Solve $\frac{x-8}{x-10}+\frac{x-4}{x-6}=\frac{x-5}{x-7}+\frac{x-7}{x-9}$ Solve $\frac{x-8}{x-10}+\frac{x-4}{x-6}=\frac{x-5}{x-7}+\frac{x-7}{x-9}$
$\Rightarrow \frac{(x-10)+2}{x-10}+\frac{(x-6)+2}{x-6}=\frac{(x-7)+2}{x-7}+\frac{(x-9)+2}{x-9} \ \ \ ...(1)$
$\Rightarrow 1+\frac{2}{x-10}+1+\frac{2}{x-6}=1+\frac{2}{x-7}+1+\frac{2}{x-9}\ \ \ ...(2)$
$\Rightarrow \frac{2}{x-10}+\frac{2}{x-6}=\frac{2}{x-7}+\frac{2}{x-9}\ \ \ ...(3)$
$\Rightarrow \frac{1}{2}(\frac{2}{x-10}+\frac{2}{x-6})=\frac{1}{2}(\frac{2}{x-7}+\frac{2}{x-9})\ \ \ ...(4)$
$\Rightarrow \frac{1}{x-10}+\frac{1}{x-6}=\frac{1}{x-7}+\frac{1}{x-9}\ \ \ ...(5)$
$\Rightarrow \frac{x-6+x-10}{(x-10)(x-6)}=\frac{x-9+x-7}{(x-7)(x-9)}\ \ \ ...(6)$
$\Rightarrow \frac{2x-16}{(x-10)(x-6)}=\frac{2x-16}{(x-7)(x-9)}\ \ \ ...(7)$
$\Rightarrow (x-10)(x-6)=(x-7)(x-9)\ \ \ ...(8)$
$\Rightarrow x^2-16x+60=x^2-16x+63\ \ \ ...(9)$
$\Rightarrow 60=63$
I feel my calculations are all correct, but 60 cannot equal 63, so something went wrong, but I can't see it. Thanks for the help.
 A: Look closesly at step $(7)$
$$\frac{2x-16}{(x-10)(x-6)}=\frac{2x-16}{(x-9)(x-7)}$$
The numerators are equal for any value of $x$, but is the same true for the denominator too i.e. for any $x\in\mathbb{R}/\{6,7,9,10\}$? No!
What can you conclude from this? One way to look at it is equating this to $0$, as mentioned in the comments. But the question is why? Let's cross-multiply and see where it gets us:
$$(2x-16)(x-9)(x-7)=(2x-16)(x-10)(x-6)$$
You have to consider two cases now (why so will be clear by the cases).
Case 1. $(2x-16)\ne0$
Then, you can divide both LHS and RHS by $(2x-16)$ and it gets cancelled out and spirals down to your steps 8 onward. However, this leads to an impossibility. You must conclude that this case is not true.
Case 2. $(2x-16)=0$
How is this going to be any different from the previous case? That lies in what we did first in case 1. Divided both sides by $(2x-16)$. However, now this would be equivalent to division by $0$, which is undefined. So, how do we get the solution?
The trick is to look at the case itself!
$$2x-16=0\Rightarrow x=8$$
Aaand.. these are exhaustive cases, so there is a sinlge solution to the original equation.
