Answer
$(4,-2,3)$
Work Step by Step
We are given the system:
$\begin{cases}
2x-y+2z=16\\
7x-3y-5z=19\\
x-3z=-5
\end{cases}$
We will use the addition method. Multiply Equation 1 by -3 and add it to Equation 2 to eliminate $y$. :
$\begin{cases}
-3(2x-y+2z)=-3(16)\\
7x-3y-5z=19\\
x-3z=-5
\end{cases}$
$\begin{cases}
-6x+3y-6z=-48\\
7x-3y-5z=19\\
x-3z=-5
\end{cases}$
$\begin{cases}
-6x+3y-6z+7x-3y-5z=-48+19\\
x-3z=-5
\end{cases}$
$\begin{cases}
x-11z=-29\\
x-3z=-5
\end{cases}$
Multiply Equation 1 by -1 and add it to Equation 2 to eliminate $x$ and determine $z$:
$\begin{cases}
-x+11z=-(-29)\\
x-3z=-5
\end{cases}$
$-x+11z+x-3z=29-5$
$8z=24$
$z=\dfrac{24}{3}$
$z=3$
Substitute the value of $z$ in the Equation $x-3z=-5$ to determine $x$:
$x-3(3)=-5$
$x-9=-5$
$x=-5+9$
$x=4$
Substitute the values of $x, z$ is Equation 1 of the given system to find $y$:
$2x-y+2z=16$
$2(4)-y+2(3)=16$
$14-y=16$
$y=14-16$
$y=-2$
The system's solution is:
$(4,-2,3)$
