Q23. - Algebra 2

Question

Solve each equation. Check your solution.

 

23. $\mathbf{12}\left|\frac{\mathbf{1}}{\mathbf{2}}\mathbf{y}\mathbf{+}\mathbf{3}\right|\mathbf{=}\mathbf{6}$

Step-by-Step Solution

Answer

The solution set of given equation is $\left\{\mathbf{-}\mathbf{5}\mathbf{,}\mathbf{-}\mathbf{7}\right\}$.

Step 1 – Description of step.

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

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

Step 2 – Description of step.

For any real numbers p and q, where $q\ge 0$, 

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

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

Step 3 – Description of step.

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

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

Step 4 – Description of step.

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

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

Step 5 – Description of step.

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.

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

Step 6 – Description of step.

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.

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

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