Question

$\frac{3(n-2)+5}{4}=\frac{11-(7-n)}{6}$
In the equation above, what is the value of n?

• A. $-\frac{1}{11}$
• B. $\frac{5}{11}$
• C. $1$
• D. $\frac{11}{7}$

Answer 1

The correct option is D $\frac{11}{7}$

$\frac{3(n-2)+5}{4}=\frac{11-(7-n)}{6}$

Multiplying both sides by $24$, we get

$\Rightarrow$ $6\left[3\right(n-2)+5]=4[11-(7-n\left)\right]$

$\Rightarrow$ $6[3n-6+5]=4[11-7+n]$

$\Rightarrow$ $6[3n-1]=4[4+n]$

$\Rightarrow$ $18n-6=16+4n$

$\Rightarrow$ $18n-4n=16+6$

$\Rightarrow$ $14n=22$

$\Rightarrow$ $n=\frac{22}{14}$

$\therefore$ $n=\frac{11}{7}$