Question
Solve the following linear equations :
$\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$.
$\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$.
✓
Answer
We have, $\frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21$
$\Rightarrow \quad \frac{6 n-9 n+10 n}{12}=21 \quad[\because$ LCM of 2,4 and 6 is 12$]$
$\Rightarrow \quad \frac{16 n-9 n}{12}=21 \Rightarrow \frac{7 n}{12}=21$
$\Rightarrow \quad \frac{7}{12} n \times \frac{12}{7}=\frac{21 \times 12}{7}\quad$ [multiplying both sides by $\frac{12}{7}$]
$\Rightarrow \quad n=\frac{21 \times 12}{7}$
$\Rightarrow n=3 \times 12=36$, which is the required solution.
$\Rightarrow \quad \frac{6 n-9 n+10 n}{12}=21 \quad[\because$ LCM of 2,4 and 6 is 12$]$
$\Rightarrow \quad \frac{16 n-9 n}{12}=21 \Rightarrow \frac{7 n}{12}=21$
$\Rightarrow \quad \frac{7}{12} n \times \frac{12}{7}=\frac{21 \times 12}{7}\quad$ [multiplying both sides by $\frac{12}{7}$]
$\Rightarrow \quad n=\frac{21 \times 12}{7}$
$\Rightarrow n=3 \times 12=36$, which is the required solution.
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