Question:
What is the value of $x$ if $-\frac23(x-5) = \frac32(x+1)$?

Answer:
First we get rid of the fractions by multiplying both sides by 6.  This gives us  \[6\left(-\frac{2}{3}(x-5)\right) = 6\left(\frac32(x+1)\right),\] so  \[-4(x-5) = 9(x+1).\] Expanding both sides gives $-4x + 20 = 9x + 9$.  Adding $4x$ to both sides and subtracting 9 from both sides gives $11 = 13x$, so $x = \boxed{\frac{11}{13}}$.