Solving $\frac12 (3y+2)-\frac58=\frac34y$ for $y$ using LCD method

Question

I am solving $$\frac12 (3y+2)-\frac58=\frac34y$$ for $y$ using LCD method.

Can't figure out what I did wrong! The answer in the back of the book is $-1/2$.

PS: In the first line that is a $1/2$ in front of the $(3y+2)$

Answer 1

The problem here is that $\frac{1}{2}(3y + 2)$ is a single term. So when you expand the $8$, you only need to multiply it with the $\frac{1}{2}$ (and not the $(3y + 2)$). Continuing from here, we find that: \begin{align*} 4(3y + 2) - 5 &= 6y \\ (12y + 8) - 5 &= 6y \\ 6y &= -3 \\ y &= -1/2 \end{align*}