Answer
$a.\qquad \left\{\begin{aligned}
x-2y+4z&=3\\
y+2z&=7\\
z&=2\end{aligned}\right.$
$b.\qquad (1,3,2)$
Work Step by Step
$a.$
$\left\{\begin{aligned}
x-2y+4z&=3\\
y+2z&=7\\
z&=2\end{aligned}\right.$
$b.$
The third equation gives $z=2$.
Back-substituting $z=2$ into the second equation,
$y+2(2)=7\quad \Rightarrow\quad y=3.$
Back-substituting $z=2 $ and $y=3$ into the first equation,
$x-2(3)+4(2)=3\Rightarrow x=1.$
The solution is $(1,3,2)$
