Precalculus (6th Edition) Blitzer

The numbers are $x=4$, $y=8$ and $z=6$.
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 $z=6$.