We are going to try and find values of \(x\), \(y\), and a \(z\) that will satisfy all three equations at the same time. We are going to use elimination to eliminate one of the variables from one of the equations and two of the variables from another of the equations. The reason for doing this will be apparent once we’ve actually done it.

The elimination method in this case will work a little differently than with two equations. As with two equations we will multiply as many equations as we need to so that if we start adding pairs of equations we can eliminate one of the variables.

In this case it looks like if we multiply the second equation by 2 it will be fairly simple to eliminate the \(y\) term from the second and third equation by adding the first equation to both of them. So, let’s first multiply the second equation by two.

\[\begin{align*}
x-2y+3z & =7 & \underrightarrow{\text{same}}\hspace{0.1in} & & x-2y+3z & =7 \\
2x+y+z & =4 & \underrightarrow{\times \,\,2}\hspace{0.1in} & & 4x+2y+2z & =8 \\
-3x+2y-2z & =-10 & \underrightarrow{\text{same}}\hspace{0.1in} & & -3x+2y-2z & =-10 \\
\end{align*}\]

Now, with this new system we will replace the second equation with the sum of the first and second equations and we will replace the third equation with the sum of the first and third equations.

Here is the resulting system of equations.

\[\begin{align*}x - 2y + 3z & = 7\\ 5x\,\,\,\,\,\,\,\,\,\,\,\, + 5z & = 15\\ - 2x\,\,\,\,\,\,\,\,\,\,\,\,\,\, + z & = - 3\end{align*}\]

So, we’ve eliminated one of the variables from two of the equations. We now need to eliminate either \(x\) or \(z\) from either the second or third equations. Again, we will use elimination to do this. In this case we will multiply the third equation by -5 since this will allow us to eliminate \(z\) from this equation by adding the second onto is.

\[\begin{align*}
x-2y+3z & =7 & \underrightarrow{\text{same}} \hspace{0.1in} & & x-2y+3z & =7 \\
5x\,\,\,\,\,\,\,\,\,\,\,\,+5z & =15 & \underrightarrow{\text{same}} \hspace{0.1in} & & 5x\,\,\,\,\,\,\,\,\,\,\,\,+5z & =15 \\
-2x\,\,\,\,\,\,\,\,\,\,\,\,\,\,+z & =-3 & \underrightarrow{\times \,\,-5} \hspace{0.1in} & & 10x\,\,\,\,\,\,\,\,\,\,\,\,-5z & =15 \\
\end{align*}\]

Now, replace the third equation with the sum of the second and third equation.

\[\begin{align*}x - 2y + 3z & = 7\\ 5x\,\,\,\,\,\,\,\,\,\,\,\, + 5z & = 15\\ 15x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, & = 30\end{align*}\]

Now, at this point notice that the third equation can be quickly solved to find that \(x = 2\). Once we know this we can plug this into the second equation and that will give us an equation that we can solve for \(z\) as follows.

\[\begin{align*}5\left( 2 \right) + 5z & = 15\\ 10 + 5z & = 15\\ 5z & = 5\\ z & = 1\end{align*}\]

Finally, we can substitute both \(x\) and \(z\) into the first equation which we can use to solve for \(y\). Here is that work.

\[\begin{align*}2 - 2y + 3\left( 1 \right) & = 7\\ - 2y + 5 & = 7\\- 2y & = 2\\ y & = - 1\end{align*}\]

So, the solution to this system is \(x = 2\), \(y = - 1\) and \(z = 1\).