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Chapter 2, continued Lesson 2.4

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Chapter 2, continued Lesson 2.4
Chapter 2,
continued
Lesson 2.4
6. The line passes through (x1, y1) 5 (22, 5) and
1. The slope is m 5 3 and the y-intercept is b 5 1. Use the
slope-intercept form to write an equation of the line.
y 5 mx 1 b
y 5 3x 1 1
2. The slope is m 5 22 and the y-intercept is b 5 24. Use
the slope-intercept form to write an equation of the line.
y 5 mx 1 b
y 5 (22)x 1 (24)
y 5 22x 2 4
3
7
3. The slope is m 5 2} and the y-intercept is b 5 }.
4
2
Use the slope-intercept form to write an equation of
the line.
y 5 mx 1 b
3
(x2, y2) 5 (4, 27). Find its slope.
y2 2 y1
2
7
4. Because you know the slope and a point on the line, use
point-slope form to write an equation of the line. Let
(x1, y1) 5 (21, 6) and m 5 4.
y 2 y1 5 m(x 2 x1)
y 2 6 5 4(x 2 (21))
y 2 6 5 4(x 1 1)
y 2 6 5 4x 1 4
y 5 4x 1 10
5. a. The given line has a slope of m1 5 3. So, a line parallel
to it has a slope of m2 5 m1 5 3. You know the slope
and a point on the line, so use the point-slope form
with (x1, y1) 5 (4, 22) to write an equation of the line.
y 2 y1 5 m2(x 2 x1)
y 2 (22) 5 3(x 2 4)
y 1 2 5 3(x 2 4)
y 1 2 5 3x 2 12
y 5 3x 2 14
b. A line perpendicular to a line with slope m1 5 3 has a
1
1
. Use point-slope form
slope of m2 5 2}
m1 5 2}
3
with (x1, y1) 5 (4, 22).
y 2 y1 5 m2(x 2 x1)
1
y 2 (22) 5 2}3 (x 2 4)
1
y 1 2 5 2}3 (x 2 4)
1
4
y 1 2 5 2}3 x 1 }3
1
2
y 5 2}3 x 2 }3
212
1
You know the slope and a point on the line, so use pointslope form with either given point to write an equation of
the line. Choose (x1, y1) 5 (22, 5).
y 2 y1 5 m(x 2 x1)
y 2 5 5 22(x 2 (22))
y 2 5 5 22(x 1 2)
y 2 5 5 22x 2 4
y 5 22x 1 1
7. The line passes through (x1, y1) 5 (6, 1) and
(x2, y2) 5 (23, 28). Find its slope.
y2 2 y1
28 2 1
29
m5}
5}
5}
51
23 2 6
29
x 2x
2
y 5 2}4 x 1 }2
27 2 5
m5}
5}
5}
5 22
x 2x
6
4 2 (22)
1
You know the slope and a point on the line, so use pointslope form with either given point to write an equation of
the line. Choose (x1, y1) 5 (23, 28).
y 2 y1 5 m(x 2 x1)
y 2 (28) 5 11 x 2 (23) 2
y 1 8 5 1(x 1 3)
y185x13
y5x25
8. The line passes through (x1, y1) 5 (21, 2) and
(x2, y2) 5 (10, 0). Find its slope.
y2 2 y1
022
22
2
m5}
5}
5}
5 2}
x 2x
11
11
10 2 (21)
2
1
You know the slope and a point on the line, so use pointslope form with either given point to write an equation of
the line. Choose (x1, y1) 5 (21, 2).
y 2 y1 5 m(x 2 x1)
2
y 2 2 5 2}
(x 2 (21))
11
2
(x 1 1)
y 2 2 5 2}
11
2
2
2
20
y 2 2 5 2}
x2}
11
11
x1}
y 5 2}
11
11
9. Let x represent the time (in years) since 1993 and let
y represent the number of participants (in millions).
