## Elementary Linear Algebra 7th Edition

Follow the steps below: Step 1: Substitute from the third equation the first one multiplied by $3$: $$3x-3x-2y-2y=8-0$$ which becomes $$-4y=8$$ and this gives $$y=-2.$$ Step 2: Substitute from the second equation the first one to eliminate $x$: $$x-x+y-2y=6-0$$ which becomes $$-y=6\Rightarrow y=-6.$$ Now since $y$ cannot be both $-6$ and $-2$ we see that this system is inconsistent and has no solutions.