I've always wondered about this type of thing, but never figured out why.
Say we have an equation $h+rx-x^3 = 0$ where $h,r$ are real-valued parameters.
If we substitute $h=0$ and then solve for $x$, the solution set is different versus if we solve the equation for $x$ and then substitute $h=0$.
Why is this? Is there any general reasoning behind this for systems of equations of arbitrary size?
The solution set should be the same. Why do you say they are different? Perhaps it's easier with a quadratic, say $h+rx-x^2=0$ where you can actually solve for $x$.
Then by the first method we substitute $h=0$ to get $rx-x^2=0$, so $x=r$ or $x=0$.
By the second method we use the quadratic equation to get $x=\frac{r\pm \sqrt{r^2+4h}}{2}$, and substituting $h=0$ we get $x=\frac{r\pm r}{2}$, i.e., $x=r$ or $x=0$.