91Ó°ÊÓ

Divide each side of the given equation by 12 and simplify.

$\begin{array}{c}\frac{12\left|\frac{1}{2}y+3\right|}{12}=\frac{6}{12}\\ \left|\frac{1}{2}y+3\right|=\frac{1}{2}\end{array}$

For any real numbers p and q, where $qâ‰\yen 0$,

Case 1) if $\left|p\right|=q$ then $p=q$ or

Case 2) if $\left|p\right|=q$ then $p=-q$

Apply case 1 mentioned in step 2 to $\left|\frac{1}{2}y+3\right|=\frac{1}{2}$ and simplify.

$\begin{array}{c}\frac{1}{2}y+3=\frac{1}{2}\\ \frac{1}{2}y+3−3=\frac{1}{2}−3\\ \frac{1}{2}y=−\frac{5}{2}\\ y=−5\end{array}$

Apply case 2 mentioned in step 2 to $\left|\frac{1}{2}y+3\right|=\frac{1}{2}$ and simplify.

$\begin{array}{c}\frac{1}{2}y+3=−\frac{1}{2}\\ \frac{1}{2}y+3−3=−\frac{1}{2}−3\\ \frac{1}{2}y=−\frac{7}{2}\\ y=−7\end{array}$

To check whether the obtained value of x is the solution of given equation, substitute $-5$ for $y$ into $12\left|\frac{1}{2}y+3\right|=6$ and simplify.

$\begin{array}{d}12\left|\frac{1}{2}\left(−5\right)+3\right|=6\\ 12\left|\frac{1}{2}\right|\stackrel{?}{=}6\\ 12â‹\dots \frac{1}{2}\stackrel{?}{=}6\\ 6\stackrel{?}{=}6\\ 6=6\end{array}$

To check whether the obtained value of x is the solution of given equation, substitute $-7$for y into $12\left|\frac{1}{2}y+3\right|=6$and simplify.

$\begin{array}{d}12\left|\frac{1}{2}\left(−7\right)+3\right|=6\\ 12\left|−\frac{1}{2}\right|\stackrel{?}{=}6\\ 12â‹\dots \frac{1}{2}\stackrel{?}{=}6\\ 6\stackrel{?}{=}6\\ 6=6\end{array}$

Therefore, the solution set of given equation is $\left\{−5,−7\right\}$.