
= b1 + b2 + (m1 + m2 ) x,
| {z }
b
so b = b1 + b2 , m = m1 + m2 .
|
{z
m
}
60. Since
√ √the lines are parallel, they have the same slope, m = −3. The new line passes through the point (x0 , y0 ) =
( 3, 8). By the point-slope formula, we have
y = y0 + m(x − x0 )
√
√
= 8 − 3(x − 3).
√
√
This can also be written in slope-intercept form as y = 8 + 3 3 − 3x.
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 51 and −6, we see that y = 15 x has the greater slope.
(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
2x
3
y=
−
4
4
1
3
y = − + x.
4
2
(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.
150
Chapter Five /SOLUTIONS
64. We begin by putting 3y = 5x − 2 into slope-intercept form by solving for y:
3y = 5x − 2
5
2
y = − + x.
3
3
(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:
y − 3 = −4(x + 2)
y − 3 = −4x − 8
y = −4x − 8 + 3
y = −5 − 4x,
and
−2x + 5y = −3
5y = −3 + 2x
2
3
y = − + x.
5
5
(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 + 2xt2 + 5 = x(3t + 2t2 ) + 5
factor out x
2
= (2t + 3t) ·x + 5,
|
{z
m
}
so m = 2t2 + 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
m1 =
0 − 12
∆y
=
= −4.
∆x
3−0
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
m2 =
∆y
∆x
5.5 SOLUTIONS
=
=
=
=
=
151
2
−0
3
17
−3
3
2
−0
3
17
− 93
3
2
3
17
− 93
3
2
3
8
3
1
2 3
· = .
3 8
4
Since this line contains the point (x0 , y0 ) = (3, 0), its equation is
y =0+
1
3
1
(x − 3) = x − .
4
4
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
1
=−
.
4
−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
m=
We know that b = 5, so the equation is
y2 − y1
5
0−5
=− .
=
x2 − x1
2−0
2
y =5−
5
x.
2
(ii) The new intercepts are (4, 0) and (0, 10), so the slope is
m=
5
0 − 10
y2 − y1
=− .
=
x2 − x1
4−0
2
Now b = 10, so the equation is
y = 10 −
5
x.
2
(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.
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
1 0−2
1
−3 − 0
1
2−3
=− ,
= − , and
=− .
0 − (−2)
2 4−0
2
10 − 4
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
1.00
1.00
1.00
1.00
1.00
∆Q
−4.99
−4.96
−4.94
−4.91
−4.89
m
−4.99
−4.96
−4.94
−4.91
−4.89
7. 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:
Q = b + 0.22t
5.79 = b + 0.22(3)
b = 5.79 − 0.66 = 5.13,
and so Q = 5.13 + 0.22t.
Table 5.3
∆t
4.00
12.00
2.00
5.00
16.00
∆Q
0.88
2.64
0.44
1.10
3.52
m
0.22
0.22
0.22
0.22
0.22
8. We compute the slope between each successive pair of points:
First pair of points:
m=
∆y
58 − 50
8
=
= = 4.
∆x
2−0
2
∆y
90 − 58
32
=
=
= 4.
∆x
10 − 2
8
130 − 90
40
∆y
=
=
= 4.
Third pair of points:
m=
∆x
20 − 10
10
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 = 50 + 4x.
Second pair of points:
m=
5.5 SOLUTIONS
153
9. (a) Let c be the number of pounds of chicken you buy and s the number of pounds of steak. Then 1.29c + 3.49s = 100.
(b) Any combination is reasonable. For example, you could buy 10 lbs of chicken, so that c = 10. This gives
1.29 · 10 + 3.49s = 100
s = (100 − 12.9)/3.49 = 24.957 ≈ 25.0.
So you would buy 25 lb of steak. Thus (10, 25) is one reasonable solution. Also, you could buy 25 lb of chicken, so
that c = 25.
1.29 · 25 + 3.49s = 100
s = (100 − 1.29 · 25)/3.49 = 19.412 ≈ 19.4.
So you would buy about 19.4 lb of steak. Thus (25, 19.4) is another reasonable solution.
(c) See Figure 5.18.
s
30
10 lb chicken, 25 lb steak
20
25 lb chicken, 19.4 lb steak
10
0
c
10
20
30
40
50
60
70
80
Figure 5.18
10. (a) If you walk for w hours and drive for d hours, then 3w + 75d = 60.
(b) If you walk for 2 hours, then w = 2,
3 · 2 + 75d = 60
d = (60 − 6)/75 = 0.72,
so you ride for 0.72 hours, or 0.72 · 60 ≈ 43 minutes. So (2, 0.72) is one reasonable solution. If you walk for 5 hours,
then w = 5, so
3 · 5 + 75d = 60
d = (60 − 15)/75 = 0.6,
so you ride for 0.6 hours, or 0.6 · 60 = 36 minutes. So another reasonable solution is (5, 0.6).
(c) See Figure 5.19.
154
Chapter Five /SOLUTIONS
d
0.8
walk 2 hours, drive 43 minutes
0.6
walk 5 hours, drive 36 minutes
0.4
0.2
w
0
5
10
15
20
Figure 5.19
11. (a) If t is the volume of titanium and i is the volume of iron (in cm3 ), then 4.5t + 7.87i = 5000. Note that the density is
given in grams and the total weight of the bicycle is given in kg, so we must convert 5 kg to 5000 g.
(b) If you use 600 cm3 of titanium, then t = 600, and
4.5 · 600 + 7.87i = 5000
i = (5000 − 4.5 · 600)/7.87 = 292.25,
3
so you would use about 292 cm of iron. So a possible solution is (600, 292). Or you could use 350 cm3 of titanium,
so
4.5 · 350 + 7.87i = 5000
i = (5000 − 4.5 · 350)/7.87 = 435.20,
so you would use about 435 cm3 of iron. So a possible solution is (350, 435).
(c) See Figure 5.20.
i
350 cm3 of titanium,
435 cm3 of iron
500
400
600 cm3 of titanium,
292 cm3 of iron
300
200
100
0
200
400
600
Figure 5.20
800
1000
t
5.5 SOLUTIONS
155
12. (a) If w is the time spent walking and c is the time spent canoeing, both in hours, then 4w + 7c = 30.
(b) If you walk for 3 hours then w = 3, so
4 · 3 + 7c = 30
c = (30 − 12)/7 = 2.57,
so you canoe for about 2.6 hours. So one possible solution is (3, 2.6). If you walk for 1 hour, then w = 1, so
4 · 1 + 7c = 30
c = (30 − 4)/7 = 3.71,
so you canoe for about 3.7 hours. So another possible solution is (1, 3.7).
(c) See Figure 5.21.
c
4
walk 1 hour, canoe 3.7 hours
3
walk 3 hours, canoe 2.6 hours
2
1
w
0
2
4
6
Figure 5.21
PROBLEMS
13. (a) Since I increases by 226.8 units for every increase of 0.5 pounds in the weight w, the instrument reading could be
linear with slope = 226.8/0.5 = 453.6. This would be consistent with the instrument measuring weight, although
not in pounds. If I is linear in w, we have I = b + mw and m = 453.6. To find b, we substitute any point from
the table, such as w = 0, I = 46 into this equation: 46 = b + (453.6)(0), so b = 46. The equation of the line is
I = 46 + 453.6w.
(b) The slope, 453.6, is the same as the number of grams in a pound. Since the reading goes up by 453.6 whenever w
goes up by 1, the slope tells us that the instrument is measuring in grams, not pounds. The vertical intercept is 46, so
this instrument has not been calibrated to weigh 0 pounds correctly. The vertical intercept represents the error in the
calibration, the amount by which the instrument overstates the weight.
14. Since the temperature, T , decreases by 13◦ C for every increase of 2000 meters, the temperature could be a linear function
with slope = −13/2000 = −0.0065. If T is a linear function of h, we have T = b + mh where m = −0.0065. To find
b, we substitute any point from the table, such as h = 0, T = 15 into this equation: 15 = b + (−0.0065)(0), so b = 15.
The equation of the line is T = 15 − 0.0065h, and this fits the data set perfectly.
15. (a) We see that f grams of fat contain 9f calories, and c grams of carbohydrates contain 4c calories. We have
Calories due
to fat
|
{z
9f
Calories due
+
}
and so the constraint equation is 9f + 4c = 2000.
= 2000
to carbohydrates
|
{z
4c
}
156
Chapter Five /SOLUTIONS
(b) The total number of allowed calories is 2000, and so the total number of allowed calories from fat is 30% of 2000
which equals 600. This means that
9f = 600
f = 66.7,
and so a 2000-calorie diet should include no more than about 67 grams of fat.
16. (a) To solve for L, we move the S term to the right and divide by the coefficient of the L term:
250S + 400L = 10,000
400L = 10,000 − 250S
5
L = 25 − S.
8
(b) Similarly, we get:
250S + 400L = 10,000
250S = 10,000 − 400L
8
S = 40 − L.
5
The form in part (a) fits best with Figure 5.25 in the text because the figure has L on the vertical axis. Notice that the
graph in the figure has a vertical drop of 25 for a horizontal run of 40, so that its slope is −25/40 = −5/8, which agrees
with the slope intercept form in part (a).
17. (a) By age 40 the hearing score is 50.
(b) When the hearing score is 40 the predicted age is 48.
(c) The graph shows that hearing ability drops by five points every four years. Thus, the slope of the line is −5/4 and
the hearing ability at birth is 100, so we have h = 100 − (5/4)a.
18. The new constraint equation is 200S + 400L = 16,000. From Figure 5.22, we see that the new constraint line is not
parallel to the old constraint line. This is because the tradeoff in floor space between large and small rooms has changed,
leading to a change in slopes. Solving for L, we see that
200S + 400L = 16,000
400L = 16,000 − 200S
L = 40 − (1/2)S.
The slope is now m = −1/2, which corresponds to the fact that every additional large room uses up enough space for 2
small rooms.
5.6 SOLUTIONS
157
L
40
0 small, 40 large
35
8 small, 36 large
30
24 small, 28 large
25
40 small, 20 large
20
15
10
64 small, 8 large
5
80 small, 0 large
0
16
32
48
64
80
S
Figure 5.22: Graph of the new motel room constraint
equation 200S + 400L = 16,000 together with the old
constraint equation 250S + 400L = 16,000
19. We see that a hectares of Arabica will yield 750a kg of beans, and that r hectares of Robusta will yield 1200r kg of
beans. We have
Amount of
amount of
+
= 1,000,000,
Arabica
Robusta
|
{z
750a
}
|
}
|
{z
}
1200r
and so the constraint equation is 750a + 1200r = 1,000,000.
20. We know that a hectares of Arabica yield 750a kg of beans worth $1.42/kg for a total dollar value of 1.42(750a) = 1065a.
Likewise, r hectares of Robusta yield 1200r kg of beans worth $0.73/kg for a total dollar value of 0.73(1200r) = 876r.
This means that
Value of
value of
+
= 1,000,000
Arabica
Robusta
|
{z
1065a
{z
876r
and so the constraint equation is 1065a + 876r = 1,000,000.
