Question
$\text{x}-\frac{\text{x}}{4}-\frac{1}{2}=3+\frac{\text{x}}{4}$

Answer

$\text{x}-\frac{\text{x}}{4}-\frac{1}{2}=3+\frac{\text{x}}{4}$
Transposing $\frac{\text{x}}{4}$ to L.H.S. and $-\frac12$ to R.H.S., we get
$=\text{x}-\frac{\text{x}}{4}-\frac{\text{x}}{4}=3+\frac12$
$=\frac{4\text{x}-\text{x}-\text{x}}{4}=\frac{6+1}{2}$
$=\frac{2\text{x}}{4}=\frac72$
Multiplying both sides by 4, we get
$=\frac{\text{2x}}{4}\times4=\frac72\times4$
$=2\text{x}=14$
Dividing both sides by 2, we get
$=\frac{\text{2x}}{2}=\frac{14}2{}$
$=\text{x}=7$
Verification:
Substituting x = 7 on both sides, we get
$7-\frac{7}{4}-\frac12=3+\frac74$
$\frac{2-7-2}{4}=\frac{12+7}{4}$
$=\frac{19}{4}=\frac{19}{4}$
$\text{L.H.S.}=\text{R.H.S.}$
Hence, verified.

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