Conditions under which $\frac{ax+b}{cx+d}$ will be rational.

Question

Suppose $x$ is an irrational number and $a,b,c,d$ are rational numbers. If we know that $$ \frac{(ax+b)}{(cx+d)} $$ is rational, then it follows that:

a.) $a=c=0$,

b.) $a=c$ and $b=d$,

c.) $a+b = c+d$,

d.) $ad = bc$

My attempt: For lack of knowledge, my initial attempt was to try out all the options one by one. When I came to the option d, I realised that it is a more general form of both option a and b, and since both option a and b were correct I marked the answer as d. However, I am searching for an approach which does not involve trying out all the options.

Answer 1

Note that if the expression has to be rational then $\exists\ r,s\in \mathbb{Z},\ s\ne 0$ with $(r,s)=1$ such that $$\frac{ax+b}{cx+d}=\frac{r}{s}\\\implies (as-cr)x=(dr-bs)\implies ad=bc$$ which comes out from the fact that $x$ is rational whereas $a,b,c,d,r,s$ are rationals.

Answer 2

$\frac{(ax+b)}{(cx+d)}=\frac ac \frac{(x+\frac ba)}{(x+\frac dc)}$

So, if $\frac ba = \frac dc$ , then it will cancel out. That means, $ad=bc$

Answer 3

Note that if $\frac{ax+b}{cx+d}$ is rational, then $ax+b=cpx+dp$, where $p$ is a rational number.

This implies that $(a-cp)x=(b-dp)$, where $x$ is irrational, implying that $a=cp$, $b=dp$. Thus the answer is $ad=bc$.