}
Solutions for Section 5.6
EXERCISES
1. Adding the two equations to eliminate y, we have
2x = 12
x = 6.
Using x = 6 in the first equation gives
6 + y = 5,
so
y = −1.
158
Chapter Five /SOLUTIONS
2. Substituting the value of y from the first equation into the second equation, we obtain
x + 2(10 − 3x) = 15
x + 20 − 6x = 15
−5x = −5
x = 1.
Now we substitute x = 1 into the first equation, obtaining 3(1) + y = 10, hence y = 7.
3. Substituting the value of y from the second equation into the first, we obtain
3x − 4(4x − 5) = 7
3x − 16x + 20 = 7
−13x = −13
x = 1.
From the second equation, we have
y = 4(1) − 5 = −1
so
y = −1.
4. Substituting the value of x from the first equation into the second equation, we obtain
4(y − 9) − y = 0
4y − 36 − y = 0
3y = 36
y = 12.
From the first equation, we have
x = 12 − 9
= 3.
5. Since the coefficients of b are the same in each of the two equations, we can add the two equations to eliminate b.
3a = 15
a = 5.
To solve for b, substitute the value for a into one of the equations:
2a + 3b = 4
2(5) + 3b = 4
10 + 3b = 4
3b = −6
b = −2.
Thus, a = 5 and b = −2 is the solution.
5.6 SOLUTIONS
159
6. We need to make the coefficients of one of the variables the same or negatives of each other. We can multiply the bottom
equation by 3 to make the coefficients of w the same. Or, we can multiply the top equation by 2 to make the coefficients
of z negatives of each other. We’ll use the latter approach. We eliminate the z and solve for w.
(
2(3w − z) = 2 · 4
w + 2z = 6
(
6w − 2z = 8
w + 2z = 6.
We now add the second equation to the first:
7w = 14
w = 2.
To solve for z, substitute the value for w into one of the equations:
w + 2z = 6
2 + 2z = 6
2z = 4
z = 2.
Thus, w = 2 and z = 2 is the solution.
7. We need to make the coefficients of one of the variables the same or negatives of each other. We can multiply the top
equation by 5 and the bottom equation by 2 to make the coefficients of p negatives of each other. Or, we can multiply the
top equation by 2 and the bottom equation by 3 to make the coefficients of r the same. We’ll use the first approach. We
eliminate the p and solve for r.
(
5(2p + 3r) = 5 · 10
2(−5p + 2r) = 2 · 13
(
10p + 15r = 50
−10p + 4r = 26
19r = 76
r = 4.
To solve for p, substitute the value for r into one of the equations:
2p + 3r = 10
2p + 3(4) = 10
2p + 12 = 10
2p = −2
p = −1.
Thus, p = −1 and r = 4 is the solution.
8. We can multiply the top equation by 4 and the bottom equation by 5 to make the coefficients of d the same. Or, we can
multiply the top equation by 5 and the bottom equation by 4 to make the coefficients of e the same. We’ll use the first
approach. We eliminate the d and solve for e.
(
4(5d + 4e) = 4 · 2
5(4d + 5e) = 5 · 7
160
Chapter Five /SOLUTIONS
(
20d + 16e = 8
20d + 25e = 35
−9e = −27
e = 3.
To solve for d, substitute the value for e into one of the equations:
5d + 4e = 2
5d + 4(3) = 2
5d + 12 = 2
5d = −10
d = −2.
Thus, d = −2 and e = 3 is the solution.
9. Solving the second equation for y gives
Substituting this into the first equation gives
y = 11 − 4x.
8x − 3 (11 − 4x) = 7
| {z }
y
8x − 3 · 11 + 3 · 4x = 7
8x + 12x = 7 + 33
20x = 40
x = 2.
This means y = 11 − 4 · 2 = 11 − 8 = 3, so (x, y) = (2, 3).
10. Solving the second equation for z gives
z = 5 + 2w.
Substituting this into the first equation gives
4w + 5 (5 + 2w) = 11
| {z }
z
4w + 5 · 5 + 5 · 2w = 11
4w + 10w = 11 − 25
14w = −14
w = −1.
This gives z = 5 + 2(−1) = 3, so (w, z) = (−1, 3).
11. Multiply the first equation by 3,
3(20n + 50m) = 3 · 15
60n + 150m = 45,
and the second by −2:
−2(70m + 30n) = −2 · 22
−140m − 60n = −44.
5.6 SOLUTIONS
Add these two equations gives
60n + 150m − 140m − 60n = 45 − 44
10m = 1
m = 0.1.
This means
20n + 50(0.1) = 15
20n + 5 = 15
20n = 10
n = 0.5,
so (m, n) = (0.1, 0.5).
12. Subtracting the second equation from the first gives
r + s − (s − 2r) = −3 − 6
r + s − s + 2r = −9
3r = −9
r = −3.
From the second equation we have
s − 2(−3) = 6
s+6 = 6
s = 0,
so (r, s) = (−3, 0).
13. Since y equals both 20 − 4x and 30 − 5x, we have
20 − 4x = 30 − 5x
5x − 4x = 30 − 20
x = 10.
Substituting x = 10 into the first equation gives
y = 20 − 4 · 10 = 20 − 40 = −20,
so (x, y) = (10, −20).
14. Multiply the first equation by 5:
5(2p + 5q) = 5 · 14
10p + 25q = 70.
Multiply the second equation by 2:
2(5p − 3q) = 2 · 4
10p − 6q = 8.
Subtract the second of these equations from the first:
10p + 25q − (10p − 6q) = 70 − 8
10p + 25q − 10p + 6q = 62
31q = 62
q = 2.
161
162
Chapter Five /SOLUTIONS
Substituting for q in the first equation gives
2p + 5 · 2 = 14
2p = 4
p = 2,
so (p, q) = (2, 2).
15. Solving by elimination, we multiply the first equation by 7 and the second equation by 9:
7(9x + 10y) = 7 · 21
multiply first equation by 7
9(7x + 11y) = 9 · 26
multiply second equation by 9
63x + 70y = 147
63x + 99y = 234.
Subtracting the first equation from the second gives
63x + 99y − (63x + 70y) = 234 − 147
29y = 87
87
= 3.
y=
29
Substituting y = 3 into the original first equation gives
9x + 10 · 3 = 21
9x + 30 = 21
9x = −9
x = −1.
We can check our answer by verifying that the point (x, y) = (−1, 3) satisfies the original second equation, 7x + 11y =
26:
7(−1) + 11 · 3 = −7 + 33 = 26.
16. Using the process of elimination, we will eliminate the variable y. We multiply the first equation by 8:
8(7x + 5y) = −1 · 8
56x + 40y = −8.
We multiply the second equation by −5:
−5(11x + 8y) = −1(−5)
−55x − 40y = 5.
Adding these two equations gives
56x + 40y + (−55x − 40y) = −8 + 5
x = −3.
Substituting this value of x into the original first equation gives:
7(−3) + 5y = −1
−21 + 5y = −1
5y = 20
y = 4.
We can verify the solution (x, y) = (−3, 4) by substituting these values into the original second equation:
11(−3) + 8(4) = −33 + 32 = −1,
as required. So the solution is (x, y) = (−3, 4).
5.6 SOLUTIONS
163
17. Solving by elimination, we multiply the first equation by 3 and the second equation by 2:
3(5x + 2y) = 3 · 1
multiply first equation by 3
15x + 6y = 3
2(2x − 3y) = 2 · 27 multiply second equation by 2
4x − 6y = 54.
Adding these two equations gives
15x + 6y + 4x − 6y = 3 + 54
19x = 57
x = 3.
Substituting x = 3 into the original first equation gives
5 · 3 + 2y = 1
15 + 2y = 1
2y = −14
y = −7.
We can check our answer by verifying that the point (x, y) = (3, −7) satisfies the original second equation, 2x−3y = 27:
2 · 3 − 3(−7) = 6 + 21 = 27.
18. Solving by elimination, we multiply the first equation by 8 and the second equation by 7:
8(11v + 7w) = 8 · 2 multiply first equation by 8
88v + 56w = 16
7(13v + 8w) = 7 · 1 multiply second equation by 7
91v + 56w = 7.
Subtracting the first equation from the second gives
91v + 56w − (88v + 56w) = 7 − 16
3v = −9
v = −3.
Substituting v = −3 into the original first equation gives
11(−3) + 7w = 2
−33 + 7w = 2
7w = 35
w = 5.
We can check our answer by verifying that (v, w) = (−3, 5) satisfies the original second equation, 13v + 8w = 1:
13(−3) + 8 · 5 = −39 + 40 = 1.
164
Chapter Five /SOLUTIONS
19. Solving by elimination, we multiply the first equation by 4 and the second equation by −3:
4(3e + 2f ) = 4 · 4
Multiply first equation by 4
12e + 8f = 16
−3(4e + 5f ) = −3(−11)
−12e − 15f = 33.
Multiply second equation by −3
Adding these two equations gives
12e + 8f − 12e − 15f = 16 + 33
−7f = 49
f = −7.
Substituting f = −7 into the first equation gives
3e + 2(−7) = 4
3e − 14 = 4
3e = 18
e = 6.
We can check our answer by verifying that (e, f ) = (6, −7) satisfies the original second equation, 4e + 5f = −11:
4 · 6 + 5(−7) = 24 − 35 = −11.
20. We will eliminate the variable p. We multiply the first equation by 3:
3(4p − 7q) = 3 · 2
12p − 21q = 6.
We multiply the second equation by 4:
4(5q − 3p) = 4(−1)
20q − 12p = −4.
Adding these two equations gives
12p − 21q + 20q − 12p = 6 + (−4)
−q = 2,
so q = −2. Substituting this value of q into the original second equation gives
5(−2) − 3p = −1
−10 − 3p = −1
−3p = 9
p = −3.
We can check our answer by substituting these values of p and q into the first equation:
4p − 7q = 4(−3) − 7(−2)
= 2.
5.6 SOLUTIONS
165
21. We set the equations y = x and y = 2 − x equal to one another.
x= 2−x
2x = 2
x=1
y = 1.
So the point of intersection is x = 1, y = 1.
22. Substituting y = x + 1 into 2x + 3y = 13 gives
2x + 3(x + 1) = 13
5x + 3 = 13
5x = 10
x = 2.
If x = 2, then y = 2 + 1 = 3. Thus, the point of intersection is x = 2, y = 3.
23. Substituting y = 2x into 2x + y = 16 gives
2x + 2x = 16
4x = 16
x = 4.
Thus, substituting x = 4 into y = 2x gives y = 8, so the point of intersection is x = 4, y = 8.
24. The point of intersection lies on the two lines
y = 2x − 3.5
and
1
y = − x + 3.5.
2
To find the point, we solve this system of equations simultaneously. Setting these two equations equal to each other and
solving for x, we have
1
2x − 3.5 = − x + 3.5
2
1
2x + x = 3.5 + 3.5 = 7
2
2x + .5x = 7
2.5x = 7
x = 2.8
Since x = 2.8, we have
y = 2x − 3.5 = 2(2.8) − 3.5 = 5.6 − 3.5 = 2.1.
Thus, the point of intersection is (2.8, 2.1).
