Question

Solve the following equation. Remember to check for extraneous solutions 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Answer 1

Answer:

-1, 2, 6

Step-by-step explanation:

We have to solve the equation as follows: 1/(x-6) + (x/(x-2)) = (4/(x²-8x+12)).

Now, we have, $\frac{1}{x-6} +\frac{x}{x-2} = \frac{4}{x^{2}-8x+12 }$

⇒$\frac{(x-2)+x(x-6)}{(x-2)(x-6)} = \frac{4}{x^{2}-8x+12 }$

⇒$\frac{x-2+x^{2}-6x }{(x-2)(x-6)} =\frac{4}{(x-2)(x-6)}$

⇒$\frac{(x-2)(x-6)}{x^{2}-5x-2 }=\frac{(x-2)(x-6)}{4}$

⇒$(x-2)(x-6)[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

⇒ $(x-2)(x-6) =0$ or, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$

If, (x-2)(x-6) =0, then x=2 or x=6

If, $[\frac{1}{x^{2} -5x-2} -\frac{1}{4} ]=0$, then $x^{2} -5x-2=4$

and (x-6)(x+1) =0

Therefore, x=6 or -1

So the solutions for x are -1, 2 6. (Answer)