Recognise an infinite family rather than one triple

The printed calculator result is $x=3+7z$, $y=-1-5z$, $z=z$. The last line means that $z$ is free, not that its value has been found. Write $z=t$, where $t\in\mathbb R$: all solutions are $(x,y,z)=(3+7t,-1-5t,t)$.

The second equation is exactly twice the first. Subtract twice the first from the third to obtain $-y-5z=1$, hence $y=-1-5z$. Substituting in the first gives $x=3+7z$. Every real $z$ gives one solution; there is no remaining independent constraint to fix it. Conversely, these elimination steps force every solution to have that form.

The speech bubble says that $z$ can be any real value and $x,y$ depend on it. Its choice $z=1$ gives $x=3+7=10$, $y=-1-5=-6$. Check $(10,-6,1)$: $10-12+3=1$, $20-24+6=2$, $20-18+1=3$. For arbitrary $t$, the three left sides simplify to $1$, $2$, $3$ respectively. Thus there are infinitely many solutions; this is a consistent system.