The initial value is 3.42. The rate of change is the slope
m. Use (x1, y1) 5 (0, 3.42) and (x2, y2) 5 (10, 3.99).
y2 2 y1
3.99 2 3.42
0.57
5}
5}
5 0.057
m5}
x 2x
10 2 0
10
2
1
Participants 5 Initial 1 Rate of +
change
number
y 5 3.42 1 0.057 + x
In slope-intercept form, a linear model is
y 5 0.057x 1 3.42
64
Algebra 2
Worked-Out Solution Key
Years
since
1993
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
2.4 Guided Practice (pp. 98–101)
Chapter 2,
10. Company A
song price
continued
Songs from
+ Company A
Company B
1 song price +
10. Let (x1, y1) 5 (3, 21) and m 5 23.
y 2 y1 5 m(x 2 x1)
Songs from
Company B 5
Your
budget
0.69 + x 1 0.89 + y 5 30
An equation for this situation is 0.69x 1 0.89y 5 30.
y 1 1 5 23(x 2 3)
y 1 1 5 23x 1 9
y 5 23x 1 8
11. Let (x1, y1) 5 (24, 3) and m 5 2.
2.4 Exercises (pp. 101–104)
y 2 y1 5 m(x 2 x1)
Skill Practice
y 2 3 5 2(x 2 (24))
1. The linear equation 6x 1 8y 5 72 is written in
standard form.
2. With two points on a line a slope can be calculated. Once
the slope is calculated, simply substitute it in the
point-slope form along with either one of the two points.
3. The slope is m 5 0 and the y-intercept is b 5 2. Use
slope-intercept form to write an equation of the line.
y 5 mx 1 b
y 5 0x 1 2
y52
4. The slope is m 5 3 and the y-intercept is b 5 24. Use
y 2 3 5 2(x 1 4)
y 2 3 5 2x 1 8
y 5 2x 1 11
12. Let (x1, y1) 5 (25, 26) and m 5 0.
y 2 y1 5 m(x 2 x1)
y 2 (26) 5 0(x 2 (25))
y 1 6 5 0(x 1 5)
y1650
y 5 26
13. Let (x1, y1) 5 (8, 13) and m 529.
slope-intercept form to write in equation of the line.
y 2 y1 5 m(x 2 x1)
y 5 mx 1 b
y 2 13 5 29(x 2 8)
y 5 3x 1 (24)
y 2 13 5 29x 1 72
y 5 3x 2 4
5. The slope is m 5 6 and the y-intercept is b 5 0. Use
slope-intercept form to write an equation of the line.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
y 2 (21) 5 23(x 2 3)
y 5 29x 1 85
3
14. Let (x1, y1) 5 (12, 0) and m 5 }.
4
y 5 mx 1 b
y 2 y1 5 m(x 2 x1)
y 5 6x 1 0
y 2 0 5 }4 (x 2 12)
y 5 6x
3
3
2
6. The slope is m 5 } and the y-intercept is b 5 4. Use
3
slope-intercept form to write an equation of the line.
y 5 mx 1 b
2
y 5 }3 x 1 4
5
7. The slope is m 5 2} and the y-intercept is b 5 7. Use
4
slope-intercept form to write an equation of the line.
y 5 mx 1 b
5
y 5 2}4 x 1 7
8. The slope is m 5 25 and the y-intercept is b 5 21. Use
slope-intercept form to write an equation of the line.
y 5 mx 1 b
y 5 }4 x 2 9
4
15. Let (x1, y1) 5 (7, 23) and m 5 2}.
7
y 2 y1 5 m(x 2 x1)
4
y 2 (23) 5 2}7 (x 2 7)
4
y 1 3 5 2}7 (x 2 7)
4
y 1 3 5 2}7 x 1 4
4
y 5 2}7 x 1 1
3
16. Let (x1, y1) 5 (24, 2) and m 5 }.
2
y 2 y1 5 m(x 2 x1)
3
y 5 25x 1 (21)
y 2 2 5 }2 (x 2 (24))
y 5 25x 2 1
y 2 2 5 }2 (x 1 4)
9. Let (x1, y1) 5 (0, 22) and m 5 4.
y 2 y1 5 m(x 2 x1)
y 2 (22) 5 4(x 2 0)
3
3
y 2 2 5 }2 x 1 6
3
y 5 }2 x 1 8
y 1 2 5 4x
y 5 4x 2 2
Algebra 2
Worked-Out Solution Key
65
continued
1
17. Let (x1, y1) 5 (9, 25) and m 5 2}.
3
y 2 y1 5 m(x 2 x1)
1
y 2 (25) 5 2}3 (x 2 9)
1
y 1 5 5 2}3 (x 2 9)
1
23. (4, 1); perpendicular to y 5 } x 1 3.
