All around us, some quantities are constant and others are variable. For instance, the number of hours in a day is constant, but the number of hours in a daylight in a day is not. now you will explore more closely certain types of relationships between variables.

1. MATHEMATICS 9

2. DIRECT VARIATION
× Sometimes changes in the values of two
variable quantities can be related. When a
change in the value of one quantity
corresponds to a predictable change in the
value of the other quantity, then we say
that the two quantities are related.

3. DIRECT VARIATION
× Let 𝑥 and 𝑦 denote two quantities.
× y varies directly with 𝑥, or 𝑦 is directly
proportional to 𝑥, if there is nonzero number
𝑘 such that:
𝒚 = 𝒌𝒙
× The number 𝒌 is called the constant of
proportionality or the constant of variation.

4. DIRECT VARIATION
STEPS IN SOLVING A DIRECT VARIATION PROBLEM:
1. Write the equation of variation: y = kx
2. Substitute known values and solve for k.
3. Replace k in the equation in STEP 1 by the value
obtained in STEP 2.
4. Solve for the desired value.

5. × EXAMPLE 1:
× The weight of an object on the
moon varies directly as its weight
on earth. An astronaut who weighs
80 kg on earth weighs 12.8 kg. on
the moon. How much would a 60-kg
person weigh on the moon?
DIRECT VARIATION

6. The weight of an
object on the moon
varies directly as
its weight on earth.
An astronaut who
weighs 80 kg on
earth weighs 12.8
kg. on the moon.
How much would a
60-kg person weigh
on the moon?
Solution:
Find the constant of variable. Let 𝑥 = weight on
earth and let 𝑦 = weight on the moon, since the
weight on the moon depends on the weight on earth.
DIRECT VARIATION
𝒚 = 𝒌𝒙
12.8 = 𝑘 ∙ 80
12.8
80
=
𝑘 ∙ 80
80
12.8
80
= 𝑘
𝟎. 𝟏𝟔 = 𝒌
Definition of direct variation
Substitute/Replace y with 12.8 and x with 80
Divide both sides by the value of x
Solve for k.
This is the constant of variation

7. The equation of variation is
Next, use the equation of variation to find y when x is 60
DIRECT VARIATION
𝒚 = 𝒌𝒙
12.8 = 𝑘 ∙ 80
12.8
80
=
𝑘 ∙ 80
80
12.8
80
= 𝑘
𝟎. 𝟏𝟔 = 𝒌
𝒚 = 𝟎. 𝟏𝟔𝒙
𝒚 = 𝒌𝒙
𝐲 = 𝟎. 𝟏𝟔 (𝟔𝟎)
𝐲 = 𝟗. 𝟔
Substitute(Replace) x with 60

8. DIRECT VARIATION
Checking:
𝟔𝟎: 𝟗. 𝟔 = 𝟖𝟎: 𝟏𝟐. 𝟖
𝟔𝟎
𝟗. 𝟔
=
𝟖𝟎
𝟏𝟐. 𝟖
𝟔𝟎
𝟗. 𝟔
=
𝟖𝟎
𝟏𝟐. 𝟖
𝟔. 𝟐𝟓 = 𝟔. 𝟐𝟓 Yes
?

9. DIRECT VARIATION
× EXAMPLE 2:
× The pressure of water on the object varies
directly with its distance from the surface.
A submarine experience a pressure of 26 lbs
per square inch at 60 feet below the
surface. How much pressure will the
submarine experience at 208 feet below the
surface?

10. × EXAMPLE 3:
× Find the constant of variation and the equation of variation in
which y varies directly as x and y=51 and x=3.
Solution:
𝑦 = 𝑘𝑥
51 = 𝑘 3
51
3
=
𝑘∙3
3
51
3
= 𝑘
17 = 𝑘
The constant of variation is 17. The equation of variation is y=17x.
DIRECT VARIATION
Substitute/Replace y with 51 and x with 3
Divide both sides by the value of x
Solve for k.
This is the constant of variation

11. × EXAMPLE 4:
× Find the constant of variation and the equation of variation in
which y varies directly as x and y=28 and x=2.
Solution:
𝑦 = 𝑘𝑥
28 = 𝑘 2
28
2
=
𝑘∙2
2
28
2
= 𝑘
14 = 𝑘
The constant of variation is 14. The equation of variation is y=14x.
DIRECT VARIATION
Substitute/Replace y with 28 and x with 2
Divide both sides by the value of x
Solve for k.
This is the constant of variation

12. × EXAMPLE 5:
× The distance (d) of a spring varies directly with the force (f) applied to
it. Suppose a spring stretches by 40 cm when a 65-kg force is applied to
it.
a. Find the equation of variation
b. Graph
Solutions (a): Let
d (y) represent the distance a spring will stretch
f (x) represent the amount of force applied. d
Since d (y) varies directly as f (x), we write d=kf
DIRECT VARIATION

13. × EXAMPLE 5:
The distance (d) of a spring
varies directly with the force
(f) applied to it. Suppose a
spring stretches by 40 cm
when a 65-kg force is applied
to it.
Solutions (a): Let: d (y)
represent the distance a spring
will stretch and f (x) represent
the amount of force applied.
DIRECT VARIATION
Since d (y) varies directly as f (x), we write d=kf
First, we solve for k.
𝑑 = 𝑘𝑓
40 = 𝑘 65
40
65
=
𝑘 (65)
65
40
65
= 𝑘
40
65
÷
5
5
= 𝑘
8
13
= 𝑘 This is the constant of variation
Substitute/Replace y with 40 and x with 65
Divide both sides by the value of x
Solve for k.
Simplify by finding the GCF of numerator and
denominator

