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.
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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