#### Answer

The statement does not make sense.

#### Work Step by Step

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,
$\begin{align}
& \frac{3x}{3}=\frac{9}{3} \\
& x=3
\end{align}$