Answer
a. The point $(x,y,z)=(0,5,1)$ is not a solution of the system.
b. The point $(x,y,z)=(0,5,-3)$ is not a solution of the system.
Work Step by Step
The given system is
$\left\{\begin{matrix}
x &+y &+z& = &-2 \\
2x& -y & +4z&= &-17 \\
x&+3y&-z&=&18
\end{matrix}\right.$
a.
Substitute $(x,y,z)=(0,5,1)$ into all three equations.
$\left\{\begin{matrix}
0 &+5 &+1& = &-2 \\
2(0)& -5 & +4(1)&= &-17 \\
0&+3(5)&-1&=&18
\end{matrix}\right.$
Simplify.
$\left\{\begin{matrix}
6& = &-2 \\
-1&= &-17 \\
14&=&18
\end{matrix}\right.$
All equations are not true, the point $(x,y,z)=(0,5,1)$ is not a solution of the system.
b.
Substitute $(x,y,z)=(0,5,-3)$ into all three equations.
$\left\{\begin{matrix}
0 &+5 &-3& = &-2 \\
2(0)& -5 & +4(-3)&= &-17 \\
0&+3(5)&-(-3)&=&18
\end{matrix}\right.$
Simplify.
$\left\{\begin{matrix}
2& = &-2 \\
-17&= &-17 \\
18&=&18
\end{matrix}\right.$
First equation is not true, the point $(x,y,z)=(0,5,-3)$ is not a solution of the system.
