what is the value of x in the equation $\frac{2}{3}(\frac{1}{2}x + 12)=\frac{1}{2}(\frac{1}{3}x +…

Explanation:

Step1: Expand both sides

First, expand $\frac{2}{3}(\frac{1}{2}x + 12)$ and $\frac{1}{2}(\frac{1}{3}x+14)-3$. $\frac{2}{3}\times\frac{1}{2}x+\frac{2}{3}\times12=\frac{1}{3}x + 8$. $\frac{1}{2}\times\frac{1}{3}x+\frac{1}{2}\times14 - 3=\frac{1}{6}x+7 - 3=\frac{1}{6}x + 4$. So the equation becomes $\frac{1}{3}x+8=\frac{1}{6}x + 4$.

Step2: Move x - terms to one side

Subtract $\frac{1}{6}x$ from both sides: $\frac{1}{3}x-\frac{1}{6}x+8=\frac{1}{6}x-\frac{1}{6}x + 4$. $\frac{2}{6}x-\frac{1}{6}x+8 = 4$, which simplifies to $\frac{1}{6}x+8 = 4$.

Step3: Move constant to the other side

Subtract 8 from both sides: $\frac{1}{6}x+8 - 8=4 - 8$. We get $\frac{1}{6}x=-4$.

Step4: Solve for x

Multiply both sides by 6: $x=-4\times6=-24$.

Answer:

-24