Answer
$\left\{\left(\dfrac{1}{3},-2,0\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
3x+2y-8z=-3\\
-x-\dfrac{2}{3}y+z=1\\
6x-3y+15z=8
\end{cases}$
Multiply the second equation by 3 and add it to the first equation to eliminate $x$; then multiply the second equation by 6 and add it to the third equation to eliminate $x$:
$\begin{cases}
3x+2y-8z+3\left(-x-\dfrac{2}{3}y+z\right)=-3+3(1)\\
6x-3y+15z+6\left(-x-\dfrac{2}{3}y+z\right)=8+6(1)
\end{cases}$
$\begin{cases}
3x+2y-8z-3x-2y+3z=0\\
6x-3y+15z-6x-4y+6z=14
\end{cases}$
$\begin{cases}
z=0\\
-7y+21z=14
\end{cases}$
$-7y+21(0)=14$
$-7y=14$
$y=-2$
$3x+2y-8z=-3$
$3x+2(-2)-8(0)=-3$
$3x-4=-3$
$3x=1$
$x=\dfrac{1}{3}$
The solution set is:
$\left\{\left(\dfrac{1}{3},-2,0\right)\right\}$
