Suppose y varies directly as x. If y = 5 when x = 8, find y when x = 64.


Suppose {eq}\,y\, {/eq} varies directly as {eq}\,x {/eq}. If {eq}\,y = 5\, {/eq} when {eq}\,x = 8 {/eq}, find {eq}\,y\, {/eq} when {eq}\,x = 64 {/eq}.

Direct Variation:

In direct variation, one variable increases (or decreases) if the other variable increases (or decreases). If y varies directly as x then {eq}y=kx {/eq}, where 'k' is a variation constant.

Answer and Explanation:

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

So by the definition of direct variation:

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

Here {eq}k {/eq} is a variation constant.

Substitute the given values {eq}y=5 \text{ and } x=8 {/eq} in (1):

$$5=k(8) \\ \text{Dividing both sides by 8}, \\ \dfrac{5}{8}=k $$

Substitute this and {eq}x=64 {/eq} in (1):

$$y= \dfrac{5}{8} (64) = \color{blue}{\boxed{\mathbf{40}}} $$

Learn more about this topic:

Solving Equations of Direct Variation

from Algebra I: High School

Chapter 17 / Lesson 9

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