Question

The solution of the equation
$\frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$ is

• A. $a=2$
• B. $a=\frac{1}{3}$
• C. $a=-2$
• D. $a=\frac{1}{2}$

Answer 1

The correct option is B $a=\frac{1}{3}$

Given, $\frac{3a-2}{3}+\frac{2a+3}{2}=a+\frac{7}{6}$
L.C.M. of denominators is $6$.
Multiplying each term by $6$, we get
$6\times \left(\frac{3a-2}{3}\right)+6\times \left(\frac{2a+3}{2}\right)=6\times a+6\times \frac{7}{6}$
$\Rightarrow 2(3a-2)+3(2a+3)=6a+7$
$\Rightarrow 6a-4+6a+9=6a+7$
$\Rightarrow 12a+5=6a+7$

$\Rightarrow 6a=2$
$\Rightarrow a=\frac{2}{6}=\frac{1}{3}$
Hence, required solution is $a=\frac{1}{3}$.