#### Answer

$\{ (\displaystyle \frac{1}{2},\frac{1}{3},-1) \}$

#### 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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First, write each equation in standard form: Ax+By+Cz=D.
$\left[\begin{array}{l}
6x+3y+5z=-1\\
2x-6y+8z=-9\\
4+4x=-3z+9y
\end{array}\right.$
$\left[\begin{array}{l}
6x+3y+5z=-1\\
2x-6y+8z=-9\\
4x-9y+3z==4
\end{array}\right.$
(1.) Perform elimination by addition method (eliminate x) by
$\left[\begin{array}{lll}
Eq.1.-3\times Eq.2\rightarrow & 21y-19z=26 & I\\
-2\times Eq.2.+Eq.3 & 3y-13z=14 & II
\end{array}\right.$ ,
(2.) ... $(Eq.I - 7\times Eq.II)$ eliminates y,
$72z=-72$
$z=-1$
Back-substitute into $21y-19z=26$
$21y-19(-1)=26$
$21y=26-19$
$y=\displaystyle \frac{7}{21}=\frac{1}{3}$
(3.) Back-substitute into one of the initial equations:
$6x+3(\displaystyle \frac{1}{3})+5(-1)=-1$
$6x-4=-1$
$6x=3$
$x=\displaystyle \frac{1}{2}$
Solution set: : $\{ (\displaystyle \frac{1}{2},\frac{1}{3},-1) \}$