## College Algebra 7th Edition

a.\qquad\left\{\begin{aligned} x+y-3z&=8\\ y-3z&=5\\ z&=-1\end{aligned}\right. $b.\qquad (3,2,-1)$
a.\left\{\begin{aligned} x+y-3z&=8\\ y-3z&=5\\ z&=-1\end{aligned}\right. $b.$ The third equation gives $z=-1$. Back-substituting $z=-1$ into the second equation, $y-3(-1)=5\quad \Rightarrow\quad y=2.$ Back-substituting $z=-1$and $y=2$ into the first equation, $x+2-3(-1)=8\Rightarrow x=3.$ The solution is $(3,2,-1)$