# variation

variation refers to the math that deals with variables. Variables are letters or symbols which represent a value

« Previous Next » There are two types of variations, namely **inverse ** and ** direct ** variation .

## Direct variation

Below are examples on how to solve direct variation

## Example 1

Y varies directly as x and z. Given that y = 9 when x = 6 and z = $\frac{1}{2}$, find

(a) The constant K

(b) The value of y when x = 4 and z = 3

(c) The value of x when y = 4$\frac{1}{2}$ and z = 5

Solutions:

(a)

we first have to come up with the formula, if y varies directly as x and z then $y=K\times x\times z$

$K=\frac{y}{\mathrm{xz}}$

Given that y = 9 when x = 6 and z = $\frac{1}{2}$ substitute it into the formula

$K=\frac{9}{\mathrm{6\; X\frac{1}{2}}}$

$K=\frac{9}{3}$

Answer: $K=3$

(b)

Find y when x = 4 and z = 3, we will use the same formula y = kxz. Therefore substituting x = 4, and z = 3 we have

$y=K\times 4\times 3$

But we already found k = 3 from question (a), hence we substitute on the variable k

$y=3\times 4\times 3$

Answer: $y=\mathrm{36}$

(c)

Finding x when y = 4$\frac{1}{2}$> and z = 5 we use the formula y = kxz. But first of all we make x the subject of the formula by dividing both sides by kz

$y=K\times x\times z$

$x=\frac{y}{\mathrm{kz}}$

substitute y = 4 $\frac{1}{2}$ z = 5 and K = 3

$x=\frac{\mathrm{4\frac{1}{2}}}{\mathrm{3\; X\; 5}}$

$x=\frac{\mathrm{4\frac{1}{2}}}{15}$

$x=\frac{9}{2}\xf7\mathrm{15}$

$x=\frac{9}{2}\times \frac{1}{15}$

$x=\frac{9}{30}$

Answer: $x=\frac{3}{10}$

### How to solve a direct variation

The variables x and y have corresponding values as shown in the table below

$\begin{array}{|llll|}\hline \mathrm{x}& 2& 3& \mathrm{a}\\ \mathrm{y}& 20& 40& 104\\ \hline\end{array}$Given that y varies direct as $({x}^{2}+1)$, find the

(a) Constant of variation K .

(b) Equation connecting y and x .

(c) Values of a .

Solution:

(a) To find K which is the constant, first you have to undertand what a constant is. A constant means a number that does change in an equation, therefore we need to come up with the equation for direct variation

Equation $y=K({x}^{2}+1)$

## Points to note:

1 The equation $y=\mathrm{7x}$ is an example of a direct variation.

2 From the equation $y=\mathrm{7x}$, 7 is a cooefficient of a variable x, if x = 2 it will make y = 14 and if x = 3 it will make y = 21 , from this the conclusion is that when every time x is increased y is equally increased with the same value hence making it a direct variation .

From the table the first set of variables are:

x = 2

y = 20

$y=K({x}^{2}+1)$

$\mathrm{20}=K({2}^{2}+1)$

$\mathrm{20}=K(4+1)$

$\mathrm{20}=K5$

$\frac{K}{5}=\frac{\mathrm{20}}{5}$

Answer: K = 4

(b) Answer: $y=4({x}^{2}+1)$

(c) Equation $y=4({x}^{2}+1)$ find the values of a , Given that: x = a and y = 104

$\mathrm{104}=4({a}^{2}+1)$$\mathrm{104}=4{a}^{2}+4$

$\mathrm{104}-4=4{a}^{2}$

$\mathrm{100}=4{a}^{2}$

$\frac{{a}^{2}}{4}=\frac{\mathrm{100}}{4}$

${a}^{2}=20$

$\sqrt{{a}^{2}}=\sqrt{20}$

$a=\pm 5$

Answer: a = + 5 and a = -5

## Inverse variation

Below is the example of solving inverse variation

## Example 2

Given that y varies inversely as x.

(a) Write an equation in x, y and k, where k is a constant.

(b) Find constant k when y = 6 and x = 2

(c) If y = 6 when x = 2, find the value of y when x = 9

Solutions:

(a)

Answer: Equation: $y=\frac{K}{\mathrm{x}}$

From the equation found the constant K is always the numerator and other values like x in this case is the denominator under an inverse variation

(b)

$y=\frac{K}{\mathrm{x}}$

Make K the subject of the formula

$k=y\times \mathrm{x}$

$k=6\times 2$

Answer: $k=12$

(c)

$y=\frac{K}{\mathrm{x}}$

$y=\frac{\mathrm{12}}{9}$

$y=\frac{4}{3}$

Answer: $y=1\frac{1}{3}$

### How to solve inverse variable

It is given that y varies inversely as the Square of x. The table below shows the values of x and corresponding values of y

$\begin{array}{|llll|}\hline \mathrm{x}& 2& \mathrm{b}& 6\\ \mathrm{y}& 9& 4& \mathrm{a}\\ \hline\end{array}$Find the

(a) value of K, the constant variation

(b) value of a

(c) value of b

Solution:

(a) Equation is $y=\frac{K}{{x}^{2}}$

## Points to note:

1 Inverse variation is the opposite of direct variation.

2 To come up with the equation of an inverse variation, the Constant value K is always the numerator while the other variables are denominators.

From the first set of variables x = 2 and y = 9

$9=\frac{K}{{2}^{2}}$$\frac{9}{1}=\frac{K}{4}$

$K=36$

Answer: K = 36

(b) value of a

x = 6

y = a

$y=\frac{\mathrm{36}}{{x}^{2}}$$a=\frac{\mathrm{36}}{{6}^{2}}$

$a=\frac{\mathrm{36}}{36}$

Answer: a = 1

(c) value of b

x = b

y = 4

$y=\frac{\mathrm{36}}{{x}^{2}}$$4=\frac{\mathrm{36}}{{b}^{2}}$

$4{b}^{2}=\mathrm{36}$

$\frac{{b}^{2}}{4}=\frac{\mathrm{36}}{4}$

${b}^{2}=9$

$\sqrt{{b}^{2}}=\sqrt{9}$

$\mathrm{b}=3$

Answer: b = 3