3
1
m1 5 }3
1
5 23
m2 5 2}
m
1
Let (x1, y1) 5 (4, 1).
1
y 2 y1 5 m2(x 2 x1)
1
y 2 1 5 23x 1 12
y 1 5 5 2}3 x 1 3
y 5 2}3 x 2 2
18. The error was made when substituting for x1. It should be
24, not 4.
y 2 y1 5 m(x 2 x1)
y 2 2 5 3(x 2 (24))
y 2 2 5 3(x 1 4)
y 2 2 5 3x 1 12
y 5 3x 1 14
19. The x1 and y1 values were transposed.
y 2 y1 5 m(x 2 x1)
y 2 1 5 22(x 2 5)
y 2 1 5 22x 1 10
y 5 22x 1 11
20. (23, 25); parallel to y 5 24x 1 1.
m1 5 24
m 2 5 m1 5 24
Let (x1, y1) 5 (23, 25).
y 2 y1 5 m 2(x 2 x1)
y 2 (25) 5 24(x 2 (23))
y 2 1 5 23(x 2 4)
y 5 23x 1 13
24. (26, 2); perpendicular to y 5 22.
m1 5 0
1
m2 5 2}
5 undefined
m
1
Because the slope is undefined, you know the line is
vertical. Because it passes through (26, 2), its equation
is x 5 26.
25. (3, 21); perpendicular to y 5 4x 1 1.
m1 5 4
1
1
Let (x1, y1) 5 (3, 21).
y 2 y1 5 m2(x 2 x1)
1
y 2 (21) 5 2}4 (x 2 3)
1
y 1 1 5 2}4 (x 2 3)
y 5 24x 2 17
21. (7, 1); parellel to y 5 2x 1 3.
m1 5 21
m 2 5 m1 5 21
Let (x1, y1) 5 (7, 1).
y 2 y1 5 m2(x 2 x1)
y 2 1 5 21(x 2 7)
y 2 1 5 2x 1 7
y 5 2x 1 8
22. (2, 8); parallel to y 5 3x 2 2.
m1 5 3
m 2 5 m1 5 3
Let (x1, y1) 5 (2, 8).
y 2 y1 5 m2(x 2 x1)
y 2 8 5 3(x 2 2)
y 2 8 5 3x 2 6
y 5 3x 1 2
66
Algebra 2
Worked-Out Solution Key
1
3
1
1
y 1 1 5 2}4 x 1 }4
y 1 5 5 24(x 1 3)
y 1 5 5 24x 2 12
1
m2 5 2}
5 2}4
m
y 5 2}4 x 2 }4
26. C; (1, 4); perpendicular to y 5 2x 2 3.
m1 5 2
1
1
m2 5 2}
5 2}2
m
1
Let (x1, y1) 5 (1, 4).
y 2 y1 5 m2(x 2 x1)
1
y 2 4 5 2}2 (x 2 1)
1
1
1
9
y 2 4 5 2}2 x 1 }2
y 5 2}2 x 1 }2
27. From the graph, you can see that the slope is m 5 22.
Choose (x1, y1) 5 (3, 0).
y 2 y1 5 m(x 2 x1)
y 2 0 5 22(x 2 3)
y 5 22x 1 6
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 2,
Chapter 2,
continued
28. From the graph, you can see that the slope is m 5 5.
Choose (x1, y1) 5 (4, 4).
y2 2 y1
527
2
22
y 2 y1 5 m(x 2 x1)
5}
5}
5 2}3
m5}
x 2x
320
3
y 2 4 5 5(x 2 4)
Choose (x1, y1) 5 (3, 5).
