Question
Solve the following equation by factorisation :
$
\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}
$
$
\frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2}
$
✓
Answer
$
\begin{aligned}
& \frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2} \\
& \Rightarrow \frac{6 x-6-2 x}{x(x-1)}=\frac{1}{x-2} \\
& \Rightarrow \frac{4 x-6}{x^2-x}=\frac{1}{x-2} \\
& \Rightarrow(4 x-6)(x-2)=x 6-x \\
& \Rightarrow 4 x^2-8 x-6 x+12=x^2-x \\
& \Rightarrow 4 x^2-14 x+12-x^2+x=0 \\
& \Rightarrow 3 x^2-13 x+12=0 \\
& \Rightarrow 3 x^2-9 x-4 x+12=0 \\
& \Rightarrow 3 x(x-3)-4(x-3)=0 \\
& \Rightarrow(x-3)(3 x-4)=0
\end{aligned}
$
Either $x-3=0$,
then $x=3$
or
$
3 x-4=0 \text {, }
$
then $3 x =4$
$
\Rightarrow x =\frac{4}{3}
$
Hence $x=3, \frac{4}{3}$.
\begin{aligned}
& \frac{6}{x}-\frac{2}{x-1}=\frac{1}{x-2} \\
& \Rightarrow \frac{6 x-6-2 x}{x(x-1)}=\frac{1}{x-2} \\
& \Rightarrow \frac{4 x-6}{x^2-x}=\frac{1}{x-2} \\
& \Rightarrow(4 x-6)(x-2)=x 6-x \\
& \Rightarrow 4 x^2-8 x-6 x+12=x^2-x \\
& \Rightarrow 4 x^2-14 x+12-x^2+x=0 \\
& \Rightarrow 3 x^2-13 x+12=0 \\
& \Rightarrow 3 x^2-9 x-4 x+12=0 \\
& \Rightarrow 3 x(x-3)-4(x-3)=0 \\
& \Rightarrow(x-3)(3 x-4)=0
\end{aligned}
$
Either $x-3=0$,
then $x=3$
or
$
3 x-4=0 \text {, }
$
then $3 x =4$
$
\Rightarrow x =\frac{4}{3}
$
Hence $x=3, \frac{4}{3}$.
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