Question
Solve for x.
$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
✓
Answer
$\frac{1}{\text{x}}+\frac{2}{2\text{x}-3}=\frac{1}{\text{x}-2},$ $\text{x}\neq0,\frac{3}{2},2$
$\Rightarrow\frac{(2\text{x}-3)+2\text{x}}{\text{x}(2\text{x}-3)}=\frac{1}{(\text{x}-2)}$
$ \Rightarrow(x-2)(4 x-3)=x(2 x-3)$
$\Rightarrow 4 x^2-11 x+6=2 x^2-3 x$
$\Rightarrow 2 x^2-8 x+6=0$
$\Rightarrow x^2-4 x+3=0$
$\Rightarrow x^2-3 x-x+3=0$
$\Rightarrow x(x-3)-(x-3)=0$
$\Rightarrow(x-1)(x-3)=0$
$\therefore x-1=0$
$\Rightarrow x=1$
$\text { and } x-3=0$
$\Rightarrow x=3$
$\Rightarrow\frac{(2\text{x}-3)+2\text{x}}{\text{x}(2\text{x}-3)}=\frac{1}{(\text{x}-2)}$
$ \Rightarrow(x-2)(4 x-3)=x(2 x-3)$
$\Rightarrow 4 x^2-11 x+6=2 x^2-3 x$
$\Rightarrow 2 x^2-8 x+6=0$
$\Rightarrow x^2-4 x+3=0$
$\Rightarrow x^2-3 x-x+3=0$
$\Rightarrow x(x-3)-(x-3)=0$
$\Rightarrow(x-1)(x-3)=0$
$\therefore x-1=0$
$\Rightarrow x=1$
$\text { and } x-3=0$
$\Rightarrow x=3$
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