Solve for $x$ in $\frac{x^2+12x+36}{x^2+4x-12}=2$ $\frac{x^2+12x+36}{x^2+4x-12}=2$
I factored top and bottom to $\frac{(x+6)(x+6)}{(x+6)(x-2)}=2$
and eliminated the common factor (x+6) for $\frac{x+6}{x-2}=2$
then $x+6=2(x-2)$  and  $x+6=2x-4$  and  $6+4=2x-x$ and finally $x = 10$
Plugging this back into the original fraction proves true
My instructor however in her exam prep test says
$\frac{x^2+12x+36}{x^2+4x-12}=2$
$$x^2+12x+36 = 2(x^2+4x-12)$$
$$x^2+12x+36 = 2x^2+8x-24$$
$$0 =2x^2-x^2 +8x -12x -24 -36$$
$$0 =x^2-4x -60$$ 
which she then factors to $(x-10)(x+6)$ saying the answer to the problem is $x=10$ or $x=-6$
But if you plug -6 back into the original fraction you get $0=2$. Where is the problem here? Thanks
 A: You're right, for (almost) the very reason that you say: $-6$ is not a solution because the left-hand side is not defined when $x=-6$.
A: In short, you are correct.
Slightly longer, the expression
$$\frac{x^2+12x+36}{x^2+4x-12}$$
is not defined for $x = {-6}$ because $x^2 + 4x -12 = 0$ when $x = -6$ and division by zero is undefined. Hence a priori to even talk about this fraction we assume that $x\neq -6$ and $x\neq2$.
A: Other people have done an excellent job of pointing out why $-6$ cannot be a solution, but let me add exactly where your professor's approach breaks down. In the first step of cross-multiplying by $x^2+4x-12$, information is being "destroyed"; because you're multiplying by something that can be zero, you might be creating a $0=0$ situation and thus creating solutions that weren't present in the initial problem. Likewise, $\frac{2x}{x} = 1$ has no solution (because the left-hand side evaluates to $2$ whenever it is defined) but $2x = x$ has the new solution $x = 0$, corresponding to the manufactured $0=0$.
In fairness, though, this is an easy mistake to make, and one I wouldn't be surprised to see made by a perfectly competent teacher. If the numbers involved were slightly different, it would have turned out fine. I would argue that the real failure on the professor's part here was failing to check; a simple examination, as you noted, shows that $-6$ can't be a solution.
