smc

The solution to $ax+b=0$ is $x=\frac{-b}{a}$, while that to $a^{\prime }x+b^{\prime }=0$ is $x=\frac{-b^{\prime }}{a^{\prime }}$. The first solution is less than the second precisely if $$\frac{-b}{a}<\frac{-b^{\prime }}{a^{\prime }},$$ and multiplying this inequality by $-1$ reversees the inequality sign, yielding $\boxed{\frac{b^{\prime }}{a^{\prime }}<\frac{b}{a}}$.