Question

Find the value of $x$ which satisfies the equation: $\frac{x^2+x-6}{x^2-8x+12}=4$.

Answer 1

We have $\frac{x^2+x-6}{x^2-8x+12}=4$

$\Rightarrow \frac{x^2+3x-2x-6}{x^2-6x-2x+12}=4$

$\Rightarrow \frac{x(x+3)-2(x+3)}{x(x-6)-2(x-6)}=4$

$\Rightarrow \frac{(x+3)(x-2)}{(x-6)(x-2)}=4$

$\Rightarrow \frac{x+3}{x-6}=4$

$\Rightarrow x+3=4x-24$

$\Rightarrow x-4x=-24-3$

$\Rightarrow -3x=-27$

$\Rightarrow x=\frac{27}{3}=9$