Answer
The solution of the system is $\left( 1,-5,-6 \right)$.
Work Step by Step
Let us consider the given equation:
$x+y=-4$ …… (I)
$y-z=1$ …… (II)
$2x+y+3z=-21$ …… (III)
Now, subtract equation (II) from equation (I):
$\begin{align}
& x+y=-4 \\
& y-z=1 \\
& \underline{-\text{ + }-} \\
& x+z=-5 \\
\end{align}$
$x+z=-5$ …… (IV)
Again, subtract equation (II) from equation (III):
$\begin{align}
& 2x+y+3z=-21 \\
& \text{ }y-z=1 \\
& \underline{\text{ - + -}} \\
& 2x+4z=-22 \\
\end{align}$
Futher simplify,
$\begin{align}
& 2\left( x+2z \right)=-22 \\
& x+2z=-11
\end{align}$ …… (V)
Subtract equation (V) from equation (IV):
$\begin{align}
& x+z=-5 \\
& x+2z=-11 \\
& \underline{-\text{ }-\text{ +}} \\
& -z=6 \\
\end{align}$
Futher simplify,
$\text{ }z=-6$
Putting the value of $z=-6$ in equation (II):
$\begin{align}
& y-\left( -6 \right)=1 \\
& y=-5 \\
\end{align}$
Putting the value of $y=-5$ in equation(I):
$\begin{align}
& x-5=-4 \\
& x=1 \\
\end{align}$
Thus, the ordered triple $\left( 1,-5,-6 \right)$ satisfies the given system.
