Answer
$\{ (1,-1,1) \}$
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)
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(1.) Perform elimination by addition method (eliminate x) by
$Eq.1-2\times Eq3 \Rightarrow (2-2)x+(1+4)y+(-2-6)z=-1-12$
$Eq.2-3\times Eq3 \Rightarrow (3-3)x+(-3+6)y+(-1-9)z=5-18$
$\left[\begin{array}{ll}
5y-8z=-13 & *\\
3y-10z=-13 &
\end{array}\right....$
(2.) ... giving us a system to solve:
Eliminate y:
multiply the first equation with $-3$,
and the second with $5 $
$\left[\begin{array}{ll}
-15y+24z=39 & \\
15y-50z=-65 &
\end{array}\right.$, and then add...
$-26z=-26$
$z=1$
Back substitute z into (*)
$5y-8(1)=-13$
$5y=-13+8$
$5y=-5$
$y=-1$
(3.) Substitute z and y into one of the initial equations:
$\mathrm{x}-2\mathrm{y}+3\mathrm{z}=6$
$x-2(-1)+3(1)=6$
$x+5=6$
$x=1$
Solution set: : $\{ (1,-1,1) \}$
