Answer
$\{ (1,0,-3) \}$
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 z):
$Eq.1$ already is without z.
$Eq.2+Eq.3 \Rightarrow$
$(3+1)x+(2+1)y+(1-1)z=4+0$
$\left[\begin{array}{ll}
2x+y=2 & I\\
4x+3y=4 & II
\end{array}\right.$ ,
(2.) ... $-3I+II$ eliminates $y$:
$(-6+4)x=-6+4$
$x=1$
Back-substitute into $2x+y=2$
$2(1)+y=2$
$y=0$
(3.) Back-substitute into one of the initial equations:
$x+y-z=4$
$1+0-z=4$
$z=-3$
Solution set: : $\{ (1,0,-3) \}$
