MATHEMATICS 9
DIRECT
VARIATION
×
Sometimes changes in the values of two
variable quantities can be related. When
a change in the value of one quantity
of one quantity corresponds to a
corresponds to a predictable change
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
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.
DIRECT
VARIATION
×
EXAMPLE 1:
The weight of an
moon
varies
weight on earth.
who weighs 80 kg
weighs 12.8 kg. on
much would a 60×
DIRECT
VARIATION
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
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. 𝒚 = 𝒌𝒙
Definition of direct variation
12.8 = 𝑘 ∙ 80
12.8 𝑘 ∙ 80
=
80
80
12.8
=𝑘
80
𝟎. 𝟏𝟔 = 𝒌
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
DIRECT
VARIATION
The equation of variation is
𝒚 = 𝒌𝒙
12.8 = 𝑘 ∙ 80
12.8 𝑘 ∙ 80
=
80
80
12.8
=𝑘
80
𝟎. 𝟏𝟔 = 𝒌
𝒚 = 𝟎. 𝟏𝟔𝒙
Next, use the equation of
𝒚 = 𝒌𝒙
𝐲 = 𝟎. 𝟏𝟔 (𝟔𝟎)
𝐲 = 𝟗. 𝟔
Substitute(Repla
ce) x with 60
DIRECT
VARIATION
Checking:
𝟔𝟎: 𝟗. 𝟔 = 𝟖𝟎: 𝟏𝟐. 𝟖
𝟔𝟎
𝟖𝟎
=
𝟗. 𝟔 𝟏𝟐. 𝟖
𝟔𝟎 ? 𝟖𝟎
=
𝟗. 𝟔 𝟏𝟐. 𝟖
𝟔. 𝟐𝟓 = 𝟔. 𝟐𝟓 Yes
DIRECT
VARIATION
×
EXAMPLE 2:
× The
pressure of water
directly
with
its
A submarine experience a
lbs per square inch at 60
surface.
How
much
submarine experience at
the surface?
DIRECT
VARIATION
×
EXAMPLE 3:
×
Find the constant of variation and
which y varies directly as x and
Solution:
𝑦 = 𝑘𝑥
51 = 𝑘 3
51
𝑘∙3
=
3
51
3
3
=𝑘
17 = 𝑘
Substitute/Replace y
with both
51 and
x with
3
Divide
sides
by the
value of x
Solve for k.
This is the constant of
variation
The constant of variation is 17. The
DIRECT
VARIATION
×
EXAMPLE 4:
×
Find the constant of variation and
which y varies directly as x and
Solution:
𝑦 = 𝑘𝑥
28 = 𝑘 2
28
𝑘∙2
=
2
28
2
2
=𝑘
14 = 𝑘
Substitute/Replace y
with both
28 and
x with
2
Divide
sides
by the
value of x
Solve for k.
This is the constant of
variation
The constant of variation is 14. The
DIRECT
VARIATION
×
EXAMPLE 5:
×
The distance (d) of a spring varies
it. Suppose a spring stretches by 40 cm
to it.
a. Find the equation of variation
b. Graph
Solutions (a): Let
d (y) represent the distance a
f (x) represent the amount of
Since d (y) varies directly as f (x),
DIRECT
VARIATION
EXAMPLE 5:
The distance (d)
varies
directly
(f) applied to it.
spring stretches
when
a
65-kg
to it.
Solutions
(a):
represent
the
spring
will
represent
the
×
Since d (y) varies directly as
f (x), we write d=kf
First, we solve for k.
Substitute/Replace y
𝑑 = 𝑘𝑓
with 40 and x with 65
40 = 𝑘 65 Divide both sides by the
40 𝑘 (65)
value of x
=
65
65
Solve for k.
40
=𝑘
Simplify by finding the GCF
65
of numerator and
40 5
denominator
÷ =𝑘
65 5
This is the constant
8
of variation
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 𝑘 (65)
=
65
65
40
=𝑘
65
40 5
÷ =𝑘
65 5
8
This is the constant
=𝑘
of variation
13
𝟖
This is the
𝒅=
𝒇
𝟏𝟑
equation of
b. To graph the relationship, let’s obtain other
possible values for f and d using the
𝟖
equation 𝒅 = 𝒇.
