Question
Solve the following equation by factorization $\frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4}$
✓
Answer
$
\begin{aligned}
& \frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4} \\
& \Rightarrow \frac{x-10+x+6}{(x+6)(x-10)}=\frac{3}{x-4} \\
& \Rightarrow \frac{2 x-4}{(x+6)(x-10)}=\frac{3}{x-4} \\
& \Rightarrow(2 x-4)(x-4)=3(x+6)(x-10) \\
& \Rightarrow 2 x^2-8 x-4 x+16=3\left(x^2-4 x-60\right) \\
& \Rightarrow 2 x^2-8 x-4 x+16=3 x^2-12 x-180 \\
& \Rightarrow 2 x^2-12 x+16-3 x^2+12 x+180=0 \\
& \Rightarrow-x^2+196=0 \\
& \Rightarrow x^2-196=0 \\
& \Rightarrow(x)^2-(14)^2=0 \\
& \Rightarrow(x+14)(x-14)=0 \\
& \text { Either } x+14=0 \\
& \text { then } x=-14
\end{aligned}
$
or
$
x-14=0 \text {, }
$
then $x =14$
$
\therefore x=14,-14 .
$
\begin{aligned}
& \frac{1}{x+6}+\frac{1}{x-10}=\frac{3}{x-4} \\
& \Rightarrow \frac{x-10+x+6}{(x+6)(x-10)}=\frac{3}{x-4} \\
& \Rightarrow \frac{2 x-4}{(x+6)(x-10)}=\frac{3}{x-4} \\
& \Rightarrow(2 x-4)(x-4)=3(x+6)(x-10) \\
& \Rightarrow 2 x^2-8 x-4 x+16=3\left(x^2-4 x-60\right) \\
& \Rightarrow 2 x^2-8 x-4 x+16=3 x^2-12 x-180 \\
& \Rightarrow 2 x^2-12 x+16-3 x^2+12 x+180=0 \\
& \Rightarrow-x^2+196=0 \\
& \Rightarrow x^2-196=0 \\
& \Rightarrow(x)^2-(14)^2=0 \\
& \Rightarrow(x+14)(x-14)=0 \\
& \text { Either } x+14=0 \\
& \text { then } x=-14
\end{aligned}
$
or
$
x-14=0 \text {, }
$
then $x =14$
$
\therefore x=14,-14 .
$
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