#### Answer

$\{ (3,1,5) \}$

#### Work Step by Step

Plan (see p.533):
1. Eliminate one variable and arrive at a system of two equations in two variables.
2. Solve the system of two equations in two variables.
3. Back-substitute the solutions of (2) to find the eliminated varable in (1)
------------
(1.) Perform elimination by addition method (eliminate $x$) by
$Eq.1-2\times Eq.2 \Rightarrow$
$(2-2)x+(-4-4)y+(3+2)z=17-0$
$Eq.3-4\times Eq.2 \Rightarrow$
$(4-4)x+(-1-8)y+(-1+4)z=6-0$
$\left[\begin{array}{ll}
-8y+5z=17 & \\
-9y+3z=6 & \div(-3)
\end{array}\right.$ ,
$\left[\begin{array}{ll}
-8y+5z=17 & I\\
3y-z=-2 & II
\end{array}\right.$ ...,
(2.) ... $I+5\times II$ eliminates $z$:
$(-8+15)z=17-10$
$7y=7$
$y=1$
Back-substitute into $3y-z=-2$
$3(3)-z=-2$
$3+2=z$
$z=5$
(3.) Back-substitute into one of the initial equations:
$x+2y-z=0$
$x+2(1)-(5)=0$
$x=3$
Solution set: : $\{ (3,1,5) \}$