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