Answer
The numbers are $x=4$, $y=8$ and $z=6$.
Work Step by Step
The system of equations is as shown below:
$\begin{align}
& \frac{x+2}{6}-\frac{y+4}{3}+\frac{z}{2}=0 \\
& \frac{x+1}{2}+\frac{y-1}{2}-\frac{z}{4}=\frac{9}{2} \\
& \frac{x-5}{4}+\frac{y+1}{3}+\frac{z-2}{2}=\frac{19}{4}
\end{align}$
Now, simplify the first equation by taking the L.C.M to get the simpler form:
$\begin{align}
& x+2-2\left( y+4 \right)+3z=0 \\
& x+2-2y-8+3z=0 \\
& x-2y+3z-6=0
\end{align}$
By adding 6 to both sides to we get:
$x-2y+3z=6$ Mark it as (I)
Simplify the second equation by taking the L.C.M; we get:
$\begin{align}
& 2\left( x+1 \right)+2\left( y-1 \right)-z=18 \\
& 2x+2+2y-2-z=18
\end{align}$
By solving the equation to we get:
$2x+2y-z=18$ Mark it as (II)
Simplify the third equation by taking the L.C.M to get:
$\begin{align}
& 3\left( x-5 \right)+4\left( y+1 \right)+6\left( z-2 \right)=57 \\
& 3x-15+4y+4+6z-12=57 \\
& 3x+4y+6z-23=57
\end{align}$
Add 23 to both sides to simplify it:
$3x+4y+6z=80$ Mark it as (III)
The system of equations is simplified and becomes:
$\begin{align}
& x-2y+3z=6 \\
& 2x+2y-z=18 \\
& 3x+4y+6z=80
\end{align}$
By multiplying equation (I) by $-2$ and eliminating x from equations (I) and (II), we get:
$\begin{align}
& 2x+2y-z=18 \\
& -2x+4y-6z=-12
\end{align}$
$6y-7z=6$ (IV)
By multiplying equation (I) by $-3$ and eliminating x from equations (I) and (II), we get:
$\begin{align}
& 3x+4y+6z=80 \\
& -3x+6y-9z=-18
\end{align}$
$10y-3z=\text{62}$ (V)
Multiply equation (IV) by 10, equation (V) by −6, and then add to get the value of z:
$\begin{align}
& 60y-70z=60 \\
& -60y+18y=-372
\end{align}$
$-52z=-312$
By dividing the equation on both sides by −52, we get
$\begin{align}
& \frac{-52z}{-52}=\frac{-312}{-52} \\
& \text{ }z=6 \\
\end{align}$
Substitute the value of z in equation (IV) to get:
$\begin{align}
& 6y-7\left( 6 \right)=6 \\
& 6y-42=6
\end{align}$
Now adding 42 to both sides to get the value of y:
$\begin{align}
& 6y-42+42=6+42 \\
& 6y=48
\end{align}$
Now, divide both sides by 6 to get:
$\begin{align}
& \frac{6y}{6}=\frac{48}{6} \\
& \text{ }y=8 \\
\end{align}$
Substitute the values of y and z in equation (I) to get:
$\begin{align}
& x-2\left( 8 \right)+3\left( 6 \right)=6 \\
& x-16+18=6 \\
& x+2=6
\end{align}$
Subtract 2 from both sides of the equation and we get:
$\begin{align}
& x+2-2=6-2 \\
& x=4
\end{align}$
Hence, the values are $x=4$, $y=8,$ and $z=6$.
