Answer
$\{ (0,0,4) \}$
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) by
$Eq.$2 already is without x.
$-3Eq.1+Eq.3 \Rightarrow$
$(-3+3)x+(-2-9)y+(9-15)z=-60+36$
$\left[\begin{array}{ll}
y-4z=-16 & I\\
-11y-6z=-24 & II
\end{array}\right.$ ,
({\it 2}.) ... $11\cdot I+II$ eliminates $y$:
$(-44-6)z=-176-24$
$-50z=-200$
$z=4$
Back-substitute into $y-4z=-16$
$y-4(4)=-16$
$y=0$
(3.) Back-substitute into one of the initial equations:
$x+3y+5z=20$
$1+0+20=20$
$x=0$
Solution set: : $\{ (0,0,4) \}$
