Question

How do you solve for s in `a=s-sr`?

Answer 1

`s=a/(1-r)`

Explanation:

`color(blue)("Shortcut method")`

By inspection, write as:`" "a=s(1-r)`

Take the `(1-r)` to the other side

`a/(1-r)=s`

'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
`color(blue)("From first principles")`

Note that:
`" "sxx1 = s`
`" "sxx (-r)=-sr`

So for the right hand side; factor out the `s`

`" "s(1-r)`

This means everything inside the bracket is multiplied by `s`

So now we have

`" "a=s(1-r)`

Divide both sides by `(1-r)` giving

`" "a/(1-r)" "=" "sxx(1-r)/(1-r)`

But `(1-r)/(1-r) = 1` so we now have

`" "a/(1-r)" "=" "sxx1`

But `sxx1 = s` so we have

`" "a/(1-r)=s`

Swap sides

`" "s=a/(1-r)`