14. DIRECT VARIATION
Since d (y) varies directly as
f (x), we write d=kf
First, we solve for k.
𝑑 = 𝑘𝑓
40 = 𝑘 65
40
65
=
𝑘 (65)
65
40
65
= 𝑘
40
65
÷
5
5
= 𝑘
8
13
= 𝑘
𝒅 =
𝟖
𝟏𝟑
𝒇
This is the constant
of variation
b. To graph the relationship, let’s obtain other
possible values for f and d using the equation
𝒅 =
𝟖
𝟏𝟑
𝒇.
For f = 0 ; 𝒅 =
𝟖
𝟏𝟑
𝟎 = 𝟎
For f = 13; 𝒅 =
𝟖
𝟏𝟑
𝟏𝟑 = 𝟖
For f = 26; 𝒅 =
𝟖
𝟏𝟑
𝟐𝟔 = 𝟏𝟔
For f = 39; 𝒅 =
𝟖
𝟏𝟑
𝟑𝟗 = 𝟐𝟒This is the equation
of variation

15. DIRECT VARIATION
b. To graph the relationship,
let’s obtain other possible
values for f and d using
the equation 𝒅 =
𝟖
𝟏𝟑
𝒇.
For f = 0 ; 𝒅 =
𝟖
𝟏𝟑
𝟎 = 𝟎
For f = 13; 𝒅 =
𝟖
𝟏𝟑
𝟏𝟑 = 𝟖
For f = 26; 𝒅 =
𝟖
𝟏𝟑
𝟐𝟔 = 𝟏𝟔
For f = 39; 𝒅 =
𝟖
𝟏𝟑
𝟑𝟗 = 𝟐𝟒
RECALL:
d (y) represent the
distance a spring will
stretch
f (x) represent the
amount of force applied.
DistanceStretched(cm)
8
16
24
32
40
y
x0
13 26 39 52 65
Force (kg)
(13, 8) (26, 16) (39,24)(0, 0)Ordered Pair (x,y):
(13, 8)
(26,16)
(39,24)
(52,32)
(65,40)

16. × EXAMPLE 6:
× If y varies directly as x, and y=24 when x=6, find the
value of y when x=15.
Solution:
Hence, y=60 when x=15
DIRECT VARIATION
Substitute the values of y with 24 and
x with 6
This is the constant of variation
This is the equation of variation
Substitute 15 for x
Solve for y
𝑦 = 𝑘𝑥
24 = 𝑘 6
4 = 𝑘
𝑦 = 4𝑥
1 𝑦 = 4𝑥
𝑦 = 4 15
𝑦 = 60
2
3

17. × EXAMPLE 7:
The number of kilograms
(n) of water in a human
body varies directly as
the total weight (t). A
person weighing 63 kg
contains 42 kg of
water. How many
kilograms of water are
in a 90-kg person.
DIRECT VARIATION
 Let: n (y) represent number of kilograms of water in a human body and t
(x) represent the total weight.
 Since n (y) varies directly as t (x), we write n=kt
𝑛 = 𝑘𝑡
63 = 𝑘 42
63
42
=
𝑘 (42)
42
63
42
= 𝑘
63
42
÷
7
7
= 𝑘
9
6
𝑜𝑟
3
2
= 𝑘
1
Solutions:
Substitute the values of n with 63 and
t with 42
This is the constant of
variation
Divide both sides by the value of x
Simplify
Find the GCF of the
numerator and
denominator
𝑛 =
3
2
𝑡
2
This is the
equation of
variation
𝑛 =
3
2
𝑡
𝑛 =
3
2
90
𝑛 =
3
2
90
1
𝑛 =
90
2
𝒏 = 𝟒𝟓
3

18. DIRECT SQUARE VARIATION
Let x and y denote two quantities:
y varies directly as the square of x, then
there is nonzero number k such that:
𝑦 = 𝑘𝑥2
The constant of variation is k.

19. × EXAMPLE 8:
× If y varies directly as the square of x, and y=432 when x=12, find the
value of y when x=20.
Solution:
Thus, y=1200 when x=20
DIRECT SQUARE VARIATION
Substitute the values of y with 432
and x with 12
This is the equation of
variation
This is the constant of variation
𝑦 = 𝑘𝑥2
432 = 𝑘(12)2
432 = 𝑘 144
432
144
=
𝑘 (144)
144
3 = 𝑘
𝑦 = 3𝑥2
1
𝑦 = 3𝑥2
y = 3(20)2
y = 3 400
𝑦 = 1200
2
3
Simplify the value of 𝑥2
Divide both sides by the value of x
Substitute 20 for 𝑥2
Simplify the value of
𝑥2
Solve for y

20. × EXAMPLE 9:
× If y varies directly as the square of x, and y=8 when x=4, find the value
of y when x=10.
Solution:
Thus, y=50 when x=10
Substitute the values of y with 8 and x
with 4
This is the equation of
variation
This is the constant of variation
𝑦 = 𝑘𝑥2
8 = 𝑘(4)2
8 = 𝑘 16
8
16
=
𝑘 (16)
16
1
2
= 𝑘
𝑦 =
1
2
𝑥2
1
𝑦 =
1
2
𝑥2
y =
1
2
(10)2
y =
1
2
100
𝑦 = 50
2
3
Simplify the value of 𝑥2
Divide both sides by the value of x
Substitute 10 for 𝑥2
Simplify the value of
𝑥2
Solve for y
DIRECT SQUARE VARIATION