## College Algebra (6th Edition)

$\{ (1,-1,1) \}$
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.) Perform elimination by addition method (eliminate x) by $Eq.1-2\times Eq3 \Rightarrow (2-2)x+(1+4)y+(-2-6)z=-1-12$ $Eq.2-3\times Eq3 \Rightarrow (3-3)x+(-3+6)y+(-1-9)z=5-18$ $\left[\begin{array}{ll} 5y-8z=-13 & *\\ 3y-10z=-13 & \end{array}\right....$ (2.) ... giving us a system to solve: Eliminate y: multiply the first equation with $-3$, and the second with $5$ $\left[\begin{array}{ll} -15y+24z=39 & \\ 15y-50z=-65 & \end{array}\right.$, and then add... $-26z=-26$ $z=1$ Back substitute z into (*) $5y-8(1)=-13$ $5y=-13+8$ $5y=-5$ $y=-1$ (3.) Substitute z and y into one of the initial equations: $\mathrm{x}-2\mathrm{y}+3\mathrm{z}=6$ $x-2(-1)+3(1)=6$ $x+5=6$ $x=1$ Solution set: : $\{ (1,-1,1) \}$