# Given y varies directly with x. When y = 20, x = 50. Find x when y = 36.

## Question:

Given {eq}y {/eq} varies directly with {eq}x {/eq}. When {eq}y = 20, \; x = 50 {/eq}. Find {eq}x {/eq} when {eq}y = 36 {/eq}.

## Direct and Inverse Variation:

Direct Variation:

• If x varies directly as y then {eq}x=ky {/eq}, where 'k' is a variation constant.

Inverse Variation:

• If x varies inversely as y then {eq}x= \dfrac{m}{y} {/eq}, where' is a variation constant.

The problem says, "{eq}y {/eq} varies directly with {eq}x {/eq}".

So, by the definition of the direct variation:

$$y = kx \,\,\,\,\,\,\, \rightarrow (1)$$

Here, {eq}k {/eq} is the proportionality constant.

Substitute the first set of given values {eq}y = 20 {/eq} and {eq}x = 50 {/eq} in (1) and solve for {eq}k {/eq}.

$$20 = k (50) \\ \text{Dividing both sides by 50}, \\ k = \dfrac{20}{50} = \dfrac{2}{5}$$

Substitute the value of k and another given value {eq}y = 36 {/eq} in (1) and solve for {eq}x {/eq}.

$$36 = \dfrac{2}{5} (x) \\ \text{Multiplying both sides by } \dfrac{5}{2}, \\ 90=x.$$

Therefore, {eq}\boxed{\mathbf{x=90}} {/eq}. 