Chapter 5, Systems of Equations and Inequalities - Section 5.2 - Systems of Linear Equations in Several Variables - 5.2 Exercises - Page 455: 17

$x=2,$ $y=1,$ $z=-3$

$\begin{cases} x-y-z=4\\ 0x+2y+z=-1\\ -x+y-2z=5 \end{cases}$ -Adding Equation 2 and Equation 3. $\begin{cases} x-y-z=4\\ -x+y-2z=5\\ -- -- -- -- \\ 0x+0y-3z=9 \end{cases}$ Thus, $z=-3$, Substituting back in to Equation 2, $2y-3=-1, y=1$. Substituting back into Equation 1, $x-1+3=4, x=2$