25. We see from Figure 5.23 that the lines cross at approximately (3, 11). To see whether x = 3, y = 11 is the solution we
need to check that they satisfy
(
y = 6x − 7
y = 3x + 2.
We see that 6x − 7 = 6(3) − 7 = 11, which is y, so the first equation is satisfied. Because 3x + 2 = 3(3) + 2 = 11,
which is y, the second equation is also satisfied, so the solution is x = 3, y = 11.
166
Chapter Five /SOLUTIONS
y
y = 6x − 7
y = 3x + 2
20
(3, 11)
10
x
2
4
6
Figure 5.23
26. We see from Figure 5.24 that the lines cross at approximately (1, 5). To see whether x = 1, y = 5 is the solution we need
to check that they satisfy
(
y = −2x + 7
y = 4x + 1.
We see that −2x + 7 = −2(1) + 7 = 5, which is y, so the first equation is satisfied. Because 4x + 1 = 4(1) + 1 = 5,
which is y, the second equation is also satisfied, so the solution is x = 1, y = 5.
y
10
y = 4x + 1
5
(1, 5)
y = −2x + 7
x
1
2
Figure 5.24
27. We see from Figure 5.25 that the lines cross at approximately (0.47, 1.21). To see whether x = 0.47, y = 1.21 is the
solution we need to check that they satisfy
(
2x + 5y = 7
−3x + 2y = 1.
We see that 2x + 5y = 2(0.47) + 5(1.21) = 6.99, which is approximately 7, so the first equation is essentially satisfied.
Because −3x + 2y = −3(0.47) + 2(1.21) = 1.01, which is approximately 1, the second equation is also essentially
satisfied. Thus, x = 0.47, y = 1.21 is an approximate solution. To find the exact solution, we need solve the system
algebraically, getting x = 9/19, y = 23/19.
5.6 SOLUTIONS
167
y
−3x + 2y = 1
2
(0.47, 1.21)
2x + 5y = 7
1
x
1
Figure 5.25
28. We see from Figure 5.26 that the lines cross at approximately (1.5, −4). To see whether x = 1.5, y = −4 is the solution
we need to check that they satisfy
(
y = 22 + 4(x − 8)
y = 11 − 2(x + 6).
We see that 22 + 4(x − 8) = 22 + 4(1.5 − 8) = −4, which is y, so the first equation is satisfied. Because 11 − 2(x + 6) =
11 − 2(1.5 + 6) = −4, which is y, the second equation is also satisfied, so the solution is x = 1.5, y = −4.
y
y = 22 + 4(x − 8)
2
x
1
2
3
(1.5, −4)
−5
y = 11 − 2(x + 6)
−10
Figure 5.26
PROBLEMS
29. The left side of the equations are not multiples of each other, so there is one solution.
30. The second equation is three times the first one, so there are infinitely many solutions.
31. The equations can be written as
(
x + 3y = 2
x − 3y = 2.
Since the left sides are not multiples of each other, there is one solution.
32. The left side of the second equation is three times the left side of the first equation; however the right sides of the two
equations are equal. Thus this system of equations has no solution.
33. Multiplying both sides of the second equation by xy gives
4
1
3
· xy − · xy =
· xy
y
x
xy
168
Chapter Five /SOLUTIONS
so the equation becomes
4x − y = 3.
Thus the two equations are the same, and there are infinitely many solutions. However, neither x nor y can equal zero, so
neither (0, −3) nor (3/4, 0) is a solution.
34. One approach is to multiply the first equation by 3 and the second by 2:
(
6x + 15y = 3
4y − 6x = 16.
Adding these two equations, we eliminate the variable x:
6x + 15y + 4y − 6x = 3 + 16
19y = 19
y = 1.
Substitution y = 1 into the first equation of the original system gives
2x + 5 · 1 = 1
2x = −4
x = −2,
so the solution to this system is (x, y) = (−2, 1).
35. One approach is to multiply the first equation by 4 and the second by 3:
4(7x − 3y) = 4 · 24
28x − 12y = 96
and
3(4y + 5x) = 3 · 11
12y + 15x = 33.
Adding these two equations gives:
28x − 12y + 12y + 15x = 96 + 33
43x = 129
129
x=
= 3.
43
Using the second equation we can solve for y:
4y + 5 · 3 = 11
4y + 15 = 11
4y = −4
y = −1,
so the solution is (x, y) = (3, −1).
36. Multiplying the first equation by 3 and the second by 7 gives
(
15x − 21y = 93
14x + 21y = −35.
5.6 SOLUTIONS
Adding these two equations gives
15x + 14x − 21y + 21yx = 93 − 35
29x = 58
x = 2.
Substituting x = 2 into the original first equation gives
5 · 2 − 7y = 31
−7y = 31 − 10
21
= −3.
y=
−7
We have (x, y) = (2, −3).
37. One approach is to multiply the first equation by 4 and the second by 7:
44α − 28β = 124
28β − 21α = 14.
Adding these two equations gives
44α − 28β + 28β − 21α = 124 + 14
23α = 138
138
= 6.
α=
23
Substituting α = 6 into the second equation gives
4β − 3 · 6 = 2
4β = 20
β = 5,
so (α, β) = (6, 5).
38. Adding the two equations gives
3β = 33
β = 11.
Substituting for β in the first equation gives
3α + 11 = 32
3α = 21
α = 7.
Thus, (α, β) = (7, 11)
Alternatively, we could multiply the first equation by 2:
2(3α + β) = 2 · 32
6α + 2β = 64.
Subtract the second equation from this:
6α + 2β − (2β − 3α) = 64 − 1
6α + 2β − 2β + 3α = 63
9α = 63
α = 7.
169
170
Chapter Five /SOLUTIONS
Substituting for α in the first equation gives
3 · 7 + β = 32
β = 32 − 31 = 11,
so (α, β) = (7, 11).
39. One approach is to multiply the first equation by 3 and the second equation by 2:
(
9x − 6y = 12
6y − 10x = −10.
Adding these equations, we eliminate the variable y:
9x − 6y + 6y − 10x = 12 − 10
9x − 10x = 2
−x = 2
x = −2.
Substituting x = −2 into the original second equation gives
3y − 5(−2) = −5
3y + 10 = −5
3y = −15
y = −5,
so the solution to the system is (x, y) = (−2, −5).
40. Solving the first equation for f gives
3(e + f ) = 5e + f + 2
3e + 3f = 5e + f + 2
3f − f = 5e − 3e + 2
2f = 2e + 2
f = e + 1.
Substituting for f in the second equation gives
4(f − e) = e + 2f − 4
4((e + 1) − e) = e + 2(e + 1) − 4
4 = e + 2e + 2 − 4
3e = 6
e = 2.
This means f = e + 1 = 2 + 1 = 3, so (e, f ) = (2, 3).
41. Multiplying the second equation by 3 gives
3(2κ + 3ψ) = 3 · 1
6κ + 9ψ = 3.
Adding this to the first equation gives
6κ + 9ψ + 7κ − 9ψ = 3 + 23
13κ = 26
κ = 2.
5.6 SOLUTIONS
171
Substituting κ = 2 into the second equation gives
2 · 2 + 3ψ = 1
3ψ = 1 − 4
ψ = −1.
So (κ, ψ) = (2, −1).
42. Builder A charges C = 3s + 500 and Builder B charges C = 2.5s + 750. When s = 200, Builder A charges 3 · 200 +
500 = 1100 dollars, and Builder B charges 2.5 · 200 + 750 = 1250 dollars, so Builder A is cheaper. For a 1000 square
foot patio, Builder A charges 3 · 1000 + 500 = 3500 dollars and Builder B charges 2.5 · 1000 + 750 = 2250 dollars, so
Builder B is cheaper. The costs are the same when we solve the system
(
C = 3s + 500
C = 2.5s + 750,
so
3s + 500 = 2.5s + 750
0.5s = 250
250
= 500.
s=
0.5
So a 500 square foot patio costs the same with either builder.
43. System II does not have the same solutions as system I since the signs in the equations have been switched.
System III has the same solutions as system I, since the first equation in III is 2 times the second equation in I.
System IV has the same solution as system I, since the first equation in IV, when multiplied by 2, is the second
equation in I. The second equation is the first equation in I if the y is put on the right.
System V does not have the same solutions as either systems I or II. In system V, only the first term on each side of I
has been multiplied by 2.
Confirmation: The solutions to the equations are:
System
I
II
III
IV
V
Solution
(17.5, 8.5)
(17.5, −8.5)
(17.5, 8.5)
(17.5, 8.5)
(17.5, 17)
44. (a) The first equation says that the sum of two numbers is four, and the second equation says that the difference of the
two numbers is two.
(b) The numbers x = 3 and y = 1 satisfy both equations. We can see this by solving the second equation for x, which
gives x = y + 2, and substituting this into the first equation, which gives
(y + 2) + y = 4
2y + 2 = 4
2y = 2
y = 1.
Putting y = 1 in x = y + 2 gives x = 3.
45. We let x and y be the two numbers, and set up two equations in two unknowns:
(
x + y = 17
x − y = 12,
and solve for x and y. Adding the equations we get
2x = 29
29
x=
,
2
172
Chapter Five /SOLUTIONS
and subtracting the second from the first we get
2y = 5
5
y= .
2
46. (a) Suppose there are S small rooms and L large rooms. Each small room uses 250 ft2 , and so S small rooms use a total
of 250S ft2 . Similarly, each large room uses 500 ft2 , and so L large rooms use a total of 500L ft2 . Since the total
square footage for rooms is 16,000 ft2 , we have
250S + 500L = 16,000.
Also, each of the S small rooms can legally hold 2 occupants for a total of 2S occupants. Similarly, each of the L
large rooms can legally hold 5 occupants for a total of 5L occupants. We have
Maximum number
Maximum number
+
of occupants in
large rooms
small rooms
|
{z
|
}
2S
{z
}
5L
Maximum total
=
of occupants in
occupancy
|
2S + 5L = 150.
{z
150
}
(b) We must solve the system of equations given by
(
250S + 500L = 16,000
2S + 5L = 150.
(equation 1)
(equation 2)
Solving for S in equation 1 gives
S + 2L = 64
dividing by 250
S = 64 − 2L.
(equation 3)
Substituting for S in equation 2 gives
2(64 − 2L) + 5L = 150
using equation 3 to substitute for S in equation 2
128 − 4L + 5L = 150
L = 22.
From equation 3, we have S = 64 − 2L = 64 − 2(22) = 20. Thus, the solution to this system is (S, L) =
(20, 22). This tells us that in order to satisfy both floor space and fire code constraints, the motel should build 20
small rooms and 22 large rooms.
47. (a) If we let f , c, and h be the price in dollars of one fish, one order of chips, and one pair of hush-puppies, respectively,
then we have
f + c + h = 2.27
2f + c + h = 3.26
f + c + 2h = 2.76.
If we subtract the first equation from the second, we find (2f + c + h) − (f + c + h) = 3.26 − 2.27, or f = 0.99.
So fish cost $0.99 a piece.
If we subtract the first equation from the third, we find (f + c + 2h) − (f + c + h) = 2.76 − 2.27, or h = 0.49.
So one pair of hush-puppies costs $0.49.