2
y 2 4 5 5x 2 20
1
29. From the graph, you can see that the slope is m 5 2}.
4
Choose (x1, y1) 5 (21, 5).
2
2
1
y 5 2}3 x 1 7
y 2 5 5 2}4 (x 2 (21))
34. Let (x1, y1) 5 (21, 2) and (x2, y2) 5 (3, 24).
1
y 2 5 5 2}4 (x 1 1)
y2 2 y1
2
6
y 2 2 5 2}2 (x 1 1)
1
Choose (x1, y1) 5 (2, 9).
y 2 9 5 2(x 2 2)
3
3
1
y 5 2}2 x 1 }2
y 2 9 5 2x 2 4
35. Let (x1, y1) 5 (25, 22) and (x2, y2) 5 (23, 8).
y 5 2x 1 5
31. Let (x1, y1) 5 (4, 21) and (x2, y2) 5 (6, 27).
26
5}
5}
5 23
m5}
x 2x
624
2
2
3
y 2 2 = 2}2 x 2 }2
y 2 y1 5 m(x 2 x1)
27 2 (21)
3
y 2 2 5 2}2 (x 2 (21))
3
5}
5 }3 5 2
m5}
x 2x
2 2 (21)
y2 2 y1
1
Choose (x1, y1) 5 (4, 21).
y 2 y1 5 m(x 2 x1)
y 2 (21) 5 23(x 2 4)
y 1 1 5 23(x 2 4)
y 1 1 5 23x 1 12
y 5 23x 1 11
y2 2 y1
8 2 (22)
2
1
Choose (x1, y1) 5 (25, 22).
y 2 y1 5 m(x 2 x1)
y 2 (22) 5 5(x 2 (25))
y 1 2 5 5(x 1 5)
y 1 2 5 5x 1 25
y 5 5x 1 23
36. Let (x1, y1) 5 (15, 20) and (x2, y2) 5 (212, 29).
29 2 20
2
Choose (x1, y1) 5 (2, 21).
y 2 20 5 2}3 (x 2 15)
2
1
y 2 y1 5 m(x 2 x1)
1
y 2 (21) 5 }2 (x 2 2)
1
y 1 1 5 }2 (x 2 2)
1
1
1
Choose (x1, y1) 5 (15, 20).
y 2 y1 5 m(x 2 x1)
2
9
5}
5}
5 2}3
m5}
x 2x
212 2 15
227
5}
5 }4 5 }2
m5}
x 2x
2 2 (22)
21 2 (23)
10
5}
5}
55
m 5}
x 2x
2
23 2 (25)
y2 2 y1
32. Let (x1, y1) 5 (22, 23) and (x2, y2) 5 (2, 21).
y2 2 y1
3
y 2 y1 5 m(x 2 x1)
30. Let (x1, y1) 5 (21, 3) and (x2, y2) 5 (2, 9).
2
26
1
Choose (x1, y1) 5 (21, 2).
19
y 5 2}4 x 1 }
4
923
24 2 2
5}
5}
5 2}2
m5}
x 2x
4
3 2 (21)
1
1
y 2 5 5 2}4 x 2 }4
y2 2 y1
2
y 2 5 5 2}3 (x 2 3)
y 2 5 5 2}3 x 1 2
y 2 y1 5 m(x 2 x1)
1
1
y 2 y1 5 m(x 2 x1)
y 5 5x 2 16
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
33. Let (x1, y1) 5 (0, 7) and (x2, y2) 5 (3, 5).
1
1
y 2 20 5 2}3 x 1 5
1
y 5 2}3 x 1 25
1
y 1 1 5 }2 x 2 1
1
y 5 }2 x 2 2
Algebra 2
Worked-Out Solution Key
67
continued
37. Let (x1, y1) 5 (3.5, 7) and (x2, y2) 5 (21, 20.5).
y2 2 y1
20.5 2 7
13.5
5}
5}
5 23
m5}
x 2x
21 2 3.5
24.5
2
1
Choose (x1, y1) 5 (21, 20.5).
