Let $a, a', b,$ and $b'$ be real numbers with $a$ and $a'$ nonzero. The solution to $ax+b=0$ is less than the solution to $a'x+b'=0$ if and only if 
$\mathrm{(A)\ } a'b<ab' \qquad \mathrm{(B) \ }ab'<a'b \qquad \mathrm{(C) \  } ab<a'b' \qquad \mathrm{(D) \  } \frac{b}{a}<\frac{b'}{a'} \qquad$
$\mathrm{(E) \  }\frac{b'}{a'}<\frac{b}{a}$
The solution to $ax+b=0$ is $\frac{-b}{a}$ and the solution to $a'x+b'=0$ is $\frac{-b'}{a'}$. The first solution is less than the second if $\frac{-b}{a}<\frac{-b'}{a'}$. We can get rid of the negative signs if we reverse the inequality sign, so we have $\frac{b'}{a'}<\frac{b}{a}$. All our steps are reversible, so our answer is $\boxed{\text{E}}$.