Question
Solve the following equation by factorization$\frac{x+2}{x+3}=\frac{2 x-3}{3 x-7}$

Answer

$
\begin{aligned}
& \frac{x+2}{x+3}=\frac{2 x-3}{3 x-7} \\
& (x+2)(3 x-7)=(2 x-3)(x+3) \\
& \Rightarrow 3 x^2-7 x+6 x-14=2 x^2+6 x-3 x-9 \\
& \Rightarrow 3 x^2-x-14=2 x^2+3 x-9 \\
& \Rightarrow 3 x^2-x-14-2 x^2-3 x+9=0 \\
& \Rightarrow x^2-4 x-5=0 \\
& \Rightarrow x^2-5 x+x-5=0 \\
& x(x-5)+1(x-5)=0 \\
& \Rightarrow(x-5)(x+1)=0
\end{aligned}
$
Either $x-5=0$,
then $x=5$
or
$
x+1=0 \text {, }
$
then $x=-1$
Hence $x=5,-1$.

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