Answer
$\left\{\left(3,-\dfrac{8}{3},\dfrac{1}{9}\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
x+4y-3z=-8\\
3x-y+3z=12\\
x+y+6z=1
\end{cases}$
Use the elimination method. Add the first equation to the second equation; then multiply the first equation by 2 and add it to the third, to eliminate $z$:
$\begin{cases}
3x-y+3z+x+4y-3z=12-8\\
x+y+6z+2(x+4y-3z)=1+2(-8)
\end{cases}$
$\begin{cases}
4x+3y=4\\
3x+9y=-15
\end{cases}$
Multiply the first equation by -3 and add it to the second to eliminate $y$ and determine $x$:
$-3(4x+3y)+3x+9y=-3(4)-15$
$-12x-9y+3x+9y=-27$
$-9x=-27$
$x=3$
Determine $y$:
$4x+3y=4$
$4(3)+3y=4$
$3y=4-12$
$3y=-8$
$y=-\dfrac{8}{3}$
Determine $z$:
$x+y+6z=1$
$3-\dfrac{8}{3}+6z=1$
$6z=1-\dfrac{1}{3}$
$6z=\dfrac{2}{3}$
$z=\dfrac{1}{9}$
The solution set of the system is:
$\left\{\left(3,-\dfrac{8}{3},\dfrac{1}{9}\right)\right\}$
