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
$2\times Eq.1-3\times Eq.2 \Rightarrow$
$(6-6)x+(4+15)y+(-6-6)z=-4+6$
$4\times Eq.1-3\times Eq3 \Rightarrow$
$(12-12)x+(8+9)y+(-12-12)z=-8-30$
$\left[\begin{array}{ll}
19y-12z=2 & I\\
17y-24z=-38 & II
\end{array}\right.$ ...,
(2.) ...$ II-2\times I$ eliminates z:
$(17-38)y=-38-4$
$-21y=-42$
$y=2$
Back-substitute into$ 19y-12z=2$
$19(2)-12z=2$
$-12z=2-38$
$z=3$
(3.) Back-substitute into one of the initial equations:
$3x+2y-3z=-2$
$3x+2(2)-3(3)=-2$
$3x-5=-2$
$3x=3$
$x=1$
Solution set: : $\{ (1,2,3) \}$
