# If y=1/5 when x=-20, how do you find y when x=-8/5 given that y varies inversely as x?

Oct 2, 2017

$y = \frac{5}{2}$

#### Explanation:

$\textcolor{b l u e}{\text{The teaching bit}}$

Inverse means '1 over'

So the inverse of $a$ is $\frac{1}{a}$
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$\textcolor{b l u e}{\text{Determining the structure of the whole equation}}$

Something changes the invers to the target value.

Let $k$ be some constant

Then we have:

y=1/x xxk color(white)("d")->color(white)("d")y=k/x" "...Equation(1)

Initial condition $\to y = \frac{k}{x} \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} \frac{1}{5} = \frac{k}{- 20}$

Multiply both sides by $\textcolor{red}{- 20} \leftarrow$ gets $k$ on its own.

color(green)(1/5=k/(-20)color(white)("ddd")->color(white)("ddd")1/5color(red)(xx(-20))=color(white)("d")k/(-20)color(red)(xx(-20))

$\textcolor{w h i t e}{\text{ddddddddddd")->color(white)("dddddd")(color(red)(-20))/color(green)(5)color(white)("ddd")=color(white)("ddd")color(green)(k)color(white)("dd")xxcolor(white)("d}} \frac{\textcolor{red}{- 20}}{\textcolor{g r e e n}{- 20}}$

But $\frac{- 20}{- 20} = + 1 \mathmr{and} k \times 1 = k$

$\textcolor{w h i t e}{\text{dddddddddddd")->color(white)("ddddd") -20/5color(white)("dd") =color(white)("d}} k$

$k = - 4$

So by substitution in $E q u a t i o n \left(1\right)$ we have:

$y = - \frac{4}{x}$
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$\textcolor{b l u e}{\text{Determine the value requested in the question}}$

Given that $x = - \frac{8}{5}$

$y = - \frac{4}{x} \textcolor{w h i t e}{\text{d")->color(white)("d}} y = - 4 \div \left(- \frac{8}{5}\right)$

Both signs are the same (negative) so the answer is positive (plus)

Turn the $\frac{8}{5}$ upside down and change divide into multiply.

$y = \left(- 4\right) \times \left(- \frac{5}{8}\right) \textcolor{w h i t e}{\text{ddd")->color(white)("ddd}} y = \left(- 5\right) \times \left(- \frac{4}{8}\right)$

Remember that the answer id positive.

$\textcolor{w h i t e}{\text{dddddddddddddddddddd")->color(white)("ddd}} y = 5 \times \frac{1}{2}$

$\textcolor{w h i t e}{\text{dddddddddddddddddddd")->color(white)("ddd}} y = \frac{5}{2}$

Oct 2, 2017

$y = 2 \frac{1}{2}$ when $x = - \frac{8}{5}$

#### Explanation:

If $\textcolor{red}{x}$ and $\textcolor{b l u e}{y}$ vary inversely then
$\textcolor{w h i t e}{\text{XXX")color(red)x xx color(blue)y=color(magenta)c color(white)("xxxx}}$for some constant $\textcolor{m a \ge n t a}{c}$

Given that $\left(\textcolor{red}{x} , \textcolor{b l u e}{y}\right) = \left(\textcolor{red}{- 20} , \textcolor{b l u e}{\frac{1}{5}}\right)$ is a solution to this relation we have
$\textcolor{w h i t e}{\text{XXX}} \textcolor{red}{- 20} \times \textcolor{b l u e}{\frac{1}{5}} = \textcolor{m a \ge n t a}{c}$

$\textcolor{w h i t e}{\text{XXX")color(brown)(-4)=color(magenta)c color(white)("xxx}}$or$\textcolor{w h i t e}{\text{xx}} \textcolor{m a \ge n t a}{c} = \textcolor{b r o w n}{- 4}$

That is
$\textcolor{w h i t e}{\text{XXX}} \textcolor{b l u e}{x} \times \textcolor{red}{y} = \textcolor{b r o w n}{- 4}$

So when $\textcolor{b l u e}{x} = \textcolor{b l u e}{- \frac{8}{5}}$
$\textcolor{w h i t e}{\text{XXX")color(blue)(} \left(- \frac{8}{5}\right)} \times \textcolor{red}{y} = \textcolor{b r o w n}{- 4}$

$\textcolor{w h i t e}{\text{XXX")color(red)y=color(brown)(""(-4)) xx color(blue)(} \left(- \frac{5}{8}\right)} = \frac{5}{2} = 2 \frac{1}{2}$