+743 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}
+399 \(xy + x = \frac{3x + 2y + z + y + 2z}{3}\)
\(xy+x=\frac{3x+3y+3z}{3}\).
\(xy+x=x+y+z\)
\(xy=y+z\)
\(x=\frac{y+z}{y}\). Can also be written as: \(x=1+\frac{z}{y}\).
+399 \(xy + x = \frac{3x + 2y + z + y + 2z}{3}\)
\(xy+x=\frac{3x+3y+3z}{3}\).
\(xy+x=x+y+z\)
\(xy=y+z\)
\(x=\frac{y+z}{y}\). Can also be written as: \(x=1+\frac{z}{y}\).
