Answer
$(8,-2,-2)$
Work Step by Step
We are given the system:
$\begin{cases}
x+2y+3z=-2\\
3x+3y+10z=-2\\
2y-5z=6
\end{cases}$
We will use the addition method. Multiply Equation 1 by -3 and add it Equation 2 to eliminate $x$:
We are given the system:
$\begin{cases}
-3(x+2y+3z)=-3(-2)\\
3x+3y+10z=-2\\
2y-5z=6
\end{cases}$
$\begin{cases}
-3x-6y-9z=6\\
3x+3y+10z=-2\\
2y-5z=6
\end{cases}$
$\begin{cases}
3x+3y+10z-3x-6y-9z=-2+6\\
2y-5z=6
\end{cases}$
$\begin{cases}
-3y+z=4\\
2y-5z=6
\end{cases}$
$\begin{cases}
5(-3y+z)=5(4)\\
2y-5z=6
\end{cases}$
$\begin{cases}
-15y+5z=20\\
2y-5z=6
\end{cases}$
$-15y+5z+2y-5z=20+6$
$-13y=26$
$y=-2$
$-3y+z=4$
$-3(-2)+z=4$
$6+z=4$
$z=-2$
$x+2y+3z=-2$
$x+2(-2)+3(-2)=-2$
$x-10=-2$
$x=8$
The system's solution is:
$(8,-2,-2)$
