Answer
$\left\{\left(\dfrac{56}{13},-\dfrac{7}{13},\dfrac{35}{13}\right)\right\}$
Work Step by Step
We are given the system of equations:
$\begin{cases}
2x+y-3z=0\\
-2x+2y+z=-7\\
3x-4y-3z=7
\end{cases}$
Use the elimination method. Multiply the second equation by 3 and add it to the first equation to eliminate $z$. Multiply the second equation by 3 and add it to the third equation to eliminate $z$.
$\begin{cases}
2x+y-3z+3(-2x+2y+z)=0+3(-7)\\
3x-4y-3z+3(-2x+2y+z)=7+3(-7)
\end{cases}$
$\begin{cases}
2x+y-3z-6x+6y+3z=-21\\
3x-4y-3z-6x+6y+3z=-14
\end{cases}$
$\begin{cases}
-4x+7y=-21\\
-3x+2y=-14
\end{cases}$
Multiply the first equation by 2, multiply the second equation by -7, and add them to eliminate $y$ and determine $x$:
$\begin{cases}
2(-4x+7y)=2(-21)\\
-7(-3x+2y)=-7(-14)
\end{cases}$
$-8x+14y+21x-14y=-42+98$
$13x=56$
$x=\dfrac{56}{13}$
Determine $y$:
$-3x+2y=-14$
$-3\left(\dfrac{56}{13}\right)+2y=-14$
$2y=-14+\dfrac{168}{13}$
$2y=-\dfrac{14}{13}$
$y=-\dfrac{7}{13}$
Determine $z$:
$-2x+2y+z=-7$
$-2\left(\dfrac{56}{13}\right)+2\left(-\dfrac{7}{13}\right)+z=-7$
$-\dfrac{126}{13}+z=-7$
$z=-7+\dfrac{126}{13}$
$z=\dfrac{35}{13}$
The solution set of the system is:
$\left\{\left(\dfrac{56}{13},-\dfrac{7}{13},\dfrac{35}{13}\right)\right\}$
