The system has an infinite number of solutions because we have 3 variables and only 2 equations:
{ 3 x − y + z = 2 2 x − z = 2 ⟹ { 5 x − y = 4 2 x − z = 2 ⟹ \begin{cases}3x-y+z=2\\
2x-z=2
\end{cases}
\implies
\begin{cases}
5x-y=4\\
2x-z=2
\end{cases}
\implies { 3 x − y + z = 2 2 x − z = 2 ⟹ { 5 x − y = 4 2 x − z = 2 ⟹ { 5 x − y = 4 x = 1 2 ( 2 + z ) ⟹ { 5 x − y = 4 x = 1 + 0.5 z ⟹ \begin{cases}
5x-y=4\\
x=\frac12(2+z)
\end{cases}
\implies\begin{cases}
5x-y=4\\
x=1+0.5z
\end{cases}\implies { 5 x − y = 4 x = 2 1 ( 2 + z ) ⟹ { 5 x − y = 4 x = 1 + 0.5 z ⟹ { 5 ( 1 + 0.5 z ) − y = 4 x = 1 + 0.5 z ⟹ { 5 + 2.5 z − y = 4 x = 1 + 0.5 z \begin{cases}
5(1+0.5z)-y=4\\
x=1+0.5z
\end{cases}\implies\begin{cases}
5+2.5z-y=4\\
x=1+0.5z
\end{cases} { 5 ( 1 + 0.5 z ) − y = 4 x = 1 + 0.5 z ⟹ { 5 + 2.5 z − y = 4 x = 1 + 0.5 z ⟹ { y = 1 + 2.5 z x = 1 + 0.5 z ⟹ \implies\begin{cases}
y=1+2.5z\\
x=1+0.5z
\end{cases}\implies ⟹ { y = 1 + 2.5 z x = 1 + 0.5 z ⟹ { x = 1 + 0.5 z y = 1 + 2.5 z z ∈ R \begin{cases}
x=1+0.5z\\
y=1+2.5z\\
z\in R
\end{cases} ⎩ ⎨ ⎧ x = 1 + 0.5 z y = 1 + 2.5 z z ∈ R
R R R - set of real numbers.
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