We can choose any one of the three variables to eliminate first. For this example, we will eliminate \(x\text{.}\) Next, we choose two of the equations, say (1) and (2), and use a linear combination: We multiply Equation (1) by \(-2\) and add the result to Equation (2) to produce Equation (4).

\begin{alignat*}{4}
-2x \amp {}-{} \amp 4y \amp {}+{} \amp 6z \amp {}={} \amp 8\amp\hphantom{blankblank} (1\text{a})\\
\underline{\hphantom{-}2x\vphantom{y}} \amp \underline{{}-{}\vphantom{y}} \amp \underline{\hphantom{-5}y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{\hphantom{6}z\vphantom{y}} \amp {}={} \amp \underline{\hphantom{1}\vphantom{y} 3}\amp\hphantom{blankblank} (2)\\
\amp \amp{-5y} \amp {}+{} \amp 7z \amp {}={} \amp 11\amp\hphantom{blankblank} (4)
\end{alignat*}

Now we have an equation involving only two variables. But we need *two* equations in two unknowns to find the solution. So we choose a different pair of equations, say (1) and (3), and eliminate \(x\) again. We multiply Equation (1) by \(-3\) and add the result to Equation (3) to obtain Equation (5).

\begin{alignat*}{4}
-3x \amp {}-{} \amp 6y \amp {}+{} \amp 9z \amp {}={} \amp 12\amp\hphantom{blankblank} (1\text{b})\\
\underline{\hphantom{-}3x\vphantom{y}} \amp \underline{{}+{}\vphantom{y}} \amp \underline{\hphantom{-}2y} \amp \underline{{}+{}\vphantom{y} }\amp \underline{\hphantom{10}z\vphantom{y}} \amp {}={} \amp \underline{\vphantom{y}10}\amp\hphantom{blankblank} (3)\\
\amp \amp{-4y} \amp {}+{} \amp 10z \amp {}={} \amp 22\amp\hphantom{blankblank} (5)
\end{alignat*}

We now form a \(2\times 2\) system with our new Equations (4) and (5).

\begin{alignat*}{2}
-5y\amp + \amp 7z \amp = 11 \hphantom{blankblank} (4)\\
-4y\amp + \amp 10z \amp = 22 \hphantom{blankblank} (5)
\end{alignat*}

Finally, we eliminate one of the remaining variables to obtain an equation in a single variable. We choose to eliminate \(y\text{,}\) so we add 4 times Equation (4) to \(-5\) times Equation (5) to obtain Equation (6).

\begin{alignat*}{2}
-20y~\amp {}+{} \amp 28z \amp =\hphantom{-1} 44 \hphantom{blankblank} (4\text{a})\\
\underline{\hphantom{-}20y~}\amp\underline{{}-{}\vphantom{{}+ y~}} \amp \underline{\hphantom{-}50z\vphantom{ -20y~}} \amp = \underline{-110\vphantom{y}} \hphantom{blankblank} (5\text{a})\\
\amp \amp -22z \amp = \hphantom{1}{-66} \hphantom{blankblankl} (6)
\end{alignat*}

Now we are ready to form a triangular system. We choose one of the original equations (in three variables), one of the equations from our \(2\times 2\) system, and our final equation in one variable. We choose Equations (1), (4), and (6).

\begin{alignat*}{4}
x \amp {}+{} \amp 2y \amp {}-{} \amp 3z \amp {}={} \amp -4\amp\hphantom{blankblank} (1)\\
\amp {}{} \amp -5 y \amp {}+{} \amp 7z \amp {}={} \amp 11\amp\hphantom{blankblank} (4)\\
\amp \amp \amp {}{} \amp -22z \amp {}={} \amp -66\amp\hphantom{blankblank} (6)
\end{alignat*}

This new system has the same solutions as the original system, and we can solveit by back-substitution. We first solve Equation (6) to find \(z = 3\text{.}\) Substituting 3 for \(z\) in Equation (4), we find

\begin{align*}
-5y+7(\alert{3}) \amp=11 \\
-5y+21 \amp=11 \\
-5y \amp=-10
\end{align*}

So \(y = 2\text{.}\) Finally, we substitute \(\alert{3}\) for \(z\) and \(\alert{2}\) for \(y\) into Equation (1) to find

\begin{align*}
x+2(\alert{2})-3(\alert{3}) \amp= -4 \\
x+4-9 \amp= -4 \\
x \amp= 1
\end{align*}

The solution to the system is the ordered triple \((1, 2, 3)\text{.}\) You should verify that this triple satisfies all three of the original equations.