Question

How do you solve `a(n-3)+8=bn` for `n`?

Answer 1

The answer is `n=(-3a+8)/(b-a)`.

Explanation:

Expand the multiplication, move the terms containing `n` to one side, factor, then isolate `n`. It ends up looking like this:

`a(n-3)+8=bn`

`an-3a+8=bn`

`-3a+8=bn-an`

`-3a+8=n(b-a)`

`color(blue)(color(black)(-3a+8)/(b-a))=color(blue)(color(black)(n(b-a))/(b-a))`

`color(blue)(color(black)(-3a+8)/(b-a))=color(blue)(color(black)(ncolor(red)cancelcolor(black)((b-a)))/color(red)cancelcolor(blue)(b-a))`

`(-3a+8)/(b-a)=n`

That's the answer. Hope this helped!