Question

What is the solution of the given equation?

$\frac{3y+4}{2}+\frac{2y-5}{3}=\frac{31}{2}$

• A. $y=1$
• B. $y=6$
• C. $y=7$
• D. $y=13$

Answer 1

The correct option is C

$y=7$

Finding the solution for the given equation:

$$\begin{gathered} \frac{3y+4}{2}+\frac{2y-5}{3}=\frac{31}{2} \\ \Rightarrow \frac{3\left(3y+4\right)+2\left(2y-5\right)}{6}=\frac{31}{2} \\ \Rightarrow \frac{9y+12+4y-10}{6}=\frac{31}{2} \\ \Rightarrow \frac{13y+2}{6}=\frac{31}{2} \\ \Rightarrow 2\left(13y+2\right)=6\left(31\right) \\ \Rightarrow 26y+4=186 \\ \Rightarrow 26y=182 \\ \Rightarrow y=\frac{182}{26} \\ \Rightarrow y=7 \end{gathered}$$

Therefore the value of $y=7$

Hence, the correct answer is option C.