Variation

Warm Up
Solve each equation.

1. 2.4 = 2 10.8
x 9
2. 1.6x = 1.8(24.8) 27.9
Determine whether each data set could
represent a linear function.
3. x 2 4 6 8 no
y 12 6 4 3

4. x –2 –1 0 1 yes
y –6 –2 2 6

In Chapter 2, you studied many types of
linear functions. One special type of linear
function is called direct variation. A direct
variation is a relationship between two
variables x and y that can be written in the
form y = kx, where k ≠ 0. In this relationship,
k is the constant of variation. For the
equation y = kx, y varies directly as x.

A direct variation equation is a
linear equation in the form y = mx
+ b, where b = 0 and the constant of
variation k is the slope. Because b =
0, the graph of a direct variation
always passes through the origin.

Example 1: Writing and Graphing Direct Variation

Given: y varies directly as x, and y = 27 when x = 6. Write
and graph the direct variation function.

y = kx y varies directly as x.
27 = k(6) Substitute 27 for y and 6 for x.

k = 4.5 Solve for the constant of variation k.
y = 4.5x Write the variation function by using
the value of k.

Example 1 Continued

Graph the direct
variation function.
The y-intercept is 0, and
the slope is 4.5.

Check Substitute the
original values of x and y
into the equation.
y = 4.5x
27 4.5(6)
27 27 

Helpful Hint
If k is positive in a direct variation, the
value of y increases as the value of x
increases.

Check It Out! Example 1
Given: y varies directly as x, and y = 6.5 when x = 13. Write
and graph the direct variation function.

y = kx y varies directly as x.
6.5 = k(13) Substitute 6.5 for y and 13 for x.

k = 0.5 Solve for the constant of variation k.
y = 0.5x Write the variation function by using
the value of k.

Check It Out! Example 1 Continued
Graph the direct
variation function.
The y-intercept is 0,
and the slope is 0.5.
Check Substitute the
original values of x and
y into the equation.
y = 0.5x
6.5 0.5(13)
6.5 6.5 

When you want to find specific values in a direct
variation problem, you can solve for k and then use
substitution or you can use the proportion derived
below.

Example 2: Solving Direct Variation Problems
The cost of an item in pesos p varies directly as the
cost of the item in dollars d, and p = 3.85 pesos when d
= $5.00. Find d when p = 10.00 pesos.

Method 1 Find k. p = kd
3.85 = k(5.00) Substitute.
0.77 = k Solve for k.
Write the variation function.
p = 0.77d Use 0.77 for k.
10.00 = 0.77d Substitute 10.00 for p.
12.99 ≈ d Solve for d.

Example 2 Continued

Method 2 Use a proportion.

p1 p2
=
d1 d2
3.85 10.00
= Substitute.
5.00 d

3.85d = 50.00 Find the cross products.

12.99 ≈ d Solve for d.

The perimeter P of a regular dodecagon varies directly as
the side length s, and P = 18 in. when s = 1.5 in. Find s
when P = 75 in.

Method 1 Find k. P = ks
18 = k(1.5) Substitute.
12 = k Solve for k.
Write the variation function.
P = 12s Use 12 for k.
75 = 12s Substitute 75 for P.
6.25 ≈ s Solve for s.


Method 2 Use a proportion.

P1 P2
=
s1 s2
18 75
= Substitute.
1.5 s
Find the cross
18s = 112.5
products.
6.25 = s Solve for s.

A joint variation is a relationship among
three variables that can be written in the
form y = kxz, where k is the constant of
variation. For the equation y = kxz, y varies
jointly as x and z.

Reading Math
The phrases “y varies directly as x” and “y
is directly proportional to x” have the same
meaning.

Example 3: Solving Joint Variation Problems

The volume V of a cone varies jointly as the area of the base B and the height h,
and
V = 12 ft3 when B = 9 ft3 and h = 4 ft. Find b when V = 24 ft3 and h = 9 ft.

Step 1 Find k. Step 2 Use the variation
function.
V = kBh
1
12 = k(9)(4) Substitute. V= Bh Use 1 for k.
3
3
1
=k Solve for k. 1
3 24 = B(9) Substitute.
3

8 = B
Solve for B.

The base is 8 ft2.


The lateral surface area L of a cone varies jointly as the area of the base
radius r and the slant height l, and L = 63 m2 when r = 3.5 m and l = 18 m.
Find r to the nearest tenth when L = 8 m2 and l = 5 m.

function.
L = krl

63 = k(3.5)(18) L = rl
Substitute. Use  for k.

=k Solve for k.
8 = r(5) Substitute.

1.6 = r
Solve for r.

A third type of variation describes a
situation in
which one quantity increases and
the other decreases. For
example, the table shows that the
time needed to
drive 600 miles decreases
as speed increases.

This type of variation is an inverse variation. An inverse variation
is a relationship between two variables x and y that can be written
k
in the form y = , where k ≠ 0. For the equation y = k , y varies
inversely as x. x x

Example 4: Writing and Graphing Inverse Variation

Given: y varies inversely as x, and y = 4 when x = 5.
Write and graph the inverse variation function.

k
y= y varies inversely as x.
x
Substitute 4 for y and
4= k
5 5 for x.
k = 20 Solve for k.
y = 20 Write the variation
x formula.

Example 4 Continued
To graph, make a table of values for both positive
and negative values of x. Plot the points, and
connect them with two smooth curves. Because
division by 0 is undefined, the function is
undefined when x = 0.

x y x y
–2 –10 2 10
–4 –5 4 5
10
–6 – 10
3
6 3
5
–8 – 25
8 2


Given: y varies inversely as x, and y = 4 when x = 10.
Write and graph the inverse variation function.

k
y= y varies inversely as x.
x
k Substitute 4 for y and
4=
10 10 for x.
k = 40 Solve for k.
40 Write the variation
y=
x formula.


