## Precalculus (6th Edition) Blitzer

The solution to the system of equations is $x=2,y=1,z=-1$
Let us consider the following system of equations: $x+2y-z=5$ $2x-y+3z=0$ $2y+z=1$ Multiply equation $x+2y-z=5$ by $2$ and equation $2x-y+3z=0$ by $1$ and now, subtract: \begin{align} & 2\left( x+2y-z \right)-1\left( 2x-y+3z \right)=2\left( 5 \right)-1\left( 0 \right) \\ & 2x+4y-2z-2x+y-3z=10-0 \\ & 5y-5z=10 \\ & y-z=2 \end{align} Add equations $2y+z=1$ and $y-z=2$ and simplify: \begin{align} & 2y+z+y-z=1+2 \\ & 3y=3 \\ & y=1 \end{align} Put the value of y in equation $y-z=2$ and get $z=-1$ Then, substitute the values of y and z in equation $x+2y-z=5$ and simplify: \begin{align} & x+2\left( 1 \right)-\left( -1 \right)=5 \\ & x+2+1=5 \\ & x=2 \end{align} Therefore, consider $x+2y-z=5$ and substitute $x=2,y=1,z=-1$ as given below: \begin{align} & \left( 2 \right)+2\left( 1 \right)-\left( -1 \right)=5 \\ & 2+2+1=5 \\ & 5=5,\text{ true}\text{.} \end{align} Therefore, consider $2x-y+3z=0$ and substitute $x=2,y=1,z=-1$ as given below: \begin{align} & 2\left( 2 \right)-\left( 1 \right)+3\left( -1 \right)=0 \\ & 4-1-3=0 \\ & 0=0,\text{ true}\text{.} \end{align} So, consider $2y+z=1$ and put $x=2,y=1,z=-1$ as given below: \begin{align} & 2\left( 1 \right)+\left( -1 \right)=1 \\ & 2-1=1 \\ & 1=1,\text{ true}\text{.} \end{align} Thus, the solution of the system of equations is $x=2,\,\,y=1,\,\,z=-1$