Answer
The solution set of equation $-3\left| 4x-7 \right|+15=0$ is $\left\{ \left. \frac{1}{2},3 \right\} \right.$.
Work Step by Step
Consider the provided equation, $-3\left| 4x-7 \right|+15=0$
The above equation can be written as
$\begin{align}
& -3\left| 4x-7 \right|+15=0 \\
& -3\left| 4x-7 \right|=-15
\end{align}$
Divide both sides by $-3$
$\left| 4x-7 \right|=5$
Apply modulus property:
$4x-7=5$
Add 7 on both sides:
$\begin{align}
& 4x-7+7=5+7 \\
& 4x=12
\end{align}$
Divide by 4 on both sides
$x=3$
Or
$4x-7=-5$
Add 7 on both sides:
$\begin{align}
& 4x-7+7=-5+7 \\
& 4x=2
\end{align}$
Divide by 4 on both sides
$\begin{align}
& x=\frac{2}{4} \\
& =\frac{1}{2}
\end{align}$
Check:
For $x=\frac{1}{2}$
Put $x=\frac{1}{2}$ in $-3\left| 4x-7 \right|+15=0$
$\begin{align}
& -3\left| 4\left( \frac{1}{2} \right)-7 \right|+15\overset{?}{\mathop{=}}\,0 \\
& -3\left| 2-7 \right|+15\overset{?}{\mathop{=}}\,0 \\
& -3\left( 5 \right)+15\overset{?}{\mathop{=}}\,0 \\
& 0=0
\end{align}$
Which is true.
For $x=3$
Put $x=3$ in $-3\left| 4x-7 \right|+15=0$
$\begin{align}
& -3\left| 4\left( 3 \right)-7 \right|+15\overset{?}{\mathop{=}}\,0 \\
& -3\left| 12-7 \right|+15\overset{?}{\mathop{=}}\,0 \\
& -3\left( 5 \right)+15\overset{?}{\mathop{=}}\,0 \\
& 0=0
\end{align}$
Which is true.
Therefore, the solution set of the equation $-3\left| 4x-7 \right|+15=0$ is $\left\{ \left. \frac{1}{2},3 \right\} \right.$.
