Simplyfying each side separately and solving linear equation

Question

I'm given an equation that first needs simplifying:

$$\frac{x-2}{x-4} - \frac{x-4}{x-6} = \frac{x-1}{x-3} - \frac{x-3}{x-5}$$

My next step is:

$$\frac{(x-2)(x-4)(x-6)}{x-4} - \frac{(x-4)(x-4)(x-6)}{x-6} = \frac{(x-1)(x-3)(x-5)}{x-3} - \frac{(x-3)(x-3)(x-5)}{x-5}$$

Tydying this up, I get:

$$(x-2)(x-6) - (x-1)(x-5) - (x-3)(x-3)$$

The next step:

$$x^2 -6x -2x +12 - x^2 + 4x - 4x +16 = x^2 - 5x -x + 5 +3x +3x -9$$

Then:

$$-8x - 4 = -4$$ $$-8x-5$$

Coming to the final solution:

$$x = \frac{5}{8}$$

which I'm not sure is the correct solution. Can somebody please take a look?

Answer 1

$$\frac{x-2}{x-4}=1+\frac{2}{x-4}$$ $$\frac{x-4}{x-6}=1+\frac{2}{x-6}$$ $$\frac{x-1}{x-3}=1+\frac{2}{x-3}$$ $$\frac{x-3}{x-5}=1+\frac{2}{x-5}$$ so $$\frac{x-2}{x-4} - \frac{x-4}{x-6} = \frac{x-1}{x-3} - \frac{x-3}{x-5}\rightarrow \frac{1}{x-4} - \frac{1}{x-3} = \frac{1}{x-6} - \frac{1}{x-5}$$ $$\frac{1}{(x-3)(x-4)}=\frac{1}{(x-5)(x-6)}$$ $$x^2-11x+30=x^2-7x+12$$ so the answer is $$x=\frac{9}{2}$$

Answer 2

Hint:

$$\dfrac{x-2}{x-4}-\dfrac{x-4}{x-6}=\dfrac{(x-2)(x-6)-(x-4)^2}{(x-4)(x-6)}=\dfrac{12-16}{(x-4)(x-6)}$$ right?

Similarly for the right hand side.

On simplification, we have $$(x-4)(x-6)=(x-3)(x-5)\iff24-10x=15-8x\iff x=?$$

Answer 3

$\frac{x-2}{x-4} - \frac{x-4}{x-6} = \frac{x-1}{x-3} - \frac{x-3}{x-5}$

put $x-3=a\Rightarrow x=a+3$ then

$\frac{a+1}{a-1}-\frac{a-1}{a-3}=\frac{a+2}{a}-\frac{a}{a-2}$

$\frac{(a+1)(a-3)-(a-1)^2}{(a-1)(a-3)}=\frac{(a^2-4)-a^2}{a(a-2)}$

$\frac{a^2-2a-3-(a^2-2a+1)}{(a-1)(a-3)}=\frac{-4}{a(a-2)}$

$\frac{-4}{(a-1)(a-3)}=\frac{-4}{a(a-2)}$

$a(a-2)=(a-1)(a-3)$

$a^2-2a=a^2-4a+3$

$2a=3\Rightarrow a=\frac{3}{2}\Rightarrow x=\frac{3}{2}+3=\frac{9}{2}$