variation

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

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There are two types of variations, namely inverse and direct variation .

Direct variation

Below are examples on how to solve direct variation

How to solve Direct variation

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

Solutions:

(a)

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

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

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

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

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

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: $y = 36$

(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

$y = K \times x \times z$

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

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

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

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

$x = \frac{9}{2} \div 15$

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

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

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}{l} x & 2 & 3 & a \\ y & 20 & 40 & 104 \end{array}

Given that y varies direct as $(x^{2} + 1)$ , find the

(a) Constant of variation K .

(b) Equation connecting y and x .

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 $y = 7x$ is an example of a direct variation.

2 From the equation $y = 7x$ , 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)

20 = K (2^{2} + 1)

20 = K (4 + 1)

20 = K 5

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

Answer: K = 4

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

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

104 = 4 a^{2} + 4

104 - 4 = 4 a^{2}

100 = 4 a^{2}

\frac{a^{2}}{4} = \frac{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

How to solve Inverse variation

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

Solutions:

(a)

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

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)

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

Make K the subject of the formula

$k = y \times x$

$k = 6 \times 2$

Answer: $k = 12$

(c)

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

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

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

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}{l} x & 2 & b & 6 \\ y & 9 & 4 & a \end{array}

Find the

(a) value of K, the constant variation

(b) value of a

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{36}{x^{2}}

a = \frac{36}{6^{2}}

a = \frac{36}{36}

Answer: a = 1

x = b

y = 4

y = \frac{36}{x^{2}}

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

4 b^{2} = 36

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

b^{2} = 9

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

b = 3

Answer: b = 3

Topics

variation

Direct variation

How to solve Direct variation

How to solve a direct variation

Points to note:

Inverse variation

How to solve Inverse variation

How to solve inverse variable

Points to note: