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## Introduction

Your first instinct when dealing with a system of equations should be to see if you can guess some likely values for each variable.  Some systems are simple enough that they can be solved by trying a few guesses until you find one that works.  When guess and check doesn’t work, however, it is time to try a more involved method of solving the system.  The two common methods for solving a system of equations are the substitution method and the elimination method.  The difference in these two methods is detailed below.

This elimination method presents an unusual concept… adding one entire equation to another entire equation to form a brand new (third) equation.  If this process is used to eliminate one of the variables, then it gives you an easy way to find the value of the remaining variable.

## Lesson

The introduction to this lesson presented a contrast between problems that should be done using the substitution method and those that should be done using the elimination method.

Substitution is the best method if you are able to isolate one of the variables in one of the equations.  If isolating the variable is difficult to do or results in some terms reducing to fractions, then elimination is a good option to solve the problem.

Elimination means just that… combine the equations to eliminate one of the variables, then solve for the remaining variable.  Example 1 shows how to do the elimination problem shown in the introduction to this lesson.

Example 1: Find the solution to the system

Example 1 was set up perfectly since the 5h and -5h had opposite signs and cancelled out when added.  This same method can be used if both equations contain the same positive term.  Simply subtract one of the equations from the other.

Example 2: Find the solution to the system

It may be tempting to try to do the subtraction in your head instead of doing the step that shows each individual negative term of the second equation here.  Some students are able to pull off doing more work in their head, but when learning the material it is important to show each step and understand it.  Remember that you the more you try to do in your head, the more likely you are to make an error.

Example 1 and 2 both contained one variable that was easy to cancel out.  For some problems, you will need to manipulate one or both equations in order to eliminate one of the terms.

Example 3: Find the solution to the system

To do example 3, take a close look at each separate variable and decide if either of them have a least common multiple that can be arrived at (with the least amount of work.)  The least common multiple of 2s and 4s is 4s, while the least common multiple of 5t and 2t is 10t.  The work shown above is done by eliminating the variable s, but you could also change both equations to eliminate the variable t.  However, eliminating the t’s is more work since you would have to multiply both of the equations by a constant in order to end up with 10t.  Some problems require that you adjust both equations in order to eliminate one of the variables.

## Try It

Find the solution to each system of equations.

1)

2)

3)

4)

1)  a = 10, b = 3

2)  c = 6, d = 0

3)  e = 2, f = 5

4)  g = 6, h = 3

Looking for a different lesson on systems of equations?  Try the links below.

Related Lessons

Before attempting to learn systems of equations, one should be comfortable solving a variety of (single) equations.

Equation Lessons

Looking for something else?  Explore our menu of general math or algebra lessons.