y 2 y1 5 m(x 2 x1)
4
43. Using m 5 } and (x1, y1) 5 (2, 3):
5
y 2 y1 5 m(x 2 x1)
4
y 2 3 5 }5 (x 2 2)
y 2 20.5 5 23(x 2 (21))
22.6 2 0.9
23.5
m5}
5}
5 21.25
3.4 2 0.6
2.8
Choose (x1, y1) 5 (0.6, 0.9).
y 2 y1 5 m(x 2 x1)
y 2 0.9 5 21.25(x 2 0.6)
y 2 0.9 5 21.25x 1 0.75
y 5 21.25x 1 1.65
39. C;
Let (x1, y1) 5 (9, 25) and m 5 26.
y 2 y1 5 m(x 2 x1)
y 2 (25) 5 26(x 2 9)
y 1 5 5 26(x 2 9)
y 1 5 5 26x 1 54
y 5 26x 1 49
So, an equation of the line is y 5 26x 1 49. The point
(7, 7) is a solution of this equation, so it lies on the line.
40. When m 5 23 and b 5 5:
y 5 mx 1 b
y 5 23x 1 5
7
24x 1 5y 5 7
44. Let (x1, y1) 5 (21, 3) and (x2, y2) 5 (26, 27).
y2 2 y1
27 2 3
210
m5}
5}
5}
52
25
x2 2 x1
26 2 (21)
Choose (x1, y1) 5 (21, 3).
y 2 y1 5 m(x 2 x1)
y 2 3 5 2(x 2 (21))
y 2 3 5 2(x 1 1)
y 2 3 5 2x 1 2
y 5 2x 1 5
22x 1 y 5 5
45. Let (x1, y1) 5 (2, 8) and (x2, y2) 5 (24, 16).
y2 2 y1
16 2 8
8
4
m5}
5}
5}
5 2}3
x 2x
24 2 2
26
2
1
Choose (x1, y1) 5 (24, 16).
y 2 y1 5 m(x 2 x1)
4
y 2 16 5 2}3 (x 2 (24))
4
y 2 16 5 2}3 (x 1 4)
4
16
4
32
y 2 16 5 2}3 x 2 }
3
y 5 2}3 x 1 }
3
3x 1 y 5 5
41. When m 5 4 and b 5 23:
y 5 mx 1 b
y 5 4x 1 (23)
y 5 4x 2 3
24x 1 y 5 23
3
42. Using m 5 2} and (x1, y1) 5 (4, 27):
2
y 2 y1 5 m(x 2 x1)
3
y 2 (27) 5 2}2 (x 2 4)
3
y 1 7 5 2}2 (x 2 4)
3
y 1 7 5 2}2 x 1 6
3
y 5 2}2 x 2 1
2y 5 23x 2 2
3x 1 2y 5 22
68
4
5y 5 4x 1 7
y 2 20.5 5 23x 2 3
38. Let (x1, y1) 5 (0.6, 0.9) and (x2, y2) 5 (3.4, 22.6).
8
y 5 }5 x 1 }5
y 2 20.5 5 23(x 1 1)
y 5 23x 1 17.5
4
y 2 3 5 }5 x 2 }5
Algebra 2
Worked-Out Solution Key
3y 5 24x 1 32
4x 1 3y 5 32
46. a. The line y 5 22 is horizontal. So, a line parallel to it
is also horizontal. Because it passes through (3, 4), its
equation is y 5 4.
b. The line y 5 22 is horizontal. So, a line perpendicular
to it is vertical. Because it passes through (3, 4), its
equation is x 5 3.
c. The line x 5 22 is vertical. So a line parallel to it
is also vertical. Because it passes through (3, 4), its
equation is x 5 3.
d. The line x 5 22 is vertical. So, a line perpendicular
to it is horizontal. Because it passes through (3, 4), its
equation is y 5 4.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Chapter 2,
Chapter 2,
continued
47. Sample answer:
49. The line has an x-intercept at point (a, 0) and a
y 5 23x 1 5; m1 5 23, b1 5 5
y-intercept at point (0, b).
y 5 2x 1 1; m2 5 2, b2 5 1
Let (x1, y1) 5 (a, 0) and (x2, y2) 5 (0, b).