If we rewrite the first equation as c = 2.27 − f − h and use f = 0.99 and h = 0.49, we find c = 2.27 − 0.99 −
0.49 = 0.79. So chips cost $0.79 an order.
Thus, the cost of two fish, two orders of chips, and one pair of hush-puppies is 2f + 2c + h = 2 · 0.99 + 2 ·
0.79 + 0.49 = 4.05, that is, $4.05.
5.6 SOLUTIONS
173
(b) By distributing the right-hand side of
2x + 2y + z = 3(x + y + z) − (x + y + 2z)
we have
3(x + y + z) − (x + y + 2z) = 3x + 3y + 3z − x − y − 2z = 2x + 2y + z.
(c) If we let x, y, and z be the price in dollars of one fish, one order of chips, and on pair of hush-puppies, respectively,
then 2x + 2y + z is the cost of two fish, two orders of chips, and one pair of hush-puppies (what we are looking
for), x + y + z is the cost of one fish, one order of chips, and one pair of hush-puppies, so x + y + z = 2.27, and
x + y + 2z is the cost of one fish, one order of chips, and two pairs of hush-puppies, so x + y + 2z = 2.76. Thus,
2x + 2y + z = 3(x + y + z) − (x + y + 2z) = 3 · 2.27 − 2.76 = 4.05.
(d) In the solution in part (c), we did not use the information that “Two fish, one order of chips, and one pair of hushpuppies costs $3.26” to solve part (a).
48. For there to be many solutions, the graphs of the two equations must be overlapping lines. This will happen if n = 10,
for then the two equations describe the same line. Otherwise, the system describes two non-overlapping parallel lines, so
there are no solutions.
49. Taking the hint, we rewrite the first two equations as
3x + 2y + 5(2x) = 11
13x + 2y = 11
and
2x − 3y + 2x = 7
4x − 3y = 7.
This gives us a new system of two equations in two variables (instead of 3 equations in 3 variables):
(
13x + 2y = 11
4x − 3y = 7.
We now multiply the first equation by 3 and the second by 2:
(
39x + 6y = 33
8x − 6y = 14.
We add these two equations to eliminate the variable y:
47x = 47
x = 1.
Given that x = 1, from the third equation of original system we have
z = 2x
= 2.
Given that x = 1, z = 2, from the second equation of the original system we have
3x + 2y + 5z = 11
3 · 1 + 2y + 5 · 2 = 11
2y = −2
y = −1,
so the solution is x = 1, y = −1, z = 2. This can also be written (x, y, z) = (1, −1, 2).
174
Chapter Five /SOLUTIONS
Solutions for Chapter 5 Review
EXERCISES
1. When n = 0, we have C = 4.29, which is the initial value. This tells us that the company charges $4.29 to rent the video
3.99
for up two days. The rate of growth (slope) is 3.99 which represents ∆∆dollars
days = 1 . Thus, for each day beyond the first
two, an additional $3.99 is charged for each additional day the video is rented.
2. The initial value of 30 tells us that the student was 30 miles from home when he noticed the formula. The rate of change
∆miles
55
(slope) is 55. Since this represents ∆
hours = 1 we see that the student’s distance from home is increasing at a rate of
55 miles per hour.
3. The initial value is 250, which represents the cost of $250 to start up the business. The rate of change (slope) is 1/36.
dollars = 1 we know that it costs $1 to produce 36 donuts.
Since this represents ∆
36
∆donuts
4. The initial value is 300, which is the number of people in the lecture hall when the lecture began. The rate of change
∆people
we see that people are leaving the lecture hall at the rate of 19
(slope) is −19/3. Since this represents ∆minutes = −19
3
people every 3 minutes.
5. This graph indicates that the temperature at midnight was 50◦ F. The temperature increased at a constant rate of 2◦ per
hour.
6. This graph indicates that the temperature at midnight was 50◦ F. The temperature decreased at a constant rate of 2◦ per
hour.
7. This graph indicates that the temperature at midnight was 50◦ F. The temperature remained a constant 50◦ throughout the
night.
8. Writing function IV in slope-intercept form, we have y = 10,000 + 200t − 2000 = 8000 + 200t. We see that investments
III and IV both begin with $8000.
9. The value of investment III is $8000, no matter what the value of t. Notice that rewriting function III as y = 8000 + 0t,
we have m = 0, which means the rate of change is $0/year.
10. Investment VI begins with $8500, which is more than the other investments.
11. We start with the fixed fee, $50 and add $45 times the number of hours, h. The total cost is 50 + 45h dollars.
12. We take the initial population, 23,400 and subtract 200 times the number of years, y. The population is 23,400 − 200y.
13. We start with the fixed cost, $2400 and subtract $500 times the number of years, y. The total value is 2400 − 500y dollars.
14. We take the initial cost, $7600 and add $3500 times the number of months, m. The total cost is 7600 + 3500m dollars.
15. The student’s score begins with an 80 and increases 2 points times p problems, giving us 80 + 2p.
16. The initial distance from home is 200. The distance increases by 50 miles per hour for h hours. The total distance from
home is 200 + 50h miles.
17. We start with the membership fee, $350 and add $30 times the number of months, m. The total cost is 350+30m dollars.
18. The starting value is b = 12,000, and the growth rate is m = 225, so h(t) = 12,000 + 225t.
19. The starting value is b = 77 and the rate of change is m = −3.2 cm/year, so d = 77 − 3.2t.
20. There are initially b = 8 million viewers. The viewership drops by 4 million in 5 episodes, or by 4/5 = 0.8 million
people per episode. This gives p(n) = 8 − 0.8n.
21. We have b = 200, m = 14.
22. Writing this as g(t) = −46 + 77t, we have b = −46, m = 77.
23. Writing this as h(t) = 0 + (1/3)t, we have b = 0, m = 1/3.
24. Writing this as 0.003 + 0 · t, we have b = 0.003, m = 0.
SOLUTIONS to Review Problems for Chapter Five
25. Writing this as
q(t) =
2t + 7
2t
7
7
2
=
+ = + t,
3
3
3
3
3
we have b = 7/3, m = 2/3.
√
√
√
√
7 + (−0.3 8)t, we have b = 7, m = −0.3 8.
26. Writing this as
27. We have,
w(−4) = 9 + 4(−4) = 9 − 16 = −7.
28. We have,
w(x − 4) = 9 + 4(x − 4)
= 9 + 4x − 16
= 4x − 7.
29. We have,
w(x + h) − w(x) = 9 + 4(x + h) − (9 + 4x)
|
{z
w(x+h)
}
| {z }
w(x)
= 9 + 4x + 4h − 9 − 4x
= 4h.
30. We have,
w(9 + 4x) = 9 + 4(9 + 4x)
= 9 + 4 · 9 + 4 · 4x
= 9 + 36 + 16x
= 45 + 16x.
31. Linear, because it can be written as −7 + 7x.
32. Not linear, because of the term in x2 .
33. Linear, because mx + b + c3 = (b + c3 ) + mx.
34. Not linear, because mx + b + c3 x2 = c3 x2 + mx + b, so there is a term in x2 .
35. Not linear. We have
P (P − b)(c − P ) = P (−P 2 + (b + c)P − bc) = −P 3 + (b + c)P 2 − bcP,
so the expression has terms in P 2 and P 3 .
36. Linear. We have
P (P − b)(c − P ) = −P 3 + (b + c)P 2 − bcP = (−P 3 + bP 2 ) + (P 2 − bP )c,
which is linear in c.
37. Linear, because P (2 + P ) − P 2 = 2P + P 2 − P 2 = 2P .
38. Not linear, because P (2 + P ) − 2P 2 = 2P + P 2 − 2P 2 = 2P − P 2 has a term in P 2 .
39. Linear, because xy + ax + by + ab = (by + ab) + (y + a)x.
40. Linear, because xy + ax + by + ab = (ax + ab) + (x + b)y.
175
176
Chapter Five /SOLUTIONS
41. Linear, because xy + ax + by + ab = (xy + by) + (x + b)a.
42. Linear, because xy + ax + by + ab = (xy + ax) + (y + a)b.
43. Positive, because 7x must be positive.
44. Positive, because 3x must be positive.
45. Positive, because 5x must be positive.
46. Negative, because −3x must be positive.
47. Negative, because −5x must be positive.
48. Positive, because, collecting like terms, we see that 5x must be positive.
49. Positive, because, collecting like terms, we see that 13x must be positive.
50. Positive, because, collecting like terms, we see that −10x must be negative.
51. Negative, because, collecting like terms, we see that 6x must be negative.
52. Negative, because, collecting like terms, 2x must be negative.
53. No solution, because the value of 8x added to 3 can’t equal the same value added to 11.
54. Zero, because, collecting like, we see that −2x must equal 0.
55. Dividing by P t gives
r=
I
.
Pt
56. We have
9
C + 32
5
9
F − 32 = C
5
5
C = (F − 32).
9
F =
57. Solving for y:
a − cy
+a = 0
b + dy
a − cy
= −a
b + dy
a − cy = −a(b + dy) = −ab − ady
ady − cy = −ab − a
(ad − c)y = −a(b + 1)
a(b + 1)
.
y=−
ad − c
58. Multiplying on both sides by C − B(1 − 2x) gives
Ax − B = 3(C − B + 2Bx)
Ax − B = 3C − 3B + 6Bx
Ax − 6Bx = 3C − 3B + B = 3C − 2B
x(A − 6B) = 3C − 2B
3C − 2B
x=
.
A − 6B
SOLUTIONS to Review Problems for Chapter Five
177
59. (a) Since the slopes are −2 and −4, we see that y = 5 − 2x has the greater slope.
(b) Since the y-intercepts are 5 and 8, we see that y = 8 − 4x has the greater y-intercept.
60. (a) Since the slopes are 3 and −10, we see that y = 7 + 3x has the greater slope.
(b) Since the y-intercepts are 7 and 8, we see that y = 8 − 10x has the greater y-intercept.
2
61. The equation y = −4 + x is in slope-intercept form. Put a point on the y-axis of −4. From there, for every 1 unit you
3
move to the right, also move up 2/3 units, which is the same as saying for every 3 units you move to the right, move up 2
units. See Figure 5.27.
y
x
3
−3
−2
−6
Figure 5.27
62. We put this in slope-intercept form, y = 2/5 − (4/5)x. So the y-intercept is at 2/5. We plot the point (0, 2/5). Then we
draw a line of slope −4/5 through this point, so we move down 4/5 for every unit increase in x, which is the same as
moving down 4 for every increase of 5 in x. See Figure 5.28.
y
14/5
x
3
−3
−2
Figure 5.28
63. We have y = −14 + 81 x, with b = −14, m = 1/8.
64. We have
y − 8 = 3(x − 5)
y − 8 = 3x − 15
y = −7 + 3x,
so b = −7 and m = 3.
65. We have
y−4
2x − 3
=
5
3
y−4
2x − 3
(5 · 3) =
(5 · 3)
5
3
3(y − 4) = 5(2x − 3)
3y − 12 = 10x − 15
3y = −3 + 10x
10
y = −1 +
x,
3
so b = −1, m = 10/3.
