## College Algebra (6th Edition)

$\{ (2,3,3) \}$
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) ------------ 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: : $\{ (2,3,3) \}$