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 \(z = 1\).

Before we get started on the solution process do not get excited about the fact that the second equation only has two variables in it. That is a fairly common occurrence when we have more than two equations in the system.

In fact, we’re going to take advantage of the fact that it only has two variables and one of them, the \(y\), has a coefficient of -1. This equation is easily solved for \(y\) to get,

\[y = 4x + 5\]

We can then substitute this into the first and third equation as follows,

\[\begin{align*}2x - 4\left( {4x + 5} \right) + 5z & = - 33\\- 2x + 2\left( {4x + 5} \right) - 3z & = 19\end{align*}\]

Now, if you think about it, this is just a system of two linear equations with two variables (\(x\) and \(z\)) and we know how to solve these kinds of systems from our work in the previous section.

First, we’ll need to do a little simplification of the system.

\[\begin{array}{*{20}{c}}{2x - 16x - 20 + 5z = - 33}\\{ - 2x + 8x + 10 - 3z = 19}\end{array}\hspace{0.25in} \to \hspace{0.25in}\begin{array}{*{20}{c}}{ - 14x + 5z = - 13}\\{6x - 3z = 9}\end{array}\]

The simplified version looks just like the systems we were solving in the previous section. Well, it’s almost the same. The variables this time are \(x\) and \(z\) instead of \(x\) and \(y\), but that really isn’t a difference. The work of solving this will be the same.

We can use either the method of substitution or the method of elimination to solve this new system of two linear equations.

If we wanted to use the method of substitution we could easily solve the second equation for \(z\) (you do see why it would be easiest to solve the second equation for \(z\) right?) and substitute that into the first equation. This would allow us to find \(x\) and we could then find both \(z\) and \(y\).

However, to make the point that often we use both methods in solving systems of three linear equations let’s use the method of elimination to solve the system of two equations. We’ll just need to multiply the first equation by 3 and the second by 5. Doing this gives,

\[\begin{align*} -14x+5z & =-13 & \underrightarrow{\times \,\,3}\hspace{0.5in} & -42x+15z=-39 \\ 6x-3z & =9 & \underrightarrow{\times \,\,5}\hspace{0.5in} & \underline{\hspace{0.25in}30x-15z=45} \\ & & & \hspace{0.5in} -12x=6 \\ \end{align*}\]

We can now easily solve for \(x\) to get \(x = - \frac{1}{2}\). The coefficients on the second equation are smaller so let’s plug this into that equation and solve for \(z\). Here is that work.

\[\begin{align*}6\left( { - \frac{1}{2}} \right) - 3z & = 9\\ - 3 - 3z & = 9\\ - 3z & = 12\\ z & = - 4\end{align*}\]

Finally, we need to determine the value of \(y\). This is very easy to do. Recall in the first step we used substitution and in that step we used the following equation.

\[y = 4x + 5\]

Since we know the value of \(x\) all we need to do is plug that into this equation and get the value of \(y\).

\[y = 4\left( { - \frac{1}{2}} \right) + 5 = 3\]

Note that in many cases where we used substitution on the very first step the equation you’ll have at this step will contain both \(x\)’s and \(z\)’s and so you will need both values to get the third variable.

Okay, to finish this example up here is the solution : \(x = - \frac{1}{2}\), \(y = 3\) and \(z = - 4\).