Question
$\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}=\frac{1}{3}$

Answer

$\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}=\frac{1}{3}$
$=\frac{5(2\text{x}-1)-3(6\text{x}-2)}{15}=\frac{1}{3}$
$=\frac{10\text{x}-5-18\text{x}+6}{15}=\frac{1}{3}$
$(L.C.M.$ of $3, 5 = 15)$
$=\frac{-8\text{x}+1}{15}=\frac{1}{3}$
$\Rightarrow(-8\text{x}+1)\times3=1\times15$ (By cross multiplication)
$\Rightarrow$ Dividing by 3 $\frac{(-8\text{x}+1)\times3}{3}=\frac{1\times15}{3}$
$\Rightarrow-8\text{x}+1=5$
Subtracting $1$ from both side, $-8\text{x}+1-1=5-1$
$\Rightarrow-8\text{x}=4$
Dividing by $-8$, $\frac{-8\text{x}}{-8}=\frac{4}{-8}$
$\Rightarrow\text{x}=\frac{1}{-2}$
$\therefore\text{x}=\frac{-1}{2}$
Verification: $\text{L.H.S.}=\frac{2\text{x}-1}{3}-\frac{6\text{x}-2}{5}$
$=\frac{2\Big(\frac{-1}{2}\Big)-1}{3}-\frac{6\Big(\frac{-1}{2}\Big)-2}{5}$
$=\frac{-1-1}{3}-\frac{-3-2}{5}$
$=\frac{-2}{3}-\frac{-5}{5}=\frac{-2}{3}+1$
$=\frac{1}{3}=\text{R.H.S.}$

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