Ok, lets examine the CDF of Y.

Step 1:

$\displaystyle G(Y) = P(Y<y) = P(aX + b < y)$

$\displaystyle G(Y) = P(Y<y) = P(X < \frac{y - b}{a})$

$\displaystyle G(Y) = P(X < \frac{y - b}{a}) = F(\frac{y-b}{a})$

Step 2:

Now, to get any pdf from a cdf, we just differenciate.

$\displaystyle g(y) = \frac{dG}{dy} = \frac{dF \left( \frac{y-b}{a} \right)}{dy}$

You can do this derivative easily using the chain rule

$\displaystyle g(y) = \frac{dF \left( \frac{y-b}{a} \right)}{dy} = \frac{1}{a} \times F' \left( \frac{y-b}{a} \right)$

Notation point: $\displaystyle F' \left( \frac{y-b}{a} \right)$ means the derivative of F with respect to x, evaluated at the point $\displaystyle \left( \frac{y-b}{a} \right)$

But by definition, the derivative of F is the pdf (ie, F' = f).

$\displaystyle g(y) = \frac{1}{a} \times f \left( \frac{y-b}{a} \right)$

To explain why F(X) = G(aX + b) you can follow the argument in step 1 again, but try to find F() instead of G(). I've done it as a spoiler in case you get stuck