178
Chapter Five /SOLUTIONS
66. We have
4x − 7y = 12
−7y = 12 − 4x
4
12
+ x,
y=−
7
7
so b = −12/7, m = 4/7.
67. Writing this as y = 90 + 0 · x, we have b = 90 and m = 0.
√
√
68. Writing this as y = 0 + 8x, we have b = 0, m = 8.
69. 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 = 3x − 2
y − 3x = −2.
Writing the variables in the more standard order, we get −3x + y = −2.
70. To put in standard form, we must gather all the variables, with their coefficients, on one side, and the constant(s) on the
other:
3x = 2y − 1
3x − 2y = −1.
71. Given the points (−2, 1) and (2, 9), we first find the slope:
m=
∆y
9−1
8
=
= = 2.
∆x
2 − (−2)
4
Using the point-slope form, we have m = 2 and (x0 , y0 ) = (2, 9), so
y = y0 + m(x − x0 )
y = 9 + 2(x − 2).
We need to convert this to slope-intercept form.
y = 9 + 2(x − 2)
y = 9 + 2x − 4
y = 5 + 2x.
Note that if we use the point (−2, 1), we obtain the same answer.
72. Given the points (−2, −7) and (2, −5), we first find the slope:
m=
−7 − (−5)
∆y
−2
1
=
=
= .
∆x
−2 − 2
−4
2
Using the point-slope form, we have m = 1/2 and (x0 , y0 ) = (−2, −7), so
y = y0 + m(x − x0 )
1
y = −7 + (x − (−2))
2
1
y = −7 + (x + 2).
2
SOLUTIONS to Review Problems for Chapter Five
We need to convert this to slope-intercept form.
1
(x + 2)
2
1
y = −7 + x + 1
2
1
y = −6 + x.
2
y = −7 +
Note that if we use the point (2, −5), we obtain the same answer.
73. Given the points (−4, 11) and (4, 5), we first find the slope:
m=
∆y
11 − 5
6
3
=
=
=− .
∆x
−4 − 4
−8
4
Using the point-slope form, we have m = −3/4 and (x0 , y0 ) = (−4, 11), so
y = y0 + m(x − x0 )
3
y = 11 − (x − (−4))
4
3
y = 11 − (x + 4).
4
We need to convert this to slope-intercept form.
3
(x + 4)
4
3
y = 11 − x − 3
4
3
y = 8 − x.
4
y = 11 −
Note that if we use the point (4, 5), we obtain the same answer.
74. Using point-slope form, we have m = 2 and (x0 , y0 ) = (4, 6), so
y = y0 + m(x − x0 )
y = 6 + 2(x − 4).
We need to convert this to slope-intercept form.
y = 6 + 2(x − 4)
y = 6 + 2x − 8
y = −2 + 2x.
75. Using point-slope form, we have m = −3 and (x0 , y0 ) = (−6, 2), so
y = y0 + m(x − x0 )
y = 2 − 3(x − (−6))
y = 2 − 3(x + 6).
We need to convert this to slope-intercept form.
y = 2 − 3(x + 6)
y = 2 − 3x − 18
y = −16 − 3x.
179
180
Chapter Five /SOLUTIONS
76. Using point-slope form, we have m = 1/2 and (x0 , y0 ) = (−4, −8), so
y = y0 + m(x − x0 )
1
y = −8 + (x − (−4))
2
1
y = −8 + (x + 4).
2
We need to convert this to slope-intercept form.
1
(x + 4)
2
1
y = −8 + x + 2
2
1
y = −6 + x.
2
y = −8 +
77. Using the point-slope form, we have m = −2/3 and (x0 , y0 ) = (9, 7), so
y = y0 + m(x − x0 )
2
y = 7 − (x − 9).
3
We need to convert this to slope-intercept form.
2
(x − 9)
3
2
y = 7− x+6
3
2
y = 13 − x.
3
y = 7−
78. We first find the slope:
8 − 12
−4
∆y
=
=
= 2.
∆x
6−8
−2
Using the point-slope form, we have m = 2 and (x0 , y0 ) = (6, 8), so
m=
y = y0 + m(x − x0 )
y = 8 + 2(x − 6).
We need to convert this to slope-intercept form.
y = 8 + 2(x − 6)
y = 8 + 2x − 12
y = −4 + 2x.
Note that if we use the point (8, 12), we obtain the same answer.
79. We first find the slope:
m=
∆y
5−4
1
=
= .
∆x
−3 − (−6)
3
Using the point-slope form, we have m = 1/3 and (x0 , y0 ) = (−3, 5), so
y = y0 + m(x − x0 )
1
y = 5 + (x − (−3))
3
1
y = 5 + (x + 3).
3
SOLUTIONS to Review Problems for Chapter Five
We need to convert this to slope-intercept form.
1
(x + 3)
3
1
y = 5+ x+1
3
1
y = 6 + x.
3
y = 5+
Note that if we use the point (−6, 4), we obtain the same answer.
80. We first find the slope:
m=
∆y
−2 − 1
−3
1
=
=
=− .
∆x
4 − (−8)
12
4
Using the point-slope form, we have m = −1/4 and (x0 , y0 ) = (4, −2), so
y = y0 + m(x − x0 )
1
y = −2 − (x − 4).
4
We need to convert this to slope-intercept form.
1
(x − 4)
4
1
y = −2 − x + 1
4
1
y = −1 − x.
4
y = −2 −
Note that if we use the point (−8, 1), we obtain the same answer.
81. We first find the slope:
6−3
3
3
∆y
=
=
=− .
∆x
5 − 10
−5
5
Using the point-slope form, we have m = −3/5 and (x0 , y0 ) = (5, 6), so
m=
y = y0 + m(x − x0 )
3
y = 6 − (x − 5).
5
We need to convert this to slope-intercept form.
3
(x − 5)
5
3
y = 6− x+3
5
3
y = 9 − x.
5
y = 6−
Note that if we use the point (10, 3), we obtain the same answer.
82. If two lines are parallel, their slopes are equal. Thus, the slope of our line is m = 1/2.
Using the point-slope form, we have m = 1/2 and (x0 , y0 ) = (4, 7), so
y = y0 + m(x − x0 )
1
y = 7 + (x − 4).
2
181
182
Chapter Five /SOLUTIONS
We need to convert this to slope-intercept form.
1
(x − 4)
2
1
y = 7+ x−2
2
1
y = 5 + x.
2
y = 7+
83. If two lines are parallel, their slopes are equal. Thus, the slope of our line is m = −4/5.
Using the point-slope form, we have m = −4/5 and (x0 , y0 ) = (−10, −5), so
y = y0 + m(x − x0 )
4
y = −5 − (x − (−10))
5
4
y = −5 − (x + 10).
5
We need to convert this to slope-intercept form.
4
(x + 10)
5
4
y = −5 − x − 8
5
4
y = −13 − x.
5
y = −5 −
84. We can write the equation in slope-intercept form
3x + 5y = 6
5y = 6 − 3x
3
6
y = − x.
5
5
The slope is −3
. Lines parallel to this line all have slope
5
to 5. So y = 5 − 53 x.
−3
.
5
Since the line passes through (0, 5), its y-intercept is equal
85. Writing 12y − 4x = 8 in slope-intercept form, we have
12y − 4x = 8
12y = 4x + 8
1
y=
(4x + 8)
12
1
2
= x+ ,
3
3
so the slope of this line is m = 1/3. Since the new line is perpendicular to this line, its slope is −1/m = −1/(1/3) = −3.
Using point-slope form with (x0 , y0 ) = (−10, −30), we have
y = −30 + (−3)(x − (−10))
= −30 − 3(x + 10)
= −30 − 3x − 30
= −60 − 3x.
86. We have y = b + mx where b = −80. The line contains that points (30, 0) and (0, −80) and so
m=
so y = 38 x − 80.
8
0 − (−80)
= ,
30 − 0
3
SOLUTIONS to Review Problems for Chapter Five
183
87. We have s(t) = b + mt where b = 8200. The growth rate is given by
m=
12,700 − 8200
= 150,
30 − 0
so s(t) = 8200 + 150t.
88. The cost is found by adding the connection fee of $2.95 to the number of minutes times the cost per minute, so C =
2.95 + 0.35n.
89. We have w(4) = 20 and w(12) = −4. This gives y = mx + b where m = (w(12) − w(4))/(12 − 4) = (−4 − 20)/8 =
−3. Solving for b, we have
w(4) = b − 3(4)
b = w(4) + 3(4) = 20 + 3(4) = 32,
so w(x) = −3x + 32.
90. We have p(−30) = 20 and p(70) = 140. This gives y = mx + b where m = (p(70) − p(−30))/(70 − (−30)) =
(140 − 20)/100 = 1.2. Solving for b, we have
p(−30) = b + 1.2(−30)
b = p(−30) − 1.2(−30) = 20 − 1.2(−30) = 56,
so p(x) = 1.2x + 56.
91. We have w(x) = b + mx where
w(0.7) − w(0.3)
0.7 − 0.3
0.01 − 0.07
=
0.4
= −0.15.
m=
We can use the point (0.3, 0.07) to solve for b:
w(0.3) = b + m(0.3)
0.07 = b + (−0.15)(0.3)
b = 0.115.
Thus, w(x) = 0.115 − 0.15x.
92. We have g(5) = 50 and g(30) = 25. This gives y = mx+b where m = (g(30)−g(5))/(30−5) = (25−50)/25 = −1.
Solving for b, we have
g(5) = b − 1(5)
b = g(5) + 1(5) = 50 + 1(5) = 55,
so g(x) = −x + 55.
93. Since the graph is parallel to the line y = 20 − 4x, the slope is the same, so m = −4. Using the point-slope form with
(x0 , y0 ) = (3, 12) we have
y = 12 − 4(x − 3)
= 12 − 4x + 12
= 24 − 4x.
184
Chapter Five /SOLUTIONS
94. We have f (−1) = 4 and f (2) = −11. This gives y = mx+b where m = (f (2)−f (−1))/(2−(−1)) = (−11−4)/3 =
−5. Solving for b, we have
f (−1) = b − 5(−1)
b = f (−1) + 5(−1) = 4 + 5(−1) = −1,
so f (x) = −5x − 1.
95. Since the function is linear, we can use any two points to find its formula. We use the form
y = b + mx
to get temperature in ◦ C, y, as a function of temperature in ◦ F, x. We use the two points, (32, 0) and (41, 5). We begin by
finding the slope, ∆y/∆x = (5 − 0)/(41 − 32) = 5/9. Next, we substitute a point into our equation using our slope of
5/9◦ C per ◦ F and solve to find b, the y-intercept. We use the point (32, 0):
0 = b+
−
5
· 32
9
160
= b.
9
Therefore,
160
5
+ x.
9
9
Traditionally, we give this formula as y = (5/9)(x − 32), which is often easier to manipulate. You might want to check
to see if the two are the same.
y=−
96. Since the function is linear, we can choose any two points (from the graph) to find its formula. We use the form
p = b + mh
to get the price of an apartment as a function of its height. We use the two points (10, 175,000) and (20, 225,000).
