Answer
there is no solution when $a=c$ but $d\ne b$.
all real numbers If $a=c$ and $d=b$.
Work Step by Step
$$ax+b = cx+d$$
$$ax-cx = d-b$$
$$(a-c)x = d-b$$
$$x = \frac{d-b}{a-c}$$
Thus there is no solution when $a=c$ but $d\ne b$.
If $a=c$ and $d=b$,
$ax+b = ax+b$, which is always true.
