## College Algebra (6th Edition)

$\{ (1,-1,2) \}$
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 y) by $Eq.2+Eq1 \Rightarrow$ $(3+1)x+(1-1)y+(-2+3)z=-2+8$ $Eq.3+4\times Eq1 \Rightarrow$ $(2+4)x+(4-4)y+(1+12)z=0+24$ $\left[\begin{array}{ll} 4x+z=6 & I\\ 6x+13z=24 & II \end{array}\right.$ ..., (2.) ...$(II-13\times I)$ eliminates z: $(6-52)x=24-78$ $-46x=-46$ $x=1$ Back-substitute into $\ \ 4x+z=6$ $4(1)+z=6$ $z=2$ (3.) Back-substitute into one of the initial equations: x-y+3z=8 $1-y+3(2)=8$ $7-8=y$ $y=-1$ Solution set: : $\{ (1,-1,2) \}$