# y varies directly as x. If x = 6 then y = 8, then find y when x = 13.

## Question:

{eq}\displaystyle{ y }{/eq} varies directly as {eq}\displaystyle{ x }{/eq}. If {eq}\displaystyle{ x = 6 }{/eq} then {eq}\displaystyle{ y = 8 }{/eq}, then find {eq}\displaystyle{ y }{/eq} when {eq}\displaystyle{ x = 13 }{/eq}.

## Proportions and Variation:

We can establish a relation between two numbers that are proportional by means of a ratio {eq}x {/eq} and {eq}y {/eq}. That is to say, a proportion shows the similarity between two ratios. When two variables are dependent, variations in the magnitude of one variable will have a proportional effect on the other. When there is an increase or decrease of a variable {eq}x {/eq} with respect to another {eq}y {/eq}, for a ratio or constant K, variations are present. In the case that we have a direct variation, it happens that when one variable increases the other increases, which can also be written as: {eq}\displaystyle \frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}} {/eq}.

{eq}\eqalign{ & {\text{In this specific case we have two values }}x\,{\text{ and }}y\,{\text{ that have a variation in directly proportional form}}{\text{. So we have:}} \cr & \,\,\,\,{y_1} = 8 \cr & \,\,\,\,{x_1} = 6 \cr & \,\,\,\,{x_2} = 13 \cr & \,\,\,\,{y_2} = ? \cr & {\text{Since}}{\text{, }}x{\text{ and }}y{\text{ vary directly}}{\text{, then}}{\text{, when }}x{\text{ increases it also increases }}y{\text{.}} \cr & {\text{ For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\, \color{red}{\frac{{{y_1}}}{{{x_1}}} = \frac{{{y_2}}}{{{x_2}}}} \cr & {\text{Now}}{\text{, solving for }}\,{y_2}{\text{:}} \cr & \,\,\,\,{y_2} = \frac{{{y_1} \cdot {x_2}}}{{{x_1}}} \cr & {\text{So}}{\text{, substituting the given values:}} \cr & \,\,\,\,{y_2} = \frac{{8 \cdot 13}}{6} = \boxed{17.3} \cr & {\text{Therefore}}{\text{, when }}x{\text{ increases from 6 to 13}}{\text{, }}y{\text{ increases from 8 to } \color{blue}{17}}{\color{blue}{\text{.3}}}{\text{.}} \cr} {/eq}