## Precalculus (6th Edition) Blitzer

The solution of the system is $\left( 1,-5,-6 \right)$.
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.