Answer
$\{(2,-5,3)\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x-2y+z=15\\
3x+y-3z=-8\\
-2x+4y-z=-27
\end{cases}$
Use the elimination method. Multiply the first equation by 3 and add it to the second. Then add the first equation to the last equation:
$\begin{cases}
3x+y-3z+3(x-2y+z)=-8+3(15)\\
-2x+4y-z+x-2y+z=-27+15
\end{cases}$
$\begin{cases}
3x+y-3z+3x-6y+3z=-8+45\\
-2x+4y-z+x-2y+z=-27+15
\end{cases}$
$\begin{cases}
6x-5y=37\\
-x+2y=-12
\end{cases}$
Multiply the second equation by 6 and add it to the first:
$\begin{cases}
6x-5y=37\\
-6x+12y=-72
\end{cases}$
$6x-5y-6x+12y=37-72$
$7y=-35$
$y=-5$
Determine $x$:
$-x+2y=-12$
$-x+2(-5)=-12$
$x=-10+12$
$x=2$
Determine $z$:
$x-2y+z=15$
$2-2(-5)+z=15$
$12+z=15$
$z=3$
The solution set is:
$\{(2,-5,3)\}$
