Let $a, b, c, d \in \mathbb{Q}$ and $r$ be an irrational number. What conditions must $a, b, c, d$ satisfy so that $\cfrac{ar + b}{cr + d}$ is rational?
I got as far as putting it in a reduced form:
$\cfrac{ar + b}{cr + d} = \cfrac{p}{q}$ with $p, q \in \mathbb{Z}$, but that didn't lead me anywhere. I'm sure I must be missing something trivial, I just don't know what.
As you thought, let
$$\frac{ar+b}{cr+d}=h\implies ar+b=hcr+hd\implies (a-hc)r=hd-b$$
Now, if $h$ is rational then it is also $a-hc$ and $hd-b$, thus the unique set of solutions that make the fraction rational are that $a=hc$ and $b=hd$ for some arbitrary $h\in\Bbb Q$.