To graph, make a table of values for both positive and negative
values of x. Plot the points, and connect them with two smooth
curves. Because division by 0 is undefined, the function is
undefined when x = 0.

x y x y
–2 –20 2 20
–4 –10 4 10
20
–6 – 20
3
6 3

–8 –5 8 5

When you want to find specific values in an inverse
variation problem, you can solve for k and then use
substitution or you can use the equation derived below.

Example 5: Sports Application
The time t needed to complete a certain race varies inversely as the runner’s
average speed s.
If a runner with an average speed of 8.82 mi/h completes the race in 2.97 h,
what is the average speed of a runner who completes the race in 3.5 h?

k
t=
Method 1 Find k. s
k
2.97 = Substitute.
8.82
k = 26.1954 Solve for k.

26.1954
t=
s Use 26.1954 for k.
26.1954
3.5 =
s Substitute 3.5 for t.
s ≈ 7.48 Solve for s.

Example 5 Continued

Method Use t1s1 = t2s2.

t 1 s1 = t 2 s2
(2.97)(8.82) = 3.5s Substitute.

26.1954 = 3.5s Simplify.
7.48 ≈ s Solve for s.

So the average speed of a runner who completes the race
in 3.5 h is approximately 7.48 mi/h.

The time t that it takes for a group of volunteers to construct a house varies
inversely as the number
of volunteers v. If 20 volunteers can build a house in 62.5 working hours,
how many working hours would it take 15 volunteers to build a house?

Method 1 Find k. k
62.5 = Substitute.
k 20
t= k = 1250 Solve for k.
v
1250
t= Use 1250 for k.
v
1250 Substitute 15 for v.
t=
15
1
t ≈ 83 Solve for t.
3


Method 2 Use t1v1 = t2v2.

t1v1 = t2v2

(62.5)(20) = 15t Substitute.

1250 = 15t Simplify.
1
83 ≈ t Solve for t.
3

So the number of working hours it would take 15
1
volunteers to build a house is approximately 83
3
hours.

You can use algebra to rewrite variation functions in
terms of k.

Notice that in direct variation, the ratio of the two
quantities is constant. In inverse variation, the product of
the two quantities is constant.

Example 6: Identifying Direct and Inverse Variation
Determine whether each data set represents a direct variation, an inverse
variation, or neither.

A. In each case xy = 52. The product is
x 6.5 13 104 constant, so this represents an
inverse variation.
y 8 4 0.5

B.
y
In each case = 6. The ratio is
x 5 8 12
x
constant, so this represents a direct
y 30 48 72 variation.

Example 6: Identifying Direct and Inverse Variation
Determine whether each data set represents a direct
variation, an inverse variation, or neither.
C. y
x 3 6 8 Since xy and are not
x
y 5 14 21 constant, this is neither a
direct variation nor an
inverse variation.

Determine whether each data set represents a direct
variation, an inverse variation, or neither.
6a.
x 3.75 15 5 In each case xy = 45. The
y 12 3 9 ratio is constant, so this
represents an inverse
variation.
6b. y
x 1 40 26 In each case = 0.2. The
x
y 0.2 8 5.2 ratio is constant, so this
represents a direct
variation.

A combined variation is a relationship
that contains both direct and inverse
variation. Quantities that vary directly
appear in the numerator, and
quantities that vary inversely appear in
the denominator.

Example 7: Chemistry Application
The change in temperature of an aluminum wire
varies inversely as its mass m and directly as the
amount of heat energy E transferred. The
temperature of an aluminum wire with a mass of
0.1 kg rises 5°C when 450 joules (J) of heat energy
are transferred to it. How much heat energy must
be transferred to an aluminum wire with a mass of
0.2 kg raise its temperature 20°C?

Example 7 Continued

function.
kE Combined
ΔT = E
m variation ΔT = Use 1 for k.
900m 900
5= k(450)
Substitute. E
0.1 20 =
1 900(0.2) Substitute.
=k Solve for k.
900
3600 = E Solve for E.

The amount of heat energy that must be transferred is
3600 joules (J).

The volume V of a gas varies inversely as the
pressure P and directly as the temperature T. A
certain gas has a volume of 10 liters (L), a
temperature of 300 kelvins (K), and a pressure of
1.5 atmospheres (atm). If the gas is heated to
400K, and has a pressure of 1 atm, what is its
volume?


function.
kT Combined
V= 0.05T
P variation V= Use 0.05 for k.
P
k(300)
10 = Substitute. 0.05(400)
1.5 V= Substitute.
(1)
0.05 = k Solve for k.
V = 20 Solve for V.

The new volume will be 20 L.

Lesson Quiz: Part I
1. The volume V of a pyramid varies jointly as the
area of the base B and the height h, and
V = 24 ft3 when B = 12 ft2 and h = 6 ft. Find B
when V = 54 ft3 and h = 9 ft.
18 ft2
2. The cost per person c of chartering a tour bus
varies inversely as the number of passengers
n. If it costs P22.50 per person to charter a
bus for 20 passengers, how much will it cost
per person to charter a bus for 36 passengers?
P12.50

Lesson Quiz: Part II

3. The pressure P of a gas varies inversely as its
volume V and directly as the temperature T. A
certain gas has a pressure of 2.7 atm, a
volume of 3.6 L, and a temperature of 324 K.
If the volume of the gas is kept constant and
the temperature is increased to 396 K, what
will the new pressure be?

3.3 atm

Variation

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