We begin by finding the slope, ∆p/∆h = (225,000 − 175,000)/(20 − 10) = 5000. Next, we substitute a point into
our equation using our slope of 5000 dollars per meter of height and solve to find b, the p-intercept. We use the point
(10, 175,000):
175,000 = b + 5000 · 10
125,000 = b.
Therefore,
p = 125,000 + 5000h.
97. Since the function is linear, we can use any two points (from the graph) to find its formula. We use the form
u = b + mn
to get the meters of shelf space used as a function of the number of different medicines stocked. We use the two points
(60, 5) and (120, 10). We begin by finding the slope, ∆u/∆n = (10 − 5)/(120 − 60) = 1/12. Next, we substitute a
point into our equation using our slope of 1/12 meters of shelf space per medicine and solve to find b, the u-intercept. We
use the point (60, 5):
5 = b + (1/12) · 60
0 = b.
Therefore,
u = (1/12)n.
The fact that b = 0 is not surprising, since we would expect that, if no medicines are stocked, they should take up no shelf
space.
SOLUTIONS to Review Problems for Chapter Five
185
98. Since the function is linear, we can use any two points (from the graph) to find its formula. We use the form
s = b + mq
to get the number of hours of sleep obtained as a function of the quantity of tea drunk. We use the two points (4, 7) and
(12, 3). We begin by finding the slope, ∆s/∆q = (3 − 7)/(12 − 4) = −0.5. Next, we substitute a point into our equation
using our slope of −0.5 hours of sleep per cup of tea and solve to find b, the s-intercept. We use the point (4, 7):
7 = b − 0.5 · 4
9 = b.
Therefore,
s = 9 − 0.5q.
99. Since we are told that the function is linear, any two points will define the line for us. We will use the form
y = b + mx
to get temperature in ◦ Rankine, y, as a function of temperature in ◦ Fahrenheit, x. (Rankine is a rarely used absolute
temperature scale.) We choose the two points, (0, 459.7) and (10, 469.7). We begin by finding the slope, ∆R/∆F =
(469.7 − 459.7)/(10 − 0) = 10/10 = 1. Next, we substitute a point into our equation using our slope of 1 ◦ R per ◦ F
and solve to find b, the y-intercept. We use the point (0, 459.7):
459.7 = b + 1 · 0
459.7 = b.
Therefore,
y = 459.7 + 1x.
Note that each ◦ R is the same as each ◦ F, but the two systems choose different starting points (zero ◦ R is absolute zero,
while zero ◦ F is an arbitrary point).
100. Since the function is linear, we can choose any two points to find its formula. We use the form
q = b + mp
to get the number of bottles sold as a function of the price per bottle. We use the two points (0.50, 1500) and (1.00, 500).
We begin by finding the slope, ∆q/∆p = (500 − 1500)/(1.00 − 0.50) = −2000. Next, we substitute a point into our
equation using our slope of −2000 bottles sold per dollar increase in price and solve to find b, the q-intercept. We use the
point (1.00, 500):
500 = b − 2000 · 1.00
2500 = b.
Therefore,
q = 2500 − 2000p.
101. As the t values increase by 1, the v values decrease by 1. Since there is a constant rate of change, this could represent a
linear function.
102. As the p values increase by 2, the q values increase by 0.5. Since there is a constant rate of change, this could represent a
linear function.
103. As the x values increase by 1, the y values increase by different amounts. Since there is not a constant rate of change, this
could not represent a linear function.
104. As the n values increase by 3, the C values first decrease by 1, and then increase by 1. Since there is not a constant rate
of change, this could not represent a linear function.
186
Chapter Five /SOLUTIONS
105. Although neither the x-values nor the y-values change by a constant amount, the y-values decrease by half the amount
the x-values increase. We can see this by calculating the slope between successive data points. We find
1 5−8
1
1−5
1
8 − 10
=− ,
= − , and
=− .
5−1
2 11 − 5
2
19 − 11
2
Thus, the data are linear with slope m = − 12 , and could represent a linear function.
106. The x-values increase by 2, 6, and then 18. The y-values decrease in steps of 5. Since there is not a constant rate of
change, this is not a linear equation.
107. We eliminate the variable b. We multiply the first equation by 4:
4(2a − 3b) = 4 · 22
8a − 12b = 88.
We multiply the second equation by 3:
3(3a + 4b) = 3 · 1
9a + 12b = −3.
Adding these two equations gives
8a − 12b + 9a + 12b = 88 − 3
17a = 85
a = 5.
Substituting this value into the original first equation gives
2 · 5 − 3b = 22
−3b = 12
b = −4.
We check our answer by substituting these values of a and b into the second equation:
3a + 4b = 3 · 5 + 4(−4)
= −1.
108. We eliminate the variable n. We multiply the first equation by 3:
3(2n + 7m) = 3 · 1
6n + 21m = 3.
We multiply the second equation by −2:
−2(3n + 10m) = −2 · 3
−6n − 20m = −6.
Adding these two equations gives
6n + 21m + (−6n − 20m) = 3 + (−6)
m = −3.
Substituting this value into the original first equation gives
2n + 7(−3) = 1
SOLUTIONS to Review Problems for Chapter Five
2n − 21 = 1
2n = 22
n = 11.
We check our answer by substituting these values of m and n into the second equation:
3n + 10m = 3 · 11 + 10(−3)
= 3.
109. We have
multiplying 1st equation by 3
3(10x + 4y) = 3(−3)
30x + 12y = −9
multiplying 2nd equation by −5
−5(6x − 5y) = −5 · 13
−30x + 25y = −65
30x + 12y + (−30x + 25y) = −9 + (−65)
adding these two equations
37y = −74
y = −2.
Substituting this into the first equation gives
10x + 4(−2) = −3
10x − 8 = −3
10x = 5
1
x= .
2
Thus, (x, y) = (0.5, −2).
110. We eliminate the variable v. We multiply the first equation by 3:
3(2v + 3w) = 3 · 11
6v + 9w = 33.
We multiply the second equation by 2:
2(2w − 3v) = 2 · 29
4w − 6v = 58.
Adding these two equations gives
6v + 9w + (4w − 6v) = 33 + 58
13w = 91
w = 7.
Substituting this value of w into the original first equation gives
2v + 3 · 7 = 11
2v + 21 = 11
2v = −10
v = −5.
187
188
Chapter Five /SOLUTIONS
We check our answer by substituting these values of v and w into the second equation:
2w − 3v = 2(7) − 3(−5)
= 29.
PROBLEMS
111. At t = 0 minutes, there are 100 + 4 · 0 = 100 gallons in the tank, so 100 represents the number of gallons initially in the
tank. After 1 minute there are 100 + 4 · 1 = 104 gallons in the tank; after 2 minutes there are 100 + 4 · 2 = 108 gallons
in the tank; and so on. Thus, 4 is the number of gallons per minute that are flowing into the tank. The tank is full when
t = 25, at which time it will hold 100 + 4 · 25 = 200 gallons.
112. Since the total cost involves multiplication of P by constants and the addition of constants, we expect the expression to
be linear in P . The facility will cost 500 + 20P + 0.20(500 + 20P ) dollars. This is linear in P and can be written as
500 + 20P + 100 + 4P = 600 + 24P .
113. Since the total cost involves multiplication of i by constants and the addition of a constant, we expect the expression to be
linear in i. The class will cost 30+12i+3i+0.07(12i) dollars. This is linear in i and can be written as 30+15i+0.84i =
30 + 15.84i.
114. The constant 200 is the starting cost, the cost of buying the printer. The constant 1.00 tells us the rate at which the total
cost goes up with each photo printed:
Cost for 1 photo = 200 + 1.00 · 1 = $201
Cost for 2 photos = 200 + 1.00 · 2 = $202
Cost for 3 photos = 200 + 1.00 · 3 = $203
..
.
115. Since r is the daily rental fee, 9r must represent the accumulated daily rental fees, which means that the skier had the skis
for 9 days. Thus 9 represents the length of the vacation in days.
116. Since the graphs of u and v intersect at x = −2, 3, we have
v(−2) = u(−2)
= 1 + (−2)3
= −7.
v(3) = u(3)
= 1 + 33
= 28.
We have v(−2) = −7 and v(3) = 28. This gives y = mx + b where m = (v(3) − v(−2))/(3 − (−2)) = (28 −
(−7))/5 = 7. Solving for b, we have
v(−2) = b + 7(−2)
b = v(−2) − 7(−2) = −7 − 7(−2) = 7,
so v(x) = 7x + 7.
117. We have
f (−2) = 0.5(−2)3 − 4 = −8
f (4) = 0.5(4)3 − 4 = 28.
SOLUTIONS to Review Problems for Chapter Five
189
This gives y = mx + b where m = (f (4) − f (−2))/(4 − (−2)) = (28 + 8)/6 = 6. Solving for b, we have
f (−2) = b + 6(−2)
b = f (−2) − 6(−2) = −8 − 6(−2) = 4,
so f (x) = 6x + 4.
118. The towns’ populations have the same, constant rate of change. The slopes of the lines describing them are each 1000
people per decade. Since the graphs are lines with the same slope, the graphs are parallel lines.
119. The bacteria populations do not have a constant rate of change, so the graphs that describe them are not lines. For example,
if one of the populations started at 100, it would have 130 bacteria after one hour and 169 bacteria after 2 hours, which is
not a constant rate of change. Since the graphs are not lines, they are not parallel lines either.
120. The countries’ populations have the same, constant rate of change. The slopes of the lines describing them are each
1 million people per decade or 100 thousand people per year (since there are 10 years in a decade, and 100 thousand
multiplied by 10 is 1 million). Since the graphs are lines with the same slope, the graphs are parallel lines.
121. The villages’ populations do not have the same constant rate of change, so the graphs that describe them are not parallel.
The graph of Village G’s population has a slope of 2 people per year, and the graph of Village H’s population has a slope
of 1 person per year.
122. The d dishes cost 3d dollars and the tip is 15% of 3d which is 0.15(3d) = 0.45d dollars. The total cost is 3d + 0.45d =
3.45d dollars which is linear in d.
123. The tip you pay is (t/100) · 15 dollars. Your total cost is 15 + (15/100)t dollars which is linear in t.
124. If your income is I dollars then your adjusted income is I−5(1000) dollars. Your tax is 0.25(I−5(1000)) = 0.25I−1250
dollars which is linear in I. Your tax is linear in your income.
125. The cost is: 9 · 2 dollars for adult tickets; 7N dollars for child tickets; 3N dollars for popcorn; and 5 dollars for parking.
The total is 9 · 2 + 7N + 3N + 5 = 23 + 10N dollars which is linear in N .
126. (a) After t hours the first person has traveled 65t miles and the second person has traveled 75(t − 1) miles. The distance
between them is 65t − 75(t − 1) miles.
(b) This is valid between when the second person departs and when the cars meet. The second person departs at t = 1
hour, and the cars meet when
65t − 75(t − 1) = 0
−10t + 75 = 0
t = 75/10 = 7.5.
Thus the expression is valid for 1 ≤ t ≤ 7.5.
127. At time d = 0 the pond is full and contains 2 · 10 = 20 ft3 water.
After 1 day (d = 1) the depth lost is 0.3 inches = 0.3/12 = 0.025 feet, so the volume lost is 10 · 0.025 ft3 , so
(20 − 10 · 0.025) ft3 remains.
