There are many possible ways to proceed in the solution process to this problem. All will give the same solution and all involve eliminating one of the variables and getting down to a system of two equations in two unknowns which you can then solve.

#### Method 1

The first solution method involves solving one of the three original equations for one of the variables. Substitute this into the other two equations. This will yield two equations in two unknowns that can be solve fairly quickly.

For this problem we’ll solve the second equation for \(x\) to get.

\[x = - z - 3y - 1\]

Plugging this into the first and third equation gives the following system of two equations.

\[\begin{align*}2\left( { - z - 3y - 1} \right) - y - 2z & = - 3\\ 5\left( { - z - 3y - 1} \right) - 4y + 3z & = 10\end{align*}\]

Or, upon simplification

\[\begin{align*} - 7y - 4z & = - 1\\ - 19y - 2z &= 15\end{align*}\]

Multiply the second equation by -2 and add.

\[\begin{align*} - 7y - 4z & = - 1\\ 38y + 4z & = - 30\\ \hline 31y & = - 31\end{align*}\]

From this we see that \(y = - 1\). Plugging this into either of the above two equations yields \(z = 2\). Finally, plugging both of these answers into \[x = - z - 3y - 1\] yields \(x = 0\).

#### Method 2

In the second method we add multiples of two equations together in such a way to eliminate one of the variables. We’ll do it using two different sets of equations eliminating the same variable in both. This will give a system of two equations in two unknowns which we can solve.

So, we’ll start by noticing that if we multiply the second equation by -2 and add it to the first equation we get.

\[\begin{align*}2x - y - 2z & = - 3\\ - 2x - 6y - 2z & = 2\\ \hline - 7y - 4z & = - 1\end{align*}\]

Next multiply the second equation by -5 and add it to the third equation. This gives

\[\begin{align*} - 5x - 15y - 5z & = 5\\ 5x - 4y + 3z & = 10\\ \hline - 19y - 2z & = 15\end{align*}\]

This gives the following system of two equations.

\[\begin{align*} - 7y - 4z & = - 1\\ - 19y - 2z & = 15\end{align*}\]

We can now solve this by multiplying the second by -2 and adding

\[\begin{align*} - 7y - 4z & = - 1\\ 38y + 4z & = - 30\\ \hline 31y & = - 31\end{align*}\]

From this we get that \(y = - 1\), the same as the first solution method. Plug this into either of the two equations involving only \(y\) and \(z\) and we’ll get that \(z = 2\). Finally plug these into any of the original three equations and we’ll get \(x = 0\).

You can use either of the two solution methods. In this case both methods involved the same basic level of work. In other cases, on may be significantly easier than the other. You’ll need to evaluate each system as you get it to determine which method will work the best.

#### Interpretation

Recall that one interpretation of the solution to a system of equations is that the solution(s) are the location(s) where the curves or surfaces (as in this case) intersect. So, the three equations in this system are the equations of planes in 3D space (you’ll learn this in Calculus II if you don’t already know this). So, from our solution we know that the three planes will intersect at the point (0,-1,2). Below is a graph of the three planes.

In this graph the red plane is the graph of \(2x - y - 2z = - 3\), the green plane is the graph of \(x + 3y + z = 1\) and the blue plane is the graph of \[5x - 4y + 3z = 10\]. You can see from this figure that the three planes do appear to intersect at a single point. It is a somewhat hard to see what the exact coordinates of this point. However, if we could zoom in and get a better graph we would see that the coordinates of this point are \((0, - 1,2)\).