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 $x = \frac{5}{2}$.
 A: The error is in the third group of parentheses in the "multiplying factors again" step. What you have as $-12x$ should be positive.
A: 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 $x$. 
A: 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.
A: Here is another way, which also illustrates how easy it is to make a mistake with these kinds of problems. If we take the negative fractions to the other side of the equation we get $$\frac 1{x-1}+\frac 1{x-4}=\frac 1 {x-3}+\frac 1{x-2}$$ which becomes $$\frac {2x-5}{(x-1)(x-4)}=\frac {2x-5}{(x-2)(x-3)}$$Now cancel and clear fractions to get $$(x-2)(x-3)=(x-1)(x-4)$$ or $$6=4$$
What went wrong? - Well in cancelling $2x-5$ I didn't check to make sure I was not dividing by zero - so the answer I want is $2x-5=0$ or $x=\frac 52$
A: The arithmetic error is already pinpointed. The result is a special case of the Theorem below
since we have $\ \overbrace{x\!-\!1\, +\, x\!-\!4}^{\Large a\ \ +\ \ b}  \, =\, \overbrace{x\!-\!2\, +\, x\!-\!3}^{\Large c\ \ +\ \ d}\ =\ \overbrace{\color{#c00}{2x\!-\!5}}^{\LARGE\color{#c00}s}\ \ $ and $\ \ \color{#0a0}{\overbrace{{x\!-\!2\neq x\!-\!1,\,x\!-\!4}}^{\Large c\ \neq\ a,\,b\quad\ \ }}\,$ 
therefore  $\quad\ \ \  \dfrac{1}{x\!-\!1}+\dfrac{1}{x\!-\!4} = \dfrac{1}{x\!-\!2}+\dfrac{1}{x\!-\!3}\!\iff\! \color{#c00}{2x\!-\!5} = 0$
Theorem $\ \ $ If  $\ a\!+\!b = c\!+\!d =: \color{#c00}s\ $ then $\ \dfrac{1}a+\dfrac{1}b =\dfrac{1}c+\dfrac{1}d \iff \color{#c00}{s = 0}\,$ or $\,\color{#0a0}{c = a}\,$ or $\,\color{#0a0}{c=b}$
$\begin{align}{\bf Proof}\qquad \dfrac{1}a+\dfrac{1}b\, &=\,\dfrac{1}c+\dfrac{1}d\\[.3em]
\iff \dfrac{s}{ab}&=\dfrac{s}{cd}\quad{\rm by} \ \ \ \color{#c00}s = a+b = c+d\\[.3em]
\iff\, \ \ 0\, &=\, s(ab-cd)\\[.3em]
&=\, s(ab-c(a+b-c))\\[.3em]
&=\, \color{#c00}s\color{#0a0}{(a-c)(b-c)}\\
\end{align}$
A: 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.
A: There is a shortcut way to evaluate $$p(x):=(x-2)(x-3)(x-4) - (x-1)(x-3)(x-4)\\ - (x-2)(x-1)(x-4) + (x-1)(x-2)(x-3).$$
It is clear that this is a polynomial of at most second degree, as all cubic term will cancel out*.
Then we perform the easy evaluations
$$p(1)=-6,\ p(2)=-2,\ p(3)=2,\ p(4)=6$$ and clearly $$p(x)=4x-10.$$

*Actually we can also find the quadratic coefficient to be $$-2-3-4+1+3+4+2+1+4-1-2-3=0,$$ but we needn't use this fact, direct evaluation is anyway required to get the linear terms.
