Question
Solve $\left(\frac{x}{x+2}\right)^2-7\left(\frac{x}{x+2}\right)+12=0 ; x \neq-2$
✓
Answer
$\left(\frac{x}{x+2}\right)^2-7\left(\frac{x}{x+2}\right)+12=0 ; x \neq-2$
Let $\frac{x}{x+2}=y$
then $y^2-7 y+12=0$
$\Rightarrow y^2-4 y-3 y+12=0 $
$\Rightarrow y(y-4)-3(y-4)=0$
then $y = 4$ and $y = 3$
$\Rightarrow \frac{x}{x+2}=4 \text { and } \frac{x}{x+2}=3$
$ \Rightarrow \frac{x}{x+2}=4 \text { and } \frac{x}{x+2}=3$
$\Rightarrow4 x +8= x \text { and } 3 x +6= x $
$\Rightarrow x=\frac{-8}{3} \text { and } x=-3$
Let $\frac{x}{x+2}=y$
then $y^2-7 y+12=0$
$\Rightarrow y^2-4 y-3 y+12=0 $
$\Rightarrow y(y-4)-3(y-4)=0$
then $y = 4$ and $y = 3$
$\Rightarrow \frac{x}{x+2}=4 \text { and } \frac{x}{x+2}=3$
$ \Rightarrow \frac{x}{x+2}=4 \text { and } \frac{x}{x+2}=3$
$\Rightarrow4 x +8= x \text { and } 3 x +6= x $
$\Rightarrow x=\frac{-8}{3} \text { and } x=-3$
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