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 $x = \frac{5}{2}$.

To avoid errors or pitfalls it can sometimes be helpful to try an initial 'lazy  approach', where you still don't drop mathematical circumspection.
When looking at the equation a third degree polynomial can be sought, but that is no fun and we are really hoping that a quadratic equation 'pops-up'. The equation is ugly but perhaps by inverting the LHS and RHS things will simplify. We check that there is no $x$ with both the LHS and RHS equal to $0$ (see next section), so we can safely invert.
For the LHS,
$\frac {1} { \frac {1}{x-1} - \frac {1}{x-2}} = \frac{(x-1)(x-2)}{x-2-(x-1)} = -(x^2-3x+2)$
For the RHS,
$\frac {1} { \frac {1}{x-3} - \frac {1}{x-4}} = \frac{(x-3)(x-4)}{x-4-(x-3)} = -(x^2-7x+12)$
Combining our work,
$\tag 1 x^2 -3x +2 = x^2 -7x +12$
OMG! It is turning into a linear equation!
$\tag 2 4x = 10$
So $x = 2.5$

Examining 
$\tag 3 \frac{1}{(x-1)} - \frac{1}{(x-2)} = \frac{1}{(x-3)} - \frac{1}{(x-4)}$
the LHS, $\frac{1}{(x-1)} - \frac{1}{(x-2)}$ is zero if and only if $x -
 1 = x - 2$, which is silly. So we can forget about looking for any $x$ solutions   that would make the LHS and RHS of (3) both equal to 0.
