Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong?

The error is in the third group of parentheses in the "multiplying factors again" step. What you have as $-12x$ should be positive.

A hint to make your life much simpler.

Always simplify your variables to start with. Here you can easily make the substitution $x-4 = y$ to get an easier equation in $y$.

Then noting that $\frac 1y - \frac 1{y+1} = \frac 1{y(y+1)}$, your can very easily see that the numerators on both sides when combining the rational expressions is just $1$ (to avoid confusion, note that I transposed the terms on each side to get a positive numerator). Taking the reciprocal on both sides and expanding, you get quadratics on both sides where the square terms immediately cancel, giving you a simple linear equation. Solve for $y$, then add $4$ to get $x$.

But as I suggested in my comment, the better approach is to first simplify each side separately . . . \begin{align*} &\frac{1}{x-1} - \frac{1}{x-2} = \frac{1}{x-3} - \frac{1}{x-4}\\[4pt] \implies\; &\frac{(x-2)-(x-1)}{(x-1)(x-2)}=\frac{(x-4)-(x-3)}{(x-3)(x-4)}\\[4pt] \implies\; &\frac{-1}{(x-1)(x-2)}=\frac{-1}{(x-3)(x-4)}\\[4pt] \implies\;&(-1)(x-3)(x-4)=(-1)(x-1)(x-2)\\[4pt] \implies\;&(x-3)(x-4)=(x-1)(x-2)\\[4pt] \implies\;&x^2-7x+12=x^2-3x+2\\[4pt] \implies\;&4x=10\\[4pt] \implies\;&x=\frac{5}{2}\\[4pt] \end{align*} Also note, as warned about in Mark Bennet's reply, we have to worry about canceling algebraic factors that might actually be equal to zero. In the steps above, the factors $$(x-1),\;(x-2),\;(x-3),\;(x-4)$$ were canceled in the cross-multiplication step, but that was safe since, based on the original equation, none of those factors had the potential to be zero.