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}
\(xy + x = \frac{3x + 2y + z + y + 2z}{3}\)
\(3xy + 3x = 3x + 2y + z + y + 2z\)
\(3xy=2y+y+2z+z\)
\(3xy=3y+3z\)
\(xy=y+z\)
\(x=1+\frac{z}{y}\)
Assuming that y is not 0.
