# For the equation -4y=8x, what is the constant of variation?

Jan 9, 2017

The constant of variation is $- 2$.

#### Explanation:

We can solve this equation for $y$ in terms of $x$, by dividing both sides by $- 4$:

$- 4 y = 8 x$

$\textcolor{w h i t e}{- 4} y = \frac{8 x}{-} 4$

$\textcolor{w h i t e}{- 4 y} = - 2 x$

Now we have an equation that says, "$y$ is always $- 2$ times as much as $x$ is". It is this $- 2$ that is our constant of variation, because every time $x$ goes up by $1$, $y$ will go "up" by $- 2$ (i.e. down by $2$).

Can we show this?

Let ${x}^{\star} = x + 1$ (i.e. ${x}^{\star}$ is one more than $x$). If ${y}^{\star}$ is in direct variation with ${x}^{\star}$, with a constant of variation of $- 2$, then

${y}^{\star} = - 2 {x}^{\star}$

Which means

${y}^{\star} = - 2 \left(x + 1\right) \text{ }$(since ${x}^{\star} = x + 1$)
$\textcolor{w h i t e}{{y}^{\star}} = - 2 x - 2$

But wait, $y = - 2 x$, so we have

${y}^{\star} = y - 2$

And there we go! When ${x}^{\star}$ is 1 more than $x$, we see ${y}^{\star}$ is $2$ less than $y$. In other words: when $x$ goes up by $1$, $y$ goes down by $2$.