## Algebra 2 (1st Edition)

there is no solution when $a=c$ but $d\ne b$. all real numbers If $a=c$ and $d=b$.
$$ax+b = cx+d$$ $$ax-cx = d-b$$ $$(a-c)x = d-b$$ $$x = \frac{d-b}{a-c}$$ 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.