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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}

 May 1, 2023
 #1
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\(xy + x = \frac{3x + 2y + z + y + 2z}{3}\)

\(3xy + 3x = 3x + 2y + z + y + 2z\)

\(3xy=2y+y+2z+z\)

\(3xy=3y+3z\)

\(xy=y+z\)

\(x=1+\frac{z}{y}\)

Assuming that y is not 0.

 May 3, 2023

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