After 2 days (d = 2), the depth lost is 2 · 0.3 inches = 2 · 0.025 feet, so the volume lost is 10 · 2 · 0.025 ft3 , so
(20 − 10 · 2 · 0.025) ft3 remains.
Thus, after d days the depth lost is d · 0.3 inches = d · 0.025 feet, so the volume lost is 10 · d · 0.025 ft3 , so
20 − 10 · d · 0.025 = 20 − 0.25d ft3 remains. This is valid from d = 0 until the pond is empty, which will occur when
20 − 0.25d = 0, that is, d = 20/0.25 = 80 days.
128. There is a net change of water of 1 − 5 = −4 gallons per minute, which is a net loss. At t = 0 there is 100 gallons, so after
1 minute there is 100 − 4 = 96 gallons, after 2 minutes there is 100 − 4 · 2 = 92 gallons. After t minutes the container
has 100 − 4t gallons. The amount of water is going down, so the container is emptying. When there are 0 gallons in the
container, it is empty. This will occur when 100 − 4t = 0, so that t = 25 minutes. Thus, there are 100 − 4t gallons in the
container for 0 ≤ t ≤ 25. The container is empty when t = 25. After that the statement of the problem no longer makes
sense, since the container is empty and cannot be losing more water.
129. (a) We first find the slope using the points (20, 68) and (25, 77).
m=
∆F
77 − 68
9
=
= .
∆C
25 − 20
5
190
Chapter Five /SOLUTIONS
Using the point-slope form, we have m = 9/5 and (C0 , F0 ) = (20, 68), so
F = F0 + m(C − C0 )
9
F = 68 + (C − 20).
5
We can rewrite the function in slope-intercept form.
9
(C − 20)
5
9
F = 68 + C − 36
5
9
F = 32 + C.
5
(i) Using the function from part (a), we can convert the temperature of 10◦ C to Fahrenheit.
F = 68 +
(b)
9
C
5
9
F = 32 + (10)
5
F = 32 + 18
F = 32 +
F = 50.
◦
◦
Thus, a temperature of 10 C is equal to 50 F.
(ii) Using the function from part (a), we can convert the temperature of 86◦ F to Celsius.
9
C
5
9
86 = 32 + C
5
9
54 = C
5
C = 30.
F = 32 +
Thus, a temperature of 86◦ F is equal to 30◦ C.
130. (a) Since the temperature drops about 3◦ F per 1000 foot rise, we know that the slope can be represented by:
∆T
3
=−
.
∆E
1000
Using the point-slope form, we have m = −3/1000 and (E0 , T0 ) = (7000, 53), so
m=
(b)
T = T0 + m(E − E0 )
3
(E − 7000)
T = 53 −
1000
3
T = 53 −
(E − 7000).
1000
(i) Using the function from part (a), we can find the average temperature at Sun Visitor Center, which is located at
9745 feet.
3
T = 53 −
(E − 7000)
1000
3
T = 53 −
(9745 − 7000)
1000
3
(2745)
T = 53 −
1000
T = 53 − 8.235
T = 44.765.
Thus, we can expect that the average temperature at Sun Visitor Center is approximately 45◦ F.
SOLUTIONS to Review Problems for Chapter Five
191
(ii) Using the function from part (a), we can approximate the average temperature at sea level (elevation = 0).
3
(E − 7000)
1000
3
T = 53 −
(0 − 7000)
1000
3
(−7000)
T = 53 −
1000
T = 53 + 21
T = 53 −
T = 74.
Thus, we can approximate the average temperature at sea level to be 74◦ F.
131. (a) We first find the slope using the points (3, 12.5) and (6, 17).
m=
17 − 12.5
4.5
∆C
=
=
= 1.5.
∆n
6−3
3
Using the point-slope form, we have m = 1.5 and (n0 , C0 ) = (3, 12.5), so
C = C0 + m(n − n0 )
C = 12.5 + 1.5(n − 3).
(b)
C = 12.5 + 1.5(n − 3)
C = 12.5 + 1.5n − 4.5
C = 8 + 1.5n.
(c) We can use the function in part (b) to determine the slope and the intercept. In the function C = 8 + 1.5n, the slope
is 1.5 and the intercept is 8.
∆C(cost)
1.5
The slope of
represents
. Thus, we can see that each additional ride costs $1.50.
1
∆n(number of rides)
The intercept of 8 tell us that the cost is $8 for no rides. Thus, the cost of admission to the carnival is $8.
132. (a) We first find the slope using the points (100, 320) and (500, 400).
400 − 320
80
∆C
=
=
= 0.2.
∆n
500 − 100
400
Using the point-slope form, we have m = 0.2 and (n0 , C0 ) = (100, 320), so
m=
C = C0 + m(n − n0 )
C = 320 + 0.2(n − 100).
(b)
C = 320 + 0.2(n − 100)
C = 320 + 0.2n − 20
C = 300 + 0.2n.
The slope is 0.2 and the intercept is 300.
∆C(cost)
0.2
The slope of
represents
. Thus, we can see that it costs $0.20 to produce each CD.
1
∆n(number of CDs)
The intercept of 300 tell us that the cost is $300 to produce 0 CDs. This $300 is the start-up cost of producing
CDs.
192
Chapter Five /SOLUTIONS
(c) To determine the cost of producing 750 CDs, we substitute 750 for n in our equation.
C = 300 + 0.2n
C = 300 + 0.2(750)
C = 300 + 150
C = 450.
Thus, it costs $450 to produce 750 CDs.
(d) To determine the number of CDs that can be produced for $500, we substitute 500 for C in our equation.
C = 300 + 0.2n
500 = 300 + 0.2n
200 = 0.2n
n = 1000.
If the band has $500, it can produce 1000 CDs.
133. The timeshare was bought at time 0, so the purchase price, including the first maintenance fee, was $3000. After 1
year the total cost was about $3,500, so the maintenance fee was $3500 − $3000 = $500. Thus, the timeshare cost
$3000 − $500 = $2500.
134. (iv), because 7x must be positive but less than 7.
135. (v), because 4x must be greater than 4.
136. (ii), because 2x must be negative but greater than −2.
137. (ii), because 11x must be negative but greater than −11.
138. (v), because 11x is greater than 11.
139. (iv), because 2x must be positive but less than 2.
140. (i), because, collecting terms, −2x equals 2.
141. (ii), because, collecting terms, 3x is negative but greater than −3.
142. (vi), because the quantity −4x added to 3 can’t equal the same quantity added to −3.
143. (iii), because, collecting terms, we find that 3x = 0.
144. (iv), because x/2 is less than 1/2.
145. (v), because 2/x is less than 2.
146. (iii), because (x + 8)/2 = 4 = (0 + 8)/2.
147. (i), because (x + 8)/3 is less than (−1 + 8)/3 = 7/3.
148. (v), because the denominator must be 10.
149. (iii), because the denominator must be 5.
150. (vi), because the numerator must equal the denominator, but 2x plus 5 can’t equal the same quantity plus 6.
151. (i), because 2x + 2 = 3x + 5.
1
152. (iv), because = 4.
x
1
153. (vi), because can’t equal 0.
x
154. (a) We need to solve F = 9C/5+32 for C. We subtract 32 from both sides and multiply by 5/9 to obtain 5(F −32)/9 =
C, so C = 5(F − 32)/9.
(b) When F = 350 we have C = 5(350−32)/9 = 176.7◦ C. When F = 450 we have C = 5(450−32)/9 = 232.2◦ C.
155. (a) Since number of degrees Fahrenheit is twice the number of degrees Celsius, we have F = 2C.
(b) We need to solve the system of linear equations
F =
9
C + 32
5
SOLUTIONS to Review Problems for Chapter Five
193
F = 2C,
for F and C. Substituting the second equation into the first gives 2C = 9C/5 + 32, so C = 160. This is equivalent
to F = 2 · 160 = 320. Thus, the two formulas agree when F = 320 and C = 160, that is, 320◦ F = 160◦ C.
(c) The equations
9
C + 32
5
F = 2C,
F =
represent two lines which intersect at C = 160, which is below the lowest temperature, 176.7◦ C, used in the oven.
Thus, the vertical distance between the two lines increases as the temperature increases. So this distance is largest
(maximum error) at the upper end of this temperature range.
F
Formulas agree
when C = 160
500
400
?
300
200
100
150
200
250
C
Figure 5.29
156. (a) When C = 21 we have F = 9 · 21/5 + 32 = 69.8, which is about 70◦ F.
(b) No. When C = 42 we have F = 9 · 42/5 + 32 = 107.6◦ F.
157. The rate of discharge is constant, so we assume f is a linear function with slope m = −0.25 mAh/s. Notice that m is
negative because the battery is discharging, so the amount of charge is going down. The initial charge is b = 4500 mAh,
so f (t) = 4500 − 0.25t.
158. The maximum charge is given by g(0) = 3200 mAh. This is 1300 mAh less, or about 29% less, than the new battery. The
battery discharges at a rate of 0.4 mAh/s. This is 0.15 mAh/s faster, or 60% faster, than the new battery.
159. We have
f (t1 ) = 0
4500 − 0.25t1 = 0
4500
= 18,000.
t1 =
0.25
g(t2 ) = 0
3200 − 0.4t2 = 0
3200
= 8000.
t2 =
0.4
This means that the new battery fully discharges after 18,000 seconds, and the old battery fully discharges after 8000
seconds. The expression
t1
18,000
= 2.5
=
t2
8, 000
tells us that the new batter lasts 2.5 times as long as the old battery.
160. The bird uses 5 calories for each hour spent singing for a total of 5s calories, and 10 calories for each hour spent defending
its territory for a total of 10d calories. We have
Calories spent
singing
+
Calories spent
defending territory
= 60
5s + 10d = 60.
194
Chapter Five /SOLUTIONS
Setting s = 0, we see that the d-intercept is d = 6, which means the bird could spend a maximum of 6 hours defending
its territory if it spends no time singing. Setting d = 0, we see that the s-intercept is s = 12, which means the bird could
spend a maximum of 12 hours singing if it spends no time defending its territory. See Figure 5.30.
d (hours)
6
5
4
3
2
1
0
3
6
9
12
s (hours)
Figure 5.30
161. The bird uses 4 calories for each hour spent singing for a total of 4s calories, and 12 calories for each hour spent defending
its territory for a total of 12d calories. We have
Calories spent
+
singing
Calories spent
= 60
defending territory
4s + 12d = 60.
Setting s = 0, we see that the d-intercept is d = 5, which means the bird could spend a maximum of 5 hours defending
its territory if it spends no time singing. Setting d = 0, we see that the s-intercept is s = 15, which means the bird could
spend a maximum of 15 hours singing if it spends no time defending its territory. See Figure 5.31.
d (hours)
5
4
3
2
1
0
3
6
9
12
Figure 5.31
162. We have a simultaneous system of linear equations:
(
5s + 10d = 60
4s + 12d = 60.
Multiplying the first equation by 4 and the second by 5 gives
(
20s + 40d = 240
20s + 60d = 300.
