## Precalculus (6th Edition) Blitzer

To check whether $z$ can be equal to $3$ or $-1$. Let us consider the system of equations as follows and mark them: $3x+9y=0$ (I) $9y+3z=0$ (II) $2x+2y-4z=-4$ (III) And eliminate y from the equations (I) and (II) to get: \begin{align} & 3x+9y\text{ }=0 \\ & \underline{\text{ }-9y-3z=0} \\ & 3x\text{ }-3z=0 \\ \end{align} Add 3z to both sides to get, \begin{align} & 3x-3z+3z=0+3z \\ & 3x=3z \end{align} And divide both sides of the equation by 3 to get, \begin{align} & \frac{3x}{3}=\frac{3z}{3} \\ & x=z \end{align} Put $x=z$ in equation (III) to get the value of y, \begin{align} & 2x+2y-4\left( x \right)=-4 \\ & 2x+2y-4x=-4 \\ & 4y=-4 \end{align} And divide both sides by 4 to get, \begin{align} & \frac{4y}{4}=\frac{-4}{4} \\ & y=-1 \end{align} . Put the value of y in equation (I) to get the value of x, \begin{align} & 3x+9\left( -1 \right)=0 \\ & 3x-9=0 \\ \end{align} Add 9 to both sides to get, \begin{align} & 3x-9+9=0+9 \\ & 3x=9 \end{align} And divide both sides by 3 to get, \begin{align} & \frac{3x}{3}=\frac{9}{3} \\ & x=3 \end{align}