Chapter 9 - Systems and Matrices - Summary Exercises on Systems of Equations - Exercises - Page 907: 18

$\{(-8z-56,z+13,z)\}$

1. There will be unlimited number of solutions. Multiply -2 to the first equation and get $-2x-10y-6z=-18$ 2. Add the above to the second equation and get $-y+z=-13$, thus $y=z+13$ 3. The first equation becomes $x+5(z+13)+3z=9$ which gives $x=-8z-56$ 4. Thus the solution set can be written as $\{(-8z-56,z+13,z)\}$