Solving $\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$. Why is my solution wrong? I'm following all hitherto me known rules for solving equations, but the result is wrong. Please explain why my approach is not correct.
We want to solve:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}\tag1$$
Moving the things in RHS to LHS:
$$\frac{1}{(x-1)} - \frac{1}{(x-2)} - \frac{1}{(x-3)} + \frac{1}{(x-4)} = 0\tag2$$
Writing everything above a common denominator:
$$\frac{1}{(x-4)(x-1)(x-2)(x-3)}\bigg[(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3)\bigg] = 0\tag3$$
Multiplying both sides with the denominator to cancel the denominator:
$$(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4) - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3) = 0\tag4$$
Multiplying the first two factors in every term:
$$(x^2-3x-2x+6)(x-4) - (x^2-3x-x+3)(x-4) - (x^2-x-2x+2)(x-4) + (x^2-2x-x+2)(x-3) = 0\tag5$$
Simplifying the first factors in every term:
$$(x^2-5x+6)(x-4) - (x^2-4x+3)(x-4) - (x^2-3x+2)(x-4) + (x^2-3x+2)(x-3) = 0\tag6$$
Multiplying factors again:
$$(x^3-4x^2-5x^2+20x+6x-24) - (x^3-4x^2-4x^2+16x+3x-12) - (x^3-4x^2-3x^2-12x+2x-8) + (x^3-3x^2-3x^2+9x+2x-6) = 0\tag7$$
Removing the parenthesis yields:
$$x^3-4x^2-5x^2+20x+6x-24 - x^3+4x^2+4x^2-16x-3x+12 - x^3+4x^2+3x^2+12x-2x+8 + x^3-3x^2-3x^2+9x+2x-6 = 0\tag8$$
Which results in:
$$28x - 10 = 0 \Rightarrow 28x = 10 \Rightarrow x = \frac{5}{14}\tag9$$
which is not correct. The correct answer is $x = \frac{5}{2}$.

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.
