Answer
$\{ (-1,-2,3) \}$
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 $x$) by
$3\times Eq.1-2\times Eq.2 \Rightarrow$
$(6-6)x+(9-4)y+(21+10)z=39+44$
$5\times Eq.1-2\times Eq3 \Rightarrow$
$(10-10)x+(15-14)y+(35+6)z=65+56$
$\left[\begin{array}{ll}
5y+31z=83 & I\\
y+41z=121 & II
\end{array}\right.$ ...,
(2.) ...$( I-5\times II)$ eliminates $y$:
$(31-205)z=83-605$
$-174z=-522$
$z=3$
Back-substitute into $\ \ y+41z=121$
$y+41(3)=121$
$y=121-123$
$y=-2$
(3.) Back-substitute into one of the initial equations:
$2x+3y+7z=13$
$2x+3(-2)+7(3)=13$
$2x=13-15$
$x=-1$
Solution set: : $\{ (-1,-2,3) \}$
