Transcript Direct Variation

ALGEBRA 1 LESSON 8-9
Direct Variation
(For help, go to Lessons 2–7 and 4–1.)
Solve each equation for the given variable.
1. nq = m; q
2. d = rt; r
3. ax + by = 0; y
Solve each proportion.
4.
7.
5 = x
8 12
7 = 35
n 50
5.
8.
4 = n
9 45
8 = 20
d 36
6. 25 = y
15
3
9. 14 = 63
18
n
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ALGEBRA 1 LESSON 8-9
Direct Variation
Solutions
1. nq = m
2. d = rt
nq = m
n
n
q=m
n
d = rt
t
t
d =r
t
r=d
t
4. 5 = x
8
12
5.
3.
ax + by = 0
ax – ax + by = 0 – ax
by = –ax
by –ax
= b
b
y = – ax
b
4
n
=
9 45
6.
25
y
=
15
3
8x = 5(12)
9n = 4(45)
15y = 25(3)
8x = 60
9n = 180
15y = 75
n = 20
y=5
x = 7.5
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ALGEBRA 1 LESSON 8-9
Direct Variation
Solutions (continued)
7.
7 35
=
n 50
35n = 7(50)
35n = 350
n = 10
8.
8 20
d = 36
20d = 8(36)
20d = 288
d = 14.4
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9.
14 63
18 = n
14n = 18(63)
14n = 1134
n = 81
ALGEBRA 1 LESSON 8-9
Direct Variation
Is each equation a direct variation? If it is, find the constant
of variation.
a. 2x – 3y = 1
–3y = 1 – 2x
y=–1 + 2x
3
3
Subtract 2x from each side.
Divide each side by –3.
The equation does not have the form y = kx. It is not a direct variation.
b. 2x – 3y = 0
–3y = –2x
y=
2
x
3
Subtract 2x from each side.
Divide each side by –3.
The equation has the form y = kx, so the equation is a direct variation.
The constant of variation is 2 .
3
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ALGEBRA 1 LESSON 8-9
Direct Variation
Write an equation for the direct variation that includes the
point (–3, 2).
y = kx
Use the general form of a direct variation.
2 = k(–3)
Substitute –3 for x and 2 for y.
2
Divide each side by –3 to solve for k.
–3=k
2
y = –3x
2
Write an equation. Substitute – 3 for k in y = kx.
2
The equation of the direct variation is y = – 3 x .
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ALGEBRA 1 LESSON 8-9
Direct Variation
The weight an object exerts on a scale varies directly with
the mass of the object. If a bowling ball has a mass of 6 kg, the scale
reads 59. Write an equation for the relationship between weight and
mass.
Relate: The weight varies directly with the mass.
When x = 6, y = 59.
Define: Let x = the mass of an object.
Let y = the weight of an object.
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ALGEBRA 1 LESSON 8-9
Direct Variation
(continued)
Write: y
=k x
Use the general form of a direct variation.
59 = k(6)
Solve for k. Substitute 6 for x and 59 for y.
59
=k
6
Divide each side by 6 to solve for k.
59
Write an equation. Substitute 59 for k in y = kx.
y= 6 x
6
The equation y = 59 x relates the weight of an object to its mass.
6
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ALGEBRA 1 LESSON 8-9
Direct Variation
For the data in each table, tell whether y varies directly with x.
If it does, write an equation for the direct variation.
x
y
–2
1
2
4
y
x
x
y
1
= –0.5
–2
–1
2
–1
–1
= –0.5
2
1
2
–2
–2
= –0.5
4
2
–4
y
x
2
= –2
–1
2
=2
1
–4
= –2
2
No, the ratio x is not the
Yes, the constant of
variation is –0.5. The
equation is y = –0.5x.
y
same for each pair of data.
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ALGEBRA 1 LESSON 8-9
Direct Variation
Suppose a windlass requires 0.75 lb of force to lift an object
that weighs 48 lb. How much force would you need to lift 210 lb?
Relate: The force of 0.75 lb lifts 48 lb. The force of n lb lifts 210 lb.
Define: Let n = the force you need to lift 210 lb.
Write:
force1
force2
=
weight1
weight2
0.75
n
=
48
210
0.75(210) = 48n
n
3.3
Use a proportion.
Substitute 0.75 for force1, 48 for
weight1, and 210 for weight2.
Use cross products.
Solve for n.
You need about 3.3 lb of force to lift 210 lb.
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ALGEBRA 1 LESSON 8-9
Direct Variation
1. Is each equation a direct variation? If it is, find the constant of variation.
a. x + 5y = 10
no
b. 3y + 8x = 0
yes; – 8
3
2. Write an equation of the direct variation that includes the point (–5, –4).
4
5
y= x
3. For each table, tell whether y varies directly with x. If it does, write an
equation for the direct variation.
a.
x
–1
0
2
3
y
3
0
–6
–9
b.
yes; y = –3x
x
–1
0
1
3
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y
–2
0
2
–6
no