15
s (hours)
SOLUTIONS to Review Problems for Chapter Five
195
Subtracting 20s + 40d = 240 from 20s + 60d = 300 gives
20d = 60
d = 3.
Solving for s using the first equation gives
5s + 10 · 3 = 60
5s = 30
s = 6.
This means that both birds can spend 6 hours singing and 3 hours defending their territories with a budget of 60 calories.
163. The total amount of Italian that can be purchased is 60/10 = 6 lbs, so this matches statement (d):
Amt. of
=
Tot. possible amt.
of Italian
K
= 6−3·
2
3
= 6 − K.
2
Italian
− 3 lbs. ×
Amt. Kenyan
2 lbs.
164. This matches statement (b):
Total − Amount spent on Italian
Price for Kenyan
60 − 10I
.
=
15
Amount of Kenyan =
165. The total amount of Kenyan that can be purchased is 4 lbs, because 60/15 = 4. Thus, this equation matches statement
(c):
Amt. of
= 1.5
Italian
Tot. possible
amt. of Kenyan
−
Amt. of Kenyan
purchased
= 1.5(4 − K).
166. This matches statement (a):
Total amount spent each week = 60
Amount spent
+
on Italian
|
167. We substitute the expression
{z
10I
}
Amount spent
= 60
on Kenyan
|
{z
15K
}
10I + 15K = 60.
3
x + 6 for y in the first equation.
4
2x + 4y
3
2x + 4
x+6
4
12
x + 24
2x +
4
5x
= 44
= 44
= 44
= 20
x = 4.
or
!
196
Chapter Five /SOLUTIONS
From the second equation, we have
3
·4+6
4
y = 9.
y=
168. One way to solve this system is by substitution. Solve the first equation for y:
3x − y = 20
−y = 20 − 3x
y = 3x − 20.
In the second equation, substitute the expression 3x − 17 for y:
−2x − 3y = 5
−2x − 3(3x − 20) = 5
−2x − 9x + 60 = 5
−11x = 5 − 60 = −55
−55
x=
= 5.
−11
Since x = 5 and y = 3x − 20, we have
y = 3(5) − 20 = 15 − 20 = −5.
Thus, the solution to the system is x = 5 and y = −5.
Check your results by substituting the values into the second equation:
−2x − 3y = 5
Substituting, we get − 2(5) − 3(−5) = 5
−10 + 15 = 5
5 = 5.
169. From the first equation, we get 2x + 2y = 5. From the second equation we get −2x − y = −9. So,
(
2x + 2y = 5
−2x − y = −9.
Adding the two equations gives y = −4, and solving for x in either equation gives x = 6.5.
170. Because we are solving for x and y, we regard b as a constant. Multiplying the first equation by b and subtracting the
second gives
b2 x + by = 2b2
x + by = 1 + b2
so, subtracting
(b2 − 1)x = b2 − 1.
Since b 6= ±1, we have b2 − 1 6= 0, so x = 1. Solving for y in the first equation gives y = 2b − b(1), so y = b.
SOLUTIONS to Review Problems for Chapter Five
197
171. Company A charges C = 10x + 40 dollars, while company B charges C = 8x + 50 dollars. Since Company A charges a
lower delivery fee, its price is lower for small quantities of sand. For example, its price for 1 cubic yard is $50, compared
with $58 for Company B. However, since Company A charges more per cubic yard, its price is eventually higher. For
example, its price for 10 cubic yards is $140, compared with $130 for Company B. Both charge the same when
(
C = 10x + 40
C = 8x + 50,
that is, when 10x + 40 = 8x + 50. The solution to this equation is x = 5 cubic yards. Thus if 0 < x < 5 use Company
A, and if 5 < x ≤ 12 use Company B. If x = 5, use either company.
172. Subtracting the second equation from the first gives
2
3
− = 2t − t = t.
4
3
Thus,
1
.
12
So a UK tablespoon is 1/12 a US cup. If we substitute this into
t=
1
2
= u+t
3
2
we find
2
1
1
= u+
,
3
2
12
or
1
1
2
−
= u,
3
12
2
giving
u=
7
.
6
So one UK cup is 7/6 a US cup.
173. Multiplying the second equation by 2 and adding it to the first gives
1+
or,
3
1
1
= u + u,
2
4
2
3
5
= u.
2
4
Thus,
6
.
5
This says that a UK cup is 6/5 a US cup. If we substitute this into
u=
1
1
= u−d
4
4
we find
1 6
1
= · − d,
4
4 5
giving
d=
This says that a UK dessertspoon is 1/20 a UK cup.
1
.
20
198
Chapter Five /SOLUTIONS
Solutions for Solving Drill
1. We have
5x + 2 = 12
5x = 10
x = 2.
2. We have
7 − 3x = 25
−3x = 18
x = −6.
3. We have
5 + 9t = 72
9t = 67
67
t=
.
9
4. We have
10 = 25 − 3r
−15 = −3r
r = 5.
5. We have
5x + 2 = 3x − 7
2x = −9
9
x = − = −4.5.
2
6. We have
7 + 5t = 10 − 3t
8t = 3
3
t = = 0.375.
8
7. We have
100 − 24w = 5w − 30
130 = 29w
130
w=
= 4.483.
29
SOLUTIONS to Review Problems for Chapter Five
8. We have
1.25 + 0.07x = 3.92
0.07x = 2.67
2.67
= 38.143.
x=
0.07
9. We have
0.5t − 13.4 = 25.8
0.5t = 39.2
39.2
= 78.4.
t=
0.5
10. We have
12.53 + 5.67x = 45.1x − 125
137.53 = 39.43x
137.53
= 3.488.
x=
39.43
11. We have
3(x − 2) + 15 = 5(x + 4)
3x − 6 + 15 = 5x + 20
3x + 9 = 5x + 20
−2x = 11
11
x=−
= −5.5.
2
12. We have
3t − 5(t + 2) = 3(2 − 4t) + 8
3t − 5t − 10 = 6 − 12t + 8
−2t − 10 = 14 − 12t
10t = 24
24
12
t=
=
= 2.4.
10
5
13. We have
5(2p − 6) + 10 = 6(p + 3) + 4(2p − 1)
10p − 30 + 10 = 6p + 18 + 8p − 4
10p − 20 = 14p + 14
−4p = 34
17
34
=−
= −8.5.
p=−
4
2
199
200
Chapter Five /SOLUTIONS
14. We have
2.3(2x + 5.9) = 0.1(24.7 + 54.2x) + 2.4x
4.6x + 13.57 = 2.47 + 5.42x + 2.4x
4.6x + 13.57 = 2.47 + 7.82x
11.1 = 3.22x
11.1
= 3.447.
x=
3.22
15. We have
10.8 − 3.5(40 − 5.1t) = 3.2t + 4.5(25.4 − 5.6t)
10.8 − 140 + 17.85t = 3.2t + 114.3 − 25.2t
−129.2 + 17.85t = 114.3 − 22t
39.85t = 243.5
243.5
t=
= 6.110.
39.85
16. We have
ax + b = c
ax = c − b
c−b
x=
.
a
17. We have
rt + s = pt + q
rt − pt = q − s
(r − p)t = q − s
q−s
t=
.
r−p
18. Since we are solving for s, we put the terms with s on one side of the equation, and the terms without s on the other side.
We have
rsw − 0.2rw + 0.1sw = 5r − 1.8s
rsw + 0.1sw + 1.8s = 5r + 0.2rw
(rw + 0.1w + 1.8)s = 5r + 0.2rw
5r + 0.2rw
s=
.
rw + 0.1w + 1.8
19. Since we are solving for r, we put the terms with r on one side of the equation, and the terms without r on the other side.
We have
rsw − 0.2rw + 0.1sw = 5r − 1.8s
rsw − 0.2rw − 5r = −0.1sw − 1.8s
(sw − 0.2w − 5)r = −(0.1sw + 1.8s)
0.1sw + 1.8s
.
r=−
sw − 0.2w − 5
SOLUTIONS to Review Problems for Chapter Five
201
20. We begin by using the distributive law. Then, since we are solving for t, we put terms with t on one side of the equation,
and terms without t on the other side. We have
r(As − Bt + Cr) = 25t + A(st + Br)
Ars − Brt + Cr 2 = 25t + Ast + ABr
−Brt − 25t − Ast = ABr − Ars − Cr 2
−(Br + As + 25)t = ABr − Ars − Cr 2
t=−
ABr − Ars − Cr 2
.
Br + As + 25
21. We begin by using the distributive law. Then, since we are solving for s, we put terms with s on one side of the equation,
and terms without s on the other side. We have
r(As − Bt + Cr) = 25t + A(st + Br)
Ars − Brt + Cr 2 = 25t + Ast + ABr
Ars − Ast = 25t + ABr + Brt − Cr 2
(Ar − At)s = 25t + ABr + Brt − Cr 2
s=
25t + ABr + Brt − Cr 2
.
Ar − At
22. We begin by using the distributive law. Then, since we are solving for P , we put terms with P on one side of the equation,
and terms without P on the other side. We have
0.2P Q2 + RQ(P − 1.5Q + 2R) = 2.5
0.2P Q2 + RP Q − 1.5RQ2 + 2R2 Q = 2.5
0.2P Q2 + RP Q = 2.5 + 1.5RQ2 − 2R2 Q
(0.2Q2 + RQ)P = 2.5 + 1.5RQ2 − 2R2 Q
P =
2.5 + 1.5RQ2 − 2R2 Q
.
0.2Q2 + RQ
23. Since we are solving for y ′ , we put terms with y ′ on one side of the equation, and terms without y ′ on the other side. We
have
x2 y 2 + 2xyy ′ + x2 y ′ − 5x + 2y ′ + 10 = 0
2xyy ′ + x2 y ′ + 2y ′ = 5x − x2 y 2 − 10
(2xy + x2 + 2)y ′ = 5x − x2 y 2 − 10
y′ =
5x − x2 y 2 − 10
.
2xy + x2 + 2
24. We first use the distributive law. Then, since we are solving for V0 , we put terms with V0 on one side of the equation, and
terms without V0 on the other side. We have
25V0 S 2 [T ] + 10(H 2 + V0 ) = A0 (3V + V0 )
25V0 S 2 [T ] + 10H 2 + 10V0 = 3A0 V + A0 V0
25V0 S 2 [T ] + 10V0 − A0 V0 = 3A0 V − 10H 2
(25S 2 [T ] + 10 − A0 )V0 = 3A0 V − 10H 2
V0 =
3A0 V − 10H 2
.
25S 2 [T ] + 10 − A0
202
Chapter Five /SOLUTIONS
25. We first use the distributive law. Then, since we are solving for [T ], we put terms with [T ] on one side of the equation,
and terms without [T ] on the other side. We have
25V0 S 2 [T ] + 10(H 2 + V0 ) = A0 (3V + V0 )
25V0 S 2 [T ] + 10H 2 + 10V0 = 3A0 V + A0 V0
25V0 S 2 [T ] = 3A0 V + A0 V0 − 10H 2 − 10V0
[T ] =
3A0 V + A0 V0 − 10H 2 − 10V0
.
25V0 S 2