1
y2 2 y1
Because m2 Þ 2}
m , the two lines are not perpendicular.
To form a right triangle line * must be perpendicular to
either y 5 23x 1 5 or y 5 2x 1 1. Because the triangle
can be any size, line * can be placed anywhere except the
point of intersection of y 5 23x 1 5 and y 5 2x 1 1.
One possible line * could be perpendicular to
y 5 23x 1 5 through the point (0, 0).
1
1
2
b
y 2 0 5 2}
(x 2 a)
a
b
y 5 2}a x 1 b
bx
a
y
x
}1}51
b
a
}1y5b
1
1
y 2 0 5 }3 (x 2 0)
1
Problem Solving
y 5 }3 x
y 5 23x 1 5
50. Let x represent the time (in months) since buying the car
y
and let y represent the total cost (in dollars).
Total
Initial
Rate of
Months
Cost 5 number 1 change + from now
y 5 2x 1 1
y5
1
x
3
b
1
Choose (x1, y1) 5 (a, 0).
m1 5 23, m2 5 2}
5 2}
5 }3
m
23
1
b20
5}
5 2}a
m5}
x 2x
02a
1
1
y
x
1
5
6500
+
350
1
x
In slope-intercept form, a linear model is
y 5 350x 1 6500.
48. Begin by finding the slope of each line.
A1x 1 B1 y 5 C1
A2 x 1 B2 y 5 C2
B1y 5 C1 2 A1x
B2 y 5 C2 2 A2x
C1
A1
C2
A2
1
1
2
2
y5}
2}
x
B
B
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
51.
A1
1
x2}
y 5 2}
B
A
1
n
1
2
A1
A2
1
2
2
A2
m1 5 2}
B
m2 5 2}
B
1
50
1
15
+
t
52. Area of plot 5 25 + 16 5 400 square feet.
C2
y 5 2}
x2}
B
A
2
5
In slope-intercept form, a linear model is n 5 15t 1 50.
y5}
2}
x
B
B
C1
Number
Initial
Rate of
Years
of houses 5 number 1 change + from now
Number of
Space of
Space of
tomato plant + tomato plants 1 pepper plant +
2
a. If A1 x 1 B1 y 5 C1 and A2 x 1 B2 y 5 C2 are
parallel, then m2 5 m1. Using the slopes you
found above, substitute for m1 and m2 as shown.
m1 5 m2
A1
A2
1
2
2}
5 2}
B
B
2A1B2 5 2A2 B1
A1B2 5 A2B1
b. If A1x 1 B1 y 5 C1 and A2 x 1 B2 y 5 C2 are
1
. Using the slopes you
perpendicular, then m2 5 2}
m1
found above, substitute for m1 and m2 as shown.
Number of
Area of
pepper plants 5
plot
8 + x 1 5 + y 5 400
An equation for this situation is 8x 1 5y 5 400.
Let x 5 15.
8(15) 1 5y 5 400
120 1 5y 5 400
5y 5 280
y 5 56
If you grow 15 tomato plants, you can grow 56 pepper
plants.
21
m2 5 }
m
1
A2
21
2}
5}
B
A
2
1 2}B 2
1
1
A2
B1
2
1
2}
5}
B
A
2A1A2 5 B1B2
0 5 A1A2 1 B1B2
Algebra 2
Worked-Out Solution Key
69
Chapter 2,
Monthly
Initial
Rate of
Years
5 number 1 change + since 1994
cost
General
Student
General
admission + tickets 1 admission +
Student
tickets
y 5 21.62 1 1.66x
In 2010, x 5 16.
Total in
5 ticket sales
y 5 21.62 1 1.66(16)
y 5 21.62 1 26.56
15 + x 1 9 + y 5 4500
y 5 48.18
An equation that models this situation is
15x 1 9y 5 4500.
