Question
$\frac{\text{x}}{2}+\frac32=\frac{2\text{x}}{5}-1$

Answer

$\frac{\text{x}}{2}+\frac{3}{2}=\frac{2\text{x}}5-1{}$
Transposing $\frac{2\text{x}}{5}$ to $L.H.S$. and $\frac32$ to $R.H.S$., we get
$=\frac{\text{x}}{2}-\frac{2\text{x}}{5}=-1-\frac32$
$=\frac{5\text{x}-4\text{x}}{10}=\frac{-2-3}{2}$
$=\frac{\text{x}}{10}=\frac{-5}{2}$
Multiplying both sides by $10,$ we get
$=\frac{\text{x}}{10}\times10=\frac{-5}{2}\times10$
$=\text{x}=-25$
Verification: Substituting $x = -25$ in the given equation, we get
$\frac{-25}{2}+\frac32=\frac{2(-25)}{5}-1$
$\frac{-22}{2}=-10-1$
$-11=-11$
$\text{L.H.S.}=\text{R.H.S.}$ Hence, verified.

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