𝟏𝟑
𝟖
For f = 0 ; 𝒅 =
𝟏𝟑
𝟖
For f = 13; 𝒅 =
𝟏𝟑
𝟖
For f = 26; 𝒅 =
𝟏𝟑
𝟖
For f = 39; 𝒅 =
𝟏𝟑
𝟎 =𝟎
𝟏𝟑 = 𝟖
𝟐𝟔 = 𝟏𝟔
𝟑𝟗 = 𝟐𝟒
b.
To
graph
the
VARIATION
relationship, let’s obtain
y
other possible values for
f and d using the
𝟖
equation 𝒅 = 𝒇.
(0,
0)
(26,
16)
(13,
8)
(39,2
4)
(65,4
0)
𝟏𝟑
𝟎 =𝟎
RECALL:
d (y) represent the
distance a spring will
stretch
f (x) represent the
amount of force applied.
40
Distance Stretched (cm)
For f = 0 ; 𝒅 =
𝟖
𝟏𝟑
Ordered Pair
(x,y):
32
(39,2
4)
24
1
6
(13,
8)
8
0
(26,1
6)
1
3
26
39
Force (kg)
(52,32
)
52
65
x
DIRECT
VARIATION
×
EXAMPLE 6:
×
If y varies directly as x, and
value of y when x=15.
Solution:
1
2
𝑦 = 𝑘𝑥
24 = 𝑘 6
4=𝑘
𝑦 = 4𝑥
Substitute the values
of y with 24 and x with
This 6
is the
3
constant of
variation
This is the
equation of
variation
Hence, y=60 when x=15
𝑦 = 4𝑥
𝑦 = 4 15
𝑦 = 60
Substitute
15 for x
Solve for y
DIRECT
VARIATION
EXAMPLE
7:
The number
kilograms (n)
in a human
directly
as
weight (t). A
weighing 63
contains 42
water. How
Solu
tions:
Let: n (y) represent number of kilograms
of water in a human body and t (x)
represent the total weight.
3

Since n (y) varies directly as t (x), we write
𝑛= 𝑡
Substitute
the
values
This 2 is
𝑛 =n=kt
𝑘𝑡
of n with 63 and t
the
with 42
63 = 𝑘 42
both sides
equatio
63 𝑘 (42) Divide
by the value of x
n
of
=
42
42
variatio
Simplif
3
63
𝑛 = 𝑡n
y
Find
the
GCF
2
=𝑘
3
42
of the
𝑛 = 90
2
63 7
numerator
3 90
÷ =𝑘
and
𝑛=
42 7
2
1
denominator
This
is
the
9 3
90
𝑛=
constant of
𝑜𝑟 = 𝑘
2
6 2
𝒏 = 𝟒𝟓
variation

×
1
2
3
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:
2
𝑦 = 𝑘𝑥
The constant of variation is k.
DIRECT SQUARE
VARIATION
×
EXAMPLE 8:
If y varies directly as the square of x,
value of y when x=20.
This is the
Solution:
2
×
1
2
𝑦 = 𝑘𝑥
432 = 𝑘(12)2
432 = 𝑘 144
432 𝑘 (144)
=
144
144
3=𝑘
2
Substitute the values
of y with 432 and x
with
12 value
Simplify
the
of 𝑥 2
Divide both sides
by the
Thisvalue
is theof x
constant of
variation
3
𝑦 = 3𝑥
𝑦 = 3𝑥 2
equation of
variation
2 Substitute
y = 3(20) 20 for 𝑥 2
the
y = 3 400 Simplify
value of 𝑥 2
𝑦 = 1200 Solve for y
Thus, y=1200 when
DIRECT SQUARE
VARIATION
×
EXAMPLE 9:
If y varies directly as the square of x,
of y when x=10.
This is the
1 2
2 𝑦= 𝑥
equation of
Solution:
2
×
1
𝑦 = 𝑘𝑥 2
8 = 𝑘(4)2
8 = 𝑘 16
8
𝑘 (16)
=
16
16
1
=𝑘
2
Substitute the values
of y with 8 and x with
Simplify 4the value
2
of 𝑥
Divide both sides
by the value of x
This is the
constant of
variation
3
variation
1 2
𝑦= 𝑥
2
1
y = (10)2
2
1
y = 100
2
𝑦 = 50
Thus,
Substitute
10 for 𝑥 2
Simplify the
value of 𝑥 2
Solve for y
y=50