jmc

We could cross-multiply, but that looks terrifying. Instead, we start by factoring each of the quadratics, hoping that we'll get some convenient cancellation. Factoring each of the 4 quadratics gives $$\frac{(r-4)(r-1)}{(r-7)(r-1)}=\frac{(r-5)(r+3)}{(r-5)(r+4)}.$$Canceling the common factors on each side gives us $$\frac{r-4}{r-7}=\frac{r+3}{r+4}.$$Cross-multiplying gives $(r-4)(r+4)=(r+3)(r-7)$. Expanding both sides gives $r^2-16=r^2-4r-21$. Solving for $r$ gives $r=\boxed{-5/4}$.