Answer
$\{ (2,-1,1) \}$
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 y) by
$Eq.2+2\times Eq1 \Rightarrow $
$(1+8)x+(2-2)y+(-1+4)z=-1+22$
$Eq.3+2\times Eq1 \Rightarrow$
$ (2+8)x+(2-2)y+(-3+4)z=-1+22$
$\left[\begin{array}{ll}
9x+3z=21 & \\
10x+z=21 &
\end{array}\right.$ ..., divide the first eq. with 3,
$\left[\begin{array}{ll}
3x+z=7 & \\
10x+z=21 &
\end{array}\right.$
(2.) ... giving us a system to solve:
Eliminate z:
Subtract the first equation from the second:
$7x=14$
$x=2$
Back substitute x=2 into $3x+z=7$
$3(2)+z=7$
$z=7-6$
$z=1$
(3.) Substitute $z$ and $x$ into one of the initial equations:
$x+2y-z=-1$
$2+2y-1=-1$
$2y=-1-1$
$y=-1$
Solution set: : $\{ (2,-1,1) \}$
