###### Itemb.iv

If $$y=0\text{,}$$ then we arrive at the system of three linear equations:

\begin{equation*} \begin{alignedat}{4} x \amp \amp \amp {}={} \amp 3 \\ \amp {}-{} \amp z \amp {}={} \amp 2 \\ 2x \amp {}+{} \amp z \amp {}={} \amp 4. \\ \end{alignedat} \end{equation*}

We have a solution when $$x=3$$ and $$z=-2\text{.}$$ Therefore, $$(x,y,z)=(3,0,-2)$$ is a solution to the original system of equations.

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