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Solve the rational equation $\frac{2}{(x-3)} + \frac{1}{x} = \frac{(x-1)}{(x-3)}$.
Given: Equation $\frac{2}{(x-3)} + \frac{1}{x} = \frac{(x-1)}{(x-3)}$.
To do: To solve $\frac{2}{(x-3)} + \frac{1}{x} = \frac{(x-1)}{(x-3)}$.
Solution:
Given equation: $\frac{2}{(x-3)} + \frac{1}{x} = \frac{(x-1)}{(x-3)}$
$\Rightarrow \frac{2x+x-3}{x( x-3)}=\frac{(x-1)}{(x-3)}$
$\Rightarrow \frac{3x-3}{x}=x-1$
$\Rightarrow \frac{3( x-1)}{x}=( x-1)$
$\Rightarrow \frac{1}{x}=1$
$\Rightarrow x=1$
Thus, $x=1$.
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