Chapter 9 - Systems and Matrices - Summary Exercises on Systems of Equations - Exercises - Page 907: 21

$\{(1,2,3)\}$

1. Multiply -2 to the first equation to get $-2x-2y+2z=0$ 2. Add the above to the third equation and get $y-2z=-4$ 3. Multiply -2 to the above equation to get $-2y+4z=8$ 4. Add the above to the second equation and get $3z=9$, thus $z=3$ 5. We then have $y=2z-4=2$ and $x=z-y=1$ 6. Thus the solution set is $\{(1,2,3)\}$