Question

Find the value of n such that $\frac{2}{3}(4n-1)-(2n-\frac{1+n}{3})=\frac{1}{3}n+\frac{4}{3}$.

• A. -$\frac{5}{4}$
• B. $\frac{5}{4}$
• C. -$\frac{5}{2}$
• D. $\frac{5}{2}$

Answer 1

The correct option is D

$\frac{5}{2}$

$\frac{2}{3}(4n-1)-(2n-\frac{1+n}{3})=\frac{1}{3}n+\frac{4}{3}$
$\frac{2}{3}(4n-1)-\left(\frac{6n-1-n}{3}\right)=\frac{1}{3}n+\frac{4}{3}$
Cross multiplying, we get
$2(4n-1)-(5n-1)=n+4$
$8n-2-5n+1=n+4$
$(8n-5n)+(-2+1)=n+4$
$3n-1=n+4$
$3n-n=4+1$
$2n=5$
$n=\frac{5}{2}$.