Answer
$(4,8,6)$
Work Step by Step
We are given the system:
$\begin{cases}
\dfrac{x+2}{6}-\dfrac{y+4}{3}+\dfrac{z}{2}=0\\
\dfrac{x+1}{2}+\dfrac{y-1}{2}-\dfrac{z}{4}=\dfrac{9}{2}\\
\dfrac{x-5}{4}+\dfrac{y+1}{3}+\dfrac{z-2}{2}=\dfrac{19}{4}
\end{cases}$
Multiply the Equation 1 by 6, Equation 2 by 4 and Equation 3 by 12 to eliminate denominators:
$\begin{cases}
6\left(\dfrac{x+2}{6}-\dfrac{y+4}{3}+\dfrac{z}{2}\right)=6(0)\\
4\left(\dfrac{x+1}{2}+\dfrac{y-1}{2}-\dfrac{z}{4}\right)=4\left(\dfrac{9}{2}\right)\\
12\left(\dfrac{x-5}{4}+\dfrac{y+1}{3}+\dfrac{z-2}{2}\right)=12\left(\dfrac{19}{4}\right)
\end{cases}$
$\begin{cases}
x+2-2(y+4)+3z=0\\
2(x+1)+2(y-1)-z=18\\
3(x-5)+4(y+1)+6(z-2)=57
\end{cases}$
$\begin{cases}
x+2-2y-8+3z=0\\
2x+2+2y-2-z=18\\
3x-15+4y+4+6z-12=57
\end{cases}$
$\begin{cases}
x-2y+3z=6\\
2x+2y-z=18\\
3x+4y+6z=80
\end{cases}$
We will use the addition method. Multiply Equation 1 by -2 and add it to Equation 2 to eliminate $x$. Also multiply Equation 1 by -3 and add it to Equation 3 to eliminate $x$:
$\begin{cases}
2x+2y-z-2x+4y-6z=18+(-2)(6)\\
3x+4y+6z-3x+6y-9z=80+(-3)(6)
\end{cases}$
$\begin{cases}
6y-7z=6\\
10y-3z=62
\end{cases}$
Multiply Equation 2 by 3, Equation 2 by -7 and add them to eliminate $z$ and determine $y$:
$\begin{cases}
3(6y-7z)=3(6)\\
-7(10y-3z)=-7(62)
\end{cases}$
$\begin{cases}
18y-21z=18\\
-70y+21z=-434
\end{cases}$
$18y-21z-70y+21z=18-434$
$-52y=-416$
$y=\dfrac{-416}{-52}$
$y=8$
Substitute the value of $y$ in the Equation $6y-7z=6$ to determine $z$:
$6(8)-7z=6$
$48-7z=6$
$7z=42$
$z=6$
Substitute the values of $y, z$ is Equation 1 of the given system to find $x$:
$x-2y+3z=6$
$x-2(8)+3(6)=6$
$x-16+18=6$
$x+2=6$
$x=4$
The system's solution is:
$(4,8,6)$
