Solve the following system of equations to find out where the intersection point is.

x_{1} |
+ | 2x_{2} |
= | 16 |

x_{1} |
+ | x_{2} |
= | 9 |

Take one of the equations and solve for one of the variables. Then substitute that into the other equation.

We'll take the second equation x_{1} + x_{2} = 9 and solve it for x_{2} = 9 - x_{1}.

Substitute that into the first equation to get x_{1} + 2(9-x_{1}) = 16.

Simplify both sides to get x_{1} + 18 - 2x_{1} = 16 or -x_{1} + 18 = 16.

Solve for x_{1} to get -x_{1} = -2 or x_{1} = 2.

Now that you know x_{1} = 2, we go back to either equation to find x_{2}.

x_{1} + x_{2} = 9

2 + x_{2} = 9

x_{2} = 7

We have the solution x_{1} = 2, x_{2} = 7.

In the addition or elimination method, you multiply one or both equations by a constant and then add the results together to eliminate one of the variables.

To eliminate the x_{2}, multiply the second equation by -2.

x_{1} |
+ | 2x_{2} |
= | 16 | stays | x_{1} |
+ | 2x_{2} |
= | 16 | ||

x_{1} |
+ | x_{2} |
= | 9 | becomes | -2x_{1} |
- | 2x_{2} |
= | -18 |

Add the resulting equations together to get -x_{1} = -2 or x_{1} = 2.

To eliminate the x_{1}, multiply the second equation by -1.

x_{1} |
+ | 2x_{2} |
= | 16 | stays | x_{1} |
+ | 2x_{2} |
= | 16 | ||

x_{1} |
+ | x_{2} |
= | 9 | becomes | -x_{1} |
- | x_{2} |
= | -9 |

Add the resulting equations together to get x_{2} = 7.

We have the solution x_{1} = 2, x_{2} = 7.

Convert the system of linear equations into an augmented matrix.

x_{1} |
+ | 2x_{2} |
= | 16 |

x_{1} |
+ | x_{2} |
= | 9 |

Each equation becomes a row and each variable becomes a column. The resulting matrix is a 2×3 matrix.

x_{1} |
x_{2} |
rhs | |||
---|---|---|---|---|---|

R_{1} |
1 | 2 | 16 | ||

R_{2} |
1 | 1 | 9 |

Perform the row operation R_{2}-R_{1} → R_{2} to clear out the 1 in the lower left.

x_{1} |
x_{2} |
rhs | |||
---|---|---|---|---|---|

R_{1} |
1 | 2 | 16 | ||

R_{2} |
0 | -1 | -7 |

Perform the row operation -R_{2} → R_{2} to change the -1 into a positive 1.

x_{1} |
x_{2} |
rhs | |||
---|---|---|---|---|---|

R_{1} |
1 | 2 | 16 | ||

R_{2} |
0 | 1 | 7 |

If we were to stop at this point, we would be performing Guassian elimination with back substitution. We might as well go ahead and completely reduce the matrix using Gauss-Jordan elimination and place the matrix into Reduced Row Echelon Form. There is no back substitution needed in this form.

Perform the row operation R_{1} - 2R_{2} → R_{1} to clear out the 2 in R_{1}C_{2}.

x_{1} |
x_{2} |
rhs | |||
---|---|---|---|---|---|

R_{1} |
1 | 0 | 2 | ||

R_{2} |
0 | 1 | 7 |

Convert the matrix back into a system of equations to get the solution x_{1} = 2, x_{2} = 7.

The TI83 and TI84 calculators have a RREF command built into them. RREF stands for Reduced Row Echelon Form and it does the entire Gauss-Jordan elimination in one step.

- Convert the system of linear equations into a matrix and enter that matrix into the calculator.
- Go to Matrix / Math / RREF(
- Select Matrix [A] or whichever matrix you used
- Close the parentheses and hit enter.

Watch the Flash video: How to row reduce a matrix using the calculator [564 KB].

When there are the same number of variables as equations, a system of linear equations can be written in matrix form as **AX** = **B**, where **A** is the coefficient matrix, **B** is the constant matrix (right hand side), and **X** is the variable matrix.

A |
= | 1 | 2 | X |
= | x_{1} |
B |
= | 16 | ||||||||

1 | 1 | x_{2} |
9 |

The solution to the equation is **X** = **A**^{-1}**B**.

**A**^{-1} can be found using the calculator.

A^{-1} |
= | -1 | 2 | ||

1 | -1 |

Then, multiplying **A**^{-1} by **B** we find the solution to the system.

X |
= | x_{1} |
= | A^{-1}B |
= | -1 | 2 | 16 | = | 2 | |||||||||

x_{2} |
1 | -1 | 9 | 7 |

We have the solution x_{1} = 2, x_{2} = 7.