# If y is 6 when x is 3 and y varies directly with x, find y when x is 8.

## Question:

If y is 6 when x is 3 and y varies directly with x, find y when x is 8.

## 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}\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 }} \cr & {\text{variation in directly proportional form}}{\text{. So we have:}} \cr & \,\,\,\,{y_1} = 6 \cr & \,\,\,\,{x_1} = 3 \cr & \,\,\,\,{x_2} = 8 \cr & \,\,\,\,{y_2} = ? \cr & {\text{Since}}{\text{, }}x{\text{ and }}y{\text{ vary directly}}{\text{, then}}{\text{, when }}x{\text{ increases it also }} \cr & {\text{increases }}y{\text{. For this reason}}{\text{, it must be satisfied that:}} \cr & \,\,\,\,\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{{6 \cdot 8}}{3} = \boxed{16} \cr & {\text{Therefore}}{\text{, when }}x{\text{ increases from 3 to 8}}{\text{, }}y{\text{ increases from 6 to 16}}{\text{.}} \cr} {/eq}