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

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

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Answer 1 · 2017-01-09T20:16:41

The constant of variation is $- 2$ $-2$ .

Explanation:

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

$- 4 y = 8 x$ $-4y=8x$

$- 4 y = \frac{8 x}{- 4}$ $color(white)(-4)y=(8x)/-4$

$- 4 y = - 2 x$ $color(white)(-4y)=-2x$

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

Can we show this?

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

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

Which means

$y^{⋆} = - 2 (x + 1)$ $y^star = -2(x+1)" "$ (since $x^{⋆} = x + 1$ $x^star = x+1$ )
$y^{⋆} = - 2 x - 2$ $color(white)(y^star) = -2x-2$

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

$y^{⋆} = y - 2$ $y^star = y - 2$

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

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

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Explanation:

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