Start at 200 on the genral admission axis and move up
until you reach the graph. Then find the point that it
corresponds to on the student admission axis. If 200
general admission tickets were sold, about 167 student
tickets were sold.
The predicted average monthly cost for basic cable in
2010 is $48.18.
56. Let y represent the tire’s pressure (in pounds per square
inch) and let x represent the air temperature (8F).
1 psi
108F
Rate of change 5 }
m 5 0.1
y
500
Let (x1, y1) 5 (55, 30).
y 2 y1 5 m(x 2 x1)
400
y 2 30 5 0.1(x 2 55)
300
y 2 30 5 0.1x 2 5.5
200
y 5 0.1x 1 24.5
Width of
1 rectangle
rectangle
100
0
54. a.
57. a. Length of
0
50
100
150
200
x
* 1 w 5 12
b. w 5 2* 1 12
w
5
Budget
16
21.75 + x 1 17 + y 5 86,000
12
21.75x 1 17y 5 86,000
8
b. Let y 5 2500.
21.75y 1 17(2500) 5 86,000
21.75x 1 42,500 5 86,000
21.75y 5 43,500
x 5 2000
2000 square feet can be rented in the first building.
c. Let x 5 y.
21.75y 1 17y 5 86,000
38.75y 5 86,000
y ø 2219
x 1 y ø 4438
The total number of square feet that can be rented is
about 4438 square feet.
55. Let y represent the average monthly cost (in dollars), and
let x represent the number of years since 1994.
Use (x1, y1) 5 (0, 21.62) and (x2, y2) 5 (10, 38.23).
Rate of change:
y2 2 y1
38.23 2 21.62
10 2 0
16.61
10
m 5 } 5 }} 5 } ø 1.66
x2 2 x1
70
Perimeter
2* 1 2w 5 24
First
Number of
Second
building rate + square feet 1 building rate +
Number of
square feet
5
Algebra 2
Worked-Out Solution Key
4
0
c.
58.
0
4
8
12
16
w
6
5
4
3
2
*
6
7
8
9
10
Total
Money
raised
5
Total
Total
Hours
fixed 1 amount + danced
amount
per hour
y 5 (15 1 35 1 20) 1 (4 1 8 1 3) + x
y 5 70 1 15x
Mixed Review for TAKS
59. C;
x 2 30 1 125 2 22 5 180
x 5 107
John had $107 in his bank account at the beginning of
the week.
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
53.
continued
Chapter 2,
continued
60. G;
Using slope-intercept form, use the point (6, 13).
measure of interior angles 5 180(n 2 2)
Problem Solving Workshop 2.4 (p. 105)
1. Find the slope of the line.
16 5 b
Choose the point (2, 15). Substitute m 5 4 and
(x, y) 5 (2, 5) into the slope-intercept form and solve
for b.
y 5 mx 1 b
15 5 4(2) 1 b
15 5 8 1 b
75b
Substitute m and b into the slope-intercept form.
y 5 4x 1 7
change in calories per hour
2. Average rate of change 5 }}
change in pounds
600 2 420
5}
172 2 120
180
52
5}
45
calories per hour per pound
5}
13
Using point-slope form, let (x1, y1) 5 (120, 420).
y 2 y1 5 m(x 2 x1)
45
y 2 420 5 }
(x 2 120)
13
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
5400
x2}
y 2 420 5 }
13
13
45
1
13 5 2}2 (6) 1 b
13 5 23 1 b
35 2 15
20
m5}
5}
54
5
722
45
y 5 mx 1 b
60
1
y 5 2}2 x 1 16
9.5 2 8
1.5
4. m 5 } 5 } 5 0.75
624
2
Using point-slope form, choose (x1, y1) 5 (4, 8).
y 2 y1 5 m(x 2 x1)
y 2 8 5 0.75(x 2 4)
y 2 8 5 0.75x 2 3
y 5 0.75x 1 5
Using slope-intercept form, use the point (6, 9.5).
y 5 mx 1 b
9.5 5 0.75(6) 1 b
9.5 5 4.5 1 b
55b
y 5 0.75x 1 5
3500 2 2600
900
225
5. m 5 } 5 } 5 }
7
54 2 26
28
Using point-slope form, choose (x1, y1) 5 (54, 3500).
