5.1 Direct Variation (Part 2)

Posted on April 24, 2020 by user

(C) Finding the value of a variable in a direct variation

1. When y varies directly as x and sufficient information is given, the value of y or x can be determined by using:

\begin{array}{l} (a) y = k x, or \\ (b) \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \end{array}

Example 2

Given that y varies directly as x and y = 24 when x = 8, find

(a) The equation relating y to x

(b) The value of y when x = 6

(c) The value of x when y = 36

Solution:

Method 1: Using y = kx

(a) y \propto x

y = kx

when y = 24, x = 8

24 = k (8)

k = 3

y = 3x

(b)
when x = 6,

y = 3 (6)

y = 18

(c)
when y = 36

36 = 3x

x =12

Method 2:

Using \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}}

(a)
Let x₁ = 8 and y₁= 24

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{y_{2}}{x_{2}} \\ \frac{3}{1} = \frac{y_{2}}{x_{2}} \to y_{2} = 3 x_{2} \\ ∴ y = 3 x \end{array}

(b)
Let x₁ = 8 and y₁= 24 and x₂= 6; find y₂.

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{y_{2}}{6} \\ y_{2} = \frac{24}{8} (6) \\ y_{2} = 18 \end{array}

(c)
Let x₁ = 8 and y₁= 24 and y₂= 36; find x₂.

\begin{array}{l} \frac{y_{1}}{x_{1}} = \frac{y_{2}}{x_{2}} \to \frac{24}{8} = \frac{36}{x_{2}} \\ 24 x_{2} = 36 \times 8 \\ x_{2} = 12 \end{array}