Answer
$\{ (2,3,3) \}$
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. Here, we have a chance to temporarily eliminate x and y by subtracting (Eq.2 - Eq.1):
resulting in:
$z=3$.
2. Substitute this z into equations 2 and 3:
$\left[\begin{array}{ll}
x+y+3(3)=14 & \\
x+2y-(3)=5 & \\
\hline & \\
x+y=14-9 & \\
x+2y=5+3 & \\
\hline & \\
x+y=5 & *\\
x+2y=8 & **
\end{array}\right]$
Subtract: $(**-*)$
$y= 3,$
Substitute $y= 3$ into $*$:
$x+3=5$
$x=5-3$
$x=2$
3. z was found in step 1 ( $z=3$ ).
Solution set: : $\{ (2,3,3) \}$
