In school, you learned how to solve simple linear systems – systems like

4x + 3y = 7
2x - 4y = 3

for example. The methods you used – substitution, elimination, etc. – can be generalized to solve more complicated systems with any number of equations and any number of variables. In this learning object, you're going to see how to solve more complicated linear systems, and solve them efficiently. Specifically, you will look at

  1. How to use augmented matrices and elementary row operations to simplify linear systems and what form a simplified system should take,
  2. How to solve a simplified system when its augmented matrix is in reduced row-echelon form,
  3. How to solve a simplified system when its augmented matrix is in row-echelon form.
A linear equation is one like

2x1 – 3x2 + 5x3 + 7x4 – 2x5 = 11

Here, the variables are x1, x2, x3, x4 and x5, and on the left, you have sums and differences of (constant)x(variable) terms – no powers or products or other complications.

Here are some equations that are not linear.

(x1)2 + 1/x2 – x3 = 5

sin(x1) + (x2 – x3)/x4 = 7x5

Linear equations generally come in linear systems - several equations in the same variables - for example, here's a linear system with four equations in five variables.
2x1 – 3x2 + 5x3 + 7x4 – 2x5
 = 11
3x1 –  x2 + 5x3 – 2x4  +  x5
= –3
x1 – 2x2 + 4x3 – 6x4  –  x5
= 7
–4x1 – 3x2 + 4x3 + 9x4 + 5x5
= 13
Prerequisites. To complete the process of solving linear systems, you should know how to row reduce a matrix. To learn how to do this, see the learning object How to Row Reduce a Matrix.
Keywords: linear system, solution to a linear system, augmented matrix, elementary row operations, reduced row-echelon form, row-echelon form, leading variables, back substitution.