A problem from Module 3 . Solve the absolute value equation or indicate that the equation has no solution. \( 3\left|6-\frac{3}{2} x\right|+6=15 \)

Solve the equation \( 3\left|6-\frac{3}{2} x\right|+6=15 \). Solve the equation by following steps: - step0: Solve for \(x\): \(3\left|6-\frac{3}{2}x\right|+6=15\) - step1: Move the expression to the left side: \(3\left|6-\frac{3}{2}x\right|+6-15=0\) - step2: Subtract the numbers: \(3\left|6-\frac{3}{2}x\right|-9=0\) - step3: Separate into possible cases: \(\begin{align}&3\left(6-\frac{3}{2}x\right)-9=0,6-\frac{3}{2}x\geq 0\\&3\left(-\left(6-\frac{3}{2}x\right)\right)-9=0,6-\frac{3}{2}x<0\end{align}\) - step4: Solve the equation: \(\begin{align}&x=2,x\leq 4\\&x=6,x>4\end{align}\) - step5: Find the intersection: \(\begin{align}&x=2\\&x=6\end{align}\) - step6: Rewrite: \(x_{1}=2,x_{2}=6\) The absolute value equation \(3\left|6-\frac{3}{2} x\right|+6=15\) has two solutions: \(x=2\) and \(x=6\).