# How to Solve a System of Linear Equations

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

 x1 + 2x2 = 16 x1 + x2 = 9

## Substitution Method

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 x1 + x2 = 9 and solve it for x2 = 9 - x1.

Substitute that into the first equation to get x1 + 2(9-x1) = 16.
Simplify both sides to get x1 + 18 - 2x1 = 16 or -x1 + 18 = 16.
Solve for x1 to get -x1 = -2 or x1 = 2.

Now that you know x1 = 2, we go back to either equation to find x2.

x1 + x2 = 9
2 + x2 = 9
x2 = 7

We have the solution x1 = 2, x2 = 7.

## Elimination Method

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 x2, multiply the second equation by -2.

 x1 + 2x2 = 16 stays x1 + 2x2 = 16 x1 + x2 = 9 becomes -2x1 - 2x2 = -18

Add the resulting equations together to get -x1 = -2 or x1 = 2.

To eliminate the x1, multiply the second equation by -1.

 x1 + 2x2 = 16 stays x1 + 2x2 = 16 x1 + x2 = 9 becomes -x1 - x2 = -9

Add the resulting equations together to get x2 = 7.

We have the solution x1 = 2, x2 = 7.

## Gauss-Jordan Elimination

Convert the system of linear equations into an augmented matrix.

 x1 + 2x2 = 16 x1 + x2 = 9

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

x1 x2 rhs 1 2 16 1 1 9

Perform the row operation R2-R1 R2 to clear out the 1 in the lower left.

x1 x2 rhs 1 2 16 0 -1 -7

Perform the row operation -R2 R2 to change the -1 into a positive 1.

x1 x2 rhs 1 2 16 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 R1 - 2R2 R1 to clear out the 2 in R1C2.

x1 x2 rhs 1 0 2 0 1 7

Convert the matrix back into a system of equations to get the solution x1 = 2, x2 = 7.

## Using RREF on Calculator (the fast way)

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.

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

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

## Matrix Inverses

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 = x1 B = 16 1 1 x2 9

The solution to the equation is X = A-1B.

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 = x1 = A-1B = -1 2 16 = 2 x2 1 -1 9 7

We have the solution x1 = 2, x2 = 7.