Solve for the variable x in terms of y and z, assuming y \neq \frac{1}{2}: xy + x = \frac{3x + 2y + z + y + 2z}{3}
\(\displaystyle xy + x = \frac{3x + 2y + z + y + 2z}{3}\\ xy + x = x + \dfrac{2y}3 + \dfrac z3 + \dfrac y3 + \dfrac{2z}3\\ xy + x = x + y + z\\ xy = y + z\\ x = \dfrac{y + z}y \)
+1247 
