## College Algebra (10th Edition)

$\{(1,3,-2)\}$
$\left\{\begin{array}{lllll} x & +y & -z & =6 & \\ 3x & -2y & +z & =-5 & /R_{2}=r_{2}+r_{1}\\ x & +3y & -2z & =14 & /R_{3}=r_{3}-2r_{1} \end{array}\right.$ ... Eliminate $z$ in eq. 2 and 3 $\left\{\begin{array}{lllll} x & +y & -z & =6 & \\ 4x & -y & & =1 & \\ -x & +y & & =2 & /R_{3}=r_{3}+r_{2} \end{array}\right.$ ... Eliminate y in equation 3 $\left\{\begin{array}{lllll} x & +y & -z & =6 & (1)\\ 4x & -y & & =1 & (2)\\ 3x & & & =3 & (3) \end{array}\right.$ $(3)\Rightarrow x=1$ $(2) \Rightarrow y=4x-1=4(1)-1=3$ $(1)\Rightarrow z=x+y-6=1+3-6=-2$ Solution set: $\{(1,3,-2)\}$