Answer
$(-3,0,2)$
Work Step by Step
We are given the system:
$\begin{cases}
\dfrac{x+3}{2}-\dfrac{y-1}{2}+\dfrac{z+2}{4}=\dfrac{3}{2}\\
\dfrac{x-5}{2}+\dfrac{y+1}{3}-\dfrac{z}{4}=-\dfrac{25}{6}\\
\dfrac{x-3}{4}-\dfrac{y+1}{2}+\dfrac{z-3}{2}=-\dfrac{5}{2}
\end{cases}$
Multiply the Equation 1 by 4, Equation 2 by 12 and Equation 3 by 4 to eliminate denominators:
$\begin{cases}
4\left(\dfrac{x+3}{2}-\dfrac{y-1}{2}+\dfrac{z+2}{4}\right)=4\left(\dfrac{3}{2}\right)\\
12\left(\dfrac{x-5}{2}+\dfrac{y+1}{3}-\dfrac{z}{4}\right)=12\left(-\dfrac{25}{6}\right)\\
4\left(\dfrac{x-3}{4}-\dfrac{y+1}{2}+\dfrac{z-3}{2}\right)=4\left(-\dfrac{5}{2}\right)
\end{cases}$
$\begin{cases}
2(x+3)-2(y-1)+z+2=2(3)\\
6(x-5)+4(y+1)-3z=2(-25)\\
x-3-2(y+1)+2(z-3)=2(-5)
\end{cases}$
$\begin{cases}
2x+6-2y+2+z+2=6\\
6x-30+4y+4-3z=-50\\
x-3-2y-2+2z-6=-10
\end{cases}$
$\begin{cases}
2x-2y+z=-4\\
6x+4y-3z=-24\\
x-2y+2z=1
\end{cases}$
We will use the addition method. Multiply Equation 1 by 2 and add it to Equation 2 to eliminate $y$. Also multiply Equation 1 by -1 and add it to Equation 3 to eliminate $y$:
$\begin{cases}
6x+4y-3z+2(2x-2y+z)=-24+2(-4)\\
x-2y+2z+(-1)(2x-2y+z)=1+(-1)(-4)
\end{cases}$
$\begin{cases}
6x+4y-3z+4x-4y+2z=-32\\
x-2y+2z-2x+2y-z=5
\end{cases}$
$\begin{cases}
10x-z=-32\\
-x+z=5
\end{cases}$
Add Equation 1 to Equation 2 to eliminate $z$ and determine $x$:
$10x-z-x+z=-32+5$
$9x=-27$
$x=-3$
Substitute the value of $x$ in the Equation $-x+z=5$ to determine $z$:
$-(-3)+z=5$
$3+z=5$
$z=2$
Substitute the values of $x, z$ is Equation 1 of the given system to find $y$:
$2x-2y+z=-4$
$2(-3)-2y+2=-4$
$-4-2y=-4$
$2y=0$
$y=0$
The system's solution is:
$(-3,0,2)$
