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Linear Combination: Definition & Examples

Instructor: Kimberly Hopkins
Linear combination is a method that is used to solve a system of linear equations. This lesson outlines the three basic ways that linear combination can be used to solve problems.

Linear Combination Definition

Have you ever wished that you could make a variable disappear? If so, then linear combination may be for you. Linear combination is the process of adding two algebraic equations so that one of the variables is eliminated.

Addition or subtraction can be used to perform a linear combination. Addition is used when the two equations have terms that are exact opposites, and subtraction is used when the two equations have terms that are the same.

The Three Basic Options

There are three basic options for linear combination.

The first option is the easiest. For this option, one of the variables already has coefficients that will cancel when added or subtracted. The first option allows for immediate linear combination.

The second option requires multiplying one of the equations by a constant in order to create a term that can be eliminated using linear combination.

The third option requires multiplying both of the equations by constants in order to combine the equations and eliminate one of the variables.

Now we are going to look at examples of each of these options to get a clearer idea of how they work.

The First Option

The first option is already in a format that can be cancelled. An example is the system:

2x + 2y = 10

3x - 2y = 20

In order to solve this system, we must first look to see if any of the coefficients and variables are exactly the same or exact opposites. 2y and -2y are exact opposites.

We add the equations by combining our like terms.

2x + 3x = 5x

2y + (-2)y = 0y

10 + 20 = 30.

Our new combined equation is 5x = 30.

Next, we solve for x by dividing both sides of the equation by 5. Our calculations reveal that x = 6.

Lastly, we can substitute 6 for x in either of the original equations. Let's substitute 6 in to the first equation.

2(6) + 2y= 10 becomes 12 + 2y = 10

Subtract 12 from both sides of the equation.

2y = -2

Divide both sides of the equation by 2.

y = -1

We have now solved the system, and the solution is (6, -1).

The Second Option

The second option requires changing one of the equations. Let's solve the system:

2x + 3y = 16

3x + 6y = 30

First, look to see if any of the terms are exactly the same or exact opposites. None of the terms fit that criteria.

Next, look to see if any of the terms are multiples of each other. Notice that 6y is a multiple of 3y.

We multiply each term in the first equation by 2.

4x + 6y = 32

Now we have terms that are exactly the same.

We use subtraction to combine the new equation with the second equation.

4x - 3x = x

6y - (6)y = 0y

32 - 30 = 2

Our new combined equation is x = 2.

Lastly, we can substitute 2 for x in either of the original equations. Let's substitute 2 in to the first equation.

2(2) + 3y= 16

4 + 3y = 16

Subtract 4 from both sides of the equation.

3y = 12

Divide both sides of the equation by 3.

y = 4

We have now solved the system, and the solution is (2, 4).

The Third Option

The third option requires changing both equations. Let's look at the system:

3x + 5y = 22

7x - 2y = 24

First, look to see if any of the coefficients and variables are exactly the same or exact opposites. None of the coefficients and variables fit that criteria.

Second, look to see if any of the terms are multiples of each other. None of the terms meet that criteria.

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