Precalculus (6th Edition) Blitzer

The solution of the system of equations is $x=0,y=1,z=2$.
Consider the system of equations, $2x-y+z=1$ $3x-3y+4z=5$ $4x-2y+3z=4$ Multiply the equation $2x-y+z=1$ by $3$ and the equation $3x-3y+4z=5$ by $2$; now, subtract as given below: \begin{align} & 3\left( 2x-y+z \right)-2\left( 3x-3y+4z \right)=3\left( 1 \right)-2\left( 5 \right) \\ & 6x-3y+3z-6x+6y-8z=3-10 \\ & 3y-5z=-7 \end{align} Multiply equation $3x-3y+4z=5$ by $4$ and equation $4x-2y+3z=4$ by $3$ and then subtract the equations as given below: \begin{align} & 4\left( 3x-3y+4z \right)-3\left( 4x-2y+3z \right)=4\left( 5 \right)-3\left( 4 \right) \\ & 12x-12y+16z-12x+6y-9z=20-12 \\ & -6y+7z=8 \end{align} Then, multiply equation $3y-5z=-7$ by 2 and add to the equation $-6y+7z=8$ as given below: \begin{align} & \not{6}y-10z-\not{6}y+7z=-14+8 \\ & -3z=-6 \\ & z=2 \end{align} Put the value of z in equation $3y-5z=-7$ and we get, $y=1$ Put the values of y and z in equation $2x-y+z=1$ and we get, $x=0$ Therefore, consider $2x-y+z=1$ and substitute $x=0,y=1,z=2$ as given below: \begin{align} & 2\left( 0 \right)-\left( 1 \right)+\left( 2 \right)=1 \\ & 0-1+2=1 \\ & 1=1 \end{align} Which is correct. So consider $3x-3y+4z=5$ and put, $x=0,y=1,z=2$ as given below: \begin{align} & 3\left( 0 \right)-3\left( 1 \right)+4\left( 2 \right)=5 \\ & 0-3+8=5 \\ & 5=5 \end{align} Which is correct. Therefore, consider $4x-2y+3z=4$ and substitute $x=0,y=1,z=2$ as given below: \begin{align} & 4\left( 0 \right)-2\left( 1 \right)+3\left( 2 \right)=4 \\ & 0-2+6=4 \\ & 4=4 \end{align} Which is also correct. Thus the solution of the system of equations is $x=0,y=1,z=2$