Question
Solve the following equation and verify the answer:$\frac{\text{n}}{4}-5=\frac{\text{n}}{6}+\frac{1}{2}$
✓
Answer
$\frac{\text{n}}{4}-5=\frac{\text{n}}{6}+\frac{1}{2}$Multiplying each term by 12, the L.C.M. of 4, 6, 2, we get
$\frac{\text{n}}{4}\times12-5\times12$
$=\frac{\text{n}}{6}\times12+\frac{1}{2}\times12$
⇒ 3n - 60 = 2n + 6
⇒ 3n - 2n = 6 + 60
(Transposing 2n to L.H.S. and -60 to R.H.S.)
⇒ n = 66
So, n = 66 is a solution of the given equation.
Check: Substituting n = 66 in the given equation, we get
$\text{L.H.S.}=\frac{66}{4}-5=\frac{33}{2}-5$
$=\frac{33-10}{2}=\frac{23}{2}$
$\text{R.H.S.}=\frac{66}{6}+\frac{1}{2}=11+\frac{1}{2}$
$=\frac{22+1}{2}=\frac{23}{2}$
$\therefore$ When n = 66, we have L.H.S. = R.H.S.
$\frac{\text{n}}{4}\times12-5\times12$
$=\frac{\text{n}}{6}\times12+\frac{1}{2}\times12$
⇒ 3n - 60 = 2n + 6
⇒ 3n - 2n = 6 + 60
(Transposing 2n to L.H.S. and -60 to R.H.S.)
⇒ n = 66
So, n = 66 is a solution of the given equation.
Check: Substituting n = 66 in the given equation, we get
$\text{L.H.S.}=\frac{66}{4}-5=\frac{33}{2}-5$
$=\frac{33-10}{2}=\frac{23}{2}$
$\text{R.H.S.}=\frac{66}{6}+\frac{1}{2}=11+\frac{1}{2}$
$=\frac{22+1}{2}=\frac{23}{2}$
$\therefore$ When n = 66, we have L.H.S. = R.H.S.
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