y 2 y1 5 m(x 2 x1)
225
y 2 3500 5 }
(x 2 54)
7
x1}
y5}
13
13
Using slope-intercept form, use the point (172, 600).
y 5 mx 1 b
45
(172) 1 b
600 5 }
13
7740
1b
600 5 }
13
60
13
}5b
225
12,150
225
12,350
x2}
y 2 3500 5 }
7
7
x1}
y5}
7
7
Using slope-intercept form, use the point (26, 2600).
y 5 mx 1 b
225
(26) 1 b
2600 5 }
7
5850
1b
2600 5 }
7
12,350
7
}5b
45
60
x1}
y5}
13
13
3. (6, 13), (20, 6)
6 2 13
27
1
m5}
5}
5 2}2
20 2 6
14
Using point-slope form, choose (x1, y1) 5 (20, 6).
y 2 y1 5 m(x 2 x1)
225
12,350
x1}
y5}
7
7
6. y 5 mx 1 b
y1 5 mx1 1 b
y1 2 mx1 5 b
b 5 y1 2 mx1
1
y 2 6 5 2}2 (x 2 20)
1
y 2 6 5 2}2 x 1 10
1
y 5 2}2 x 1 16
Algebra 2
Worked-Out Solution Key
71
Chapter 2,
continued
Mixed Review for TEKS (p.106)
3. y 5 ax
1. B;
v 5 50,000 1 1200t
4.
y 5 ax
3 5 a(5)
22 5 a(6)
3
5
2}3 5 a
1
}5a
2. J;
3
1
So, y 5 2}3x.
So, y 5 }5x.
1
4
} y 2 3x 5 5
y
1
4
} y 5 3x 1 5
y
1
y 5 12x 1 20
1
x
22
x
21
The slope is 12.
3. A;
5. If the radius is 0.6 inches, then the diameter is
x 1 3y 5 12
2(0.6) 5 1.2 inches. When d 5 1.2:
3y 5 2x 1 12
d 5 0.0625t
1
y 5 2}3 x 1 4
1.2 5 0.0625t
19.2 5 t
1
The slope must be 2}3 and the intercept must be negative.
1
y 5 2}3x 2 4
4. H;
227,818 2 219,478
2000 2 1990
}} 5 834
Tooth length, t (cm)
1.8
2.4
Body mass, m (kg)
80
220 375 730 1690 3195
5. D;
} ø 92
730
3.6
} ø 360
} ø 203
7g 1 4s 5 11,200
3.6
4.7
5.8
220
2.4
} ø 129
375
2.9
1690
4.7
} ø 551
3195
5.8
8x 1 4y 5 512
2.5 Exercises (pp. 109 –111)
3
023
7. slope of given line: } 5 2}
5
520
Skill Practice
5
slope of > line: }3 5 1.67
1. The constant of variation is the nonzero constant a
in the equation y 5 ax, for two variables x and y that
vary directly.
Lesson 2.5
2. In a table of ordered pairs (x, y), find the ratio of y to x
2.5 Guided Practice (pp. 107–109)
2.
y 5 ax
for each term in the table. If the ratios are approximately
equal, the data show direct variation.
y 5 ax
4 5 a(27)
29 5 a(3)
3. y 5 ax
4
2}7 5 a
23 5 a
4.
6 5 a(2)
12 5 a(23)
35a
4
So, y 5 23x.
So, y 5 2}7x.
y
y
y 5 ax
24 5 a
So, y 5 3x.
So, y 5 24x.
y
3
y
2
22
21
72
Algebra 2
Worked-Out Solution Key
x
1
x
21
x
21
x
Copyright © by McDougal Littell, a division of Houghton Mifflin Company.
Because the ratios are not approximately equal, tooth
length and body mass do not show direct variation.
6. F;
3
2.9
80
1.8
} ø 44
The population increases by about 834 people per year.
1.
The hailstone has been forming for 19.2 minutes.
6.
Fly UP