Solve for x: (6)/(x−4) = (x+4)/(x)

The equation is:

On the left side:

On the right side, use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

So the equation becomes:

Subtract $6x$ from both sides:

We need two numbers that multiply to $-16$ and add to $-6$. These numbers are 2 and $-8$.

The original denominators are $x-4$ and $x$.
If $x=4$, $x-4=0$, which is undefined.
If $x=0$, $x=0$, which is undefined.
Neither of our solutions, $x=-2$ or $x=8$, makes the denominators zero. Therefore, both solutions are valid.