###### Example 6.71

Use elimination to solve the system of equations.

We first eliminate \(c\) from the system by combining the equations in pairs. We can add \(-1\) times Equation (2) to Equation (1) to get a new equation in two variables:

\(3a\) | \(+\) | \(2b\) | \(+\) | \(c\) | \(=\) | \(-1\) | \(\hphantom{blank}\) | \((1)\) |

\(-a\) | \(+\) | \(2b\) | \(-\) | \(c\) | \(=\) | \(3\) | \(\hphantom{blank}\) | \(-1\text{ times Equation }(2)\) |

\(2a\) | \(+\) | \(4b\) | \(\) | \(\) | \(=\) | \(2\) | \(\hphantom{blank}\) | \((4)\) |

Next, we add \(-1\) times Equation (2) to Equation (3) to get a second equation in two variables:

\(2a\) | \(+\) | \(3b\) | \(+\) | \(c\) | \(=\) | \(-4\) | \(\hphantom{blank}\) | \((3)\) |

\(-a\) | \(+\) | \(2b\) | \(-\) | \(c\) | \(=\) | \(3\) | \(\hphantom{blank}\) | \(-1\times (2)\) |

\(a\) | \(+\) | \(5b\) | \(\) | \(\) | \(=\) | \(7\) | \(\hphantom{blank}\) | \((5)\) |

By combining Equations (4) and (5), we have a \(2 \times 2\) linear system, which we can solve as usual.

To eliminate \(a\text{,}\) we add \(-2\) times Equation (5) to Equation (4):

\(2a\) | \(+\) | \(4b\) | \(=\) | \(2\) | \(\hphantom{blank}\) | \((4)\) |

\(-2a\) | \(-\) | \(10b\) | \(=\) | \(-14\) | \(\) | \(-2 \times(5)\) |

\(\) | \(\) | \(-6b\) | \(=\) | \(-12\) | \(\) | \(\) |

Solving this last equation gives us \(b = 2\text{.}\) Then we substitute \(b = 2\) into either of Equations (4) or (5) to find \(a = -3\text{.}\) Finally, we substitute both values into one of the three original equations to find \(c = 4\text{.}\) The solution of the system is \(a = -3\text{,}\) \(b = 2\text{,}\) \(c = 4\text{.}\)