A crucial step

mark

How do I show that
[dmath]
(ax+b) - \frac{b}{1-a} = a(x-\frac{b}{1-a})?
[/dmath]
I mean, WolframAlpha says it's true , but how do I derive it??

theoldernoah

We can see with a little bit of algebra that WolframAlpha is right:

[dmath] (ax+b)-\frac{b}{1-a} = ax + \left(\frac{b(1-a)}{1-a}-\frac{b}{1-a}\right)[/dmath]

[dmath] = ax+\left(\frac{b-ab}{1-a} - \frac{b}{1-a}\right) =
ax-\frac{ab}{1-a} = a\left(x-\frac{b}{1-a}\right) [/dmath]