Question

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

Answer 1

`y=5/2`

Explanation:

`color(blue)("The teaching bit")`

Inverse means '1 over'

So the inverse of `a` is `1/a`
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`color(blue)("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 `-> y=k/xcolor(white)("ddd")->color(white)("ddd")1/5=k/(-20)`

Multiply both sides by `color(red)(-20)larr` 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))`

`color(white)("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")(color(red)(-20))/(color(green)(-20))`

But `(-20)/(-20)=+1 and kxx1=k`

`color(white)("dddddddddddd")->color(white)("ddddd") -20/5color(white)("dd") =color(white)("d") k`

`k=-4`

So by substitution in `Equation(1)` we have:

`y=-4/x`
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`color(blue)("Determine the value requested in the question")`

Given that `x=-8/5`

`y=-4/xcolor(white)("d")->color(white)("d")y=-4-:(-8/5)`

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

Turn the `8/5` upside down and change divide into multiply.

`y=(-4)xx(-5/8)color(white)("ddd")->color(white)("ddd")y=(-5)xx(-4/8)`

Remember that the answer id positive.

`color(white)("dddddddddddddddddddd")->color(white)("ddd")y=5xx1/2`

`color(white)("dddddddddddddddddddd")->color(white)("ddd")y=5/2`

Answer 2

`y=2 1/2` when `x=-8/5`

Explanation:

If `color(red)x` and `color(blue)y` vary inversely then
`color(white)("XXX")color(red)x xx color(blue)y=color(magenta)c color(white)("xxxx")`for some constant `color(magenta)c`

Given that `(color(red)x,color(blue)y)=(color(red)(-20),color(blue)(1/5))` is a solution to this relation we have
`color(white)("XXX")color(red)(-20) xx color(blue)(1/5)=color(magenta)c`

`color(white)("XXX")color(brown)(-4)=color(magenta)c color(white)("xxx")`or`color(white)("xx")color(magenta)c=color(brown)(-4)`

That is
`color(white)("XXX")color(blue)x xx color(red)y=color(brown)(-4)`

So when `color(blue)x=color(blue)(-8/5)`
`color(white)("XXX")color(blue)(""(-8/5))xxcolor(red)y = color(brown)(-4)`

`color(white)("XXX")color(red)y=color(brown)(""(-4)) xx color(blue)(""(-5/8))=5/2=2 1/2`