How to Solve With Elimination – We have solved systems of linear equations by graphing and by substitution. Graphing works well when the variable coefficients are small and the solution has integer values. Substitution works well when we can easily solve one equation for one of the variables and not have too many fractions in the resulting expression.

The third method of solving systems of linear equations is called the Elimination Method. When we solved a system by substitution, we started with two equations and two variables and reduced it to one equation with one variable. This is what we’ll do with the elimination method, too, but we’ll have a different way to get there.

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# Solve a System of Equations by Elimination

The **Elimination Method** is based on the Addition Property of Equality. The Addition Property of Equality says that when you add the same quantity to both sides of an equation, you still have equality. We will extend the Addition Property of Equality to say that when you add equal quantities to both sides of an equation, the results are equal.

For any expressions *a*, *b*, *c*, and *d*, if and then a=bc=da+c=b+d(5.3.1)(5.3.1) if a=b and c=d then a+c=b+d

To solve a system of equations by elimination, we start with both equations in standard form. Then we decide which variable will be easiest to eliminate. How do we decide? We want to have the coefficients of one variable be opposites, so that we can add the equations together and eliminate that variable.

Notice how that works when we add these two equations together:3x+y=52x−y=0–––––––––––5x=5(5.3.2)(5.3.2)3x+y=52x−y=0_5x=5

The *y*’s add to zero and we have one equation with one variable.

Let’s try another one:{x+4y=22x+5y=−2(5.3.3)(5.3.3){x+4y=22x+5y=−2

This time we don’t see a variable that can be immediately eliminated if we add the equations.

But if we multiply the first equation by −2, we will make the coefficients of *x* opposites. We must multiply every term on both sides of the equation by −2.

Now we see that the coefficients of the *x* terms are opposites, so *x* will be eliminated when we add these two equations.

Add the equations yourself—the result should be −3*y* = −6. And that looks easy to solve, doesn’t it? Here is what it would look like.

We’ll do one more:{4x−3y=103x+5y=−7(5.3.4)(5.3.4){4x−3y=103x+5y=−7

It doesn’t appear that we can get the coefficients of one variable to be opposites by multiplying one of the equations by a constant, unless we use fractions. So instead, we’ll have to multiply both equations by a constant.

We can make the coefficients of *x* be opposites if we multiply the first equation by 3 and the second by −4, so we get 12*x* and −12*x*.

This gives us these two new equations:{12x−9y−12x−20y=30=28{12x−9y=30−12x−20y=28

When we add these equations,\[{12x−9y=30−12x−20y=28–––––––––––––––––−29y=58(5.3.5)(5.3.5)\[{12x−9y=30−12x−20y=28_−29y=58\]

the *x*’s are eliminated and we just have −29*y* = 58.

Once we get an equation with just one variable, we solve it. Then we substitute that value into one of the original equations to solve for the remaining variable. And, as always, we check our answer to make sure it is a solution to both of the original equations.

Now we’ll see how to use elimination to solve the same system of equations we solved by graphing and by substitution.

Exercise 5.3.15.3.1: How to Solve a System of Equations by Elimination

Solve the system by elimination. {2x+y=7x−2y=6{2x+y=7x−2y=6**Answer**

Exercise 5.3.25.3.2

Solve the system by elimination. {3x+y=52x−3y=7{3x+y=52x−3y=7**Answer**

Exercise 5.3.35.3.3

Solve the system by elimination. {4x+y=−5−2x−2y=−2{4x+y=−5−2x−2y=−2**Answer**

The steps are listed below for easy reference.

HOW TO SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION.

- Write both equations in standard form. If any coefficients are fractions, clear them.
- Make the coefficients of one variable opposites.
- Decide which variable you will eliminate.
- Multiply one or both equations so that the coefficients of that variable are opposites.

- Add the equations resulting from Step 2 to eliminate one variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Write the solution as an ordered pair.
- Check that the ordered pair is a solution to
**both**original equations.

First we’ll do an example where we can eliminate one variable right away.

Exercise 5.3.45.3.4

Solve the system by elimination. {x+y=10x−y=12{x+y=10x−y=12**Answer**

Exercise 5.3.55.3.5

Solve the system by elimination. {2x+y=5x−y=4{2x+y=5x−y=4**Answer**

Exercise 5.3.65.3.6

Solve the system by elimination.{x+y=3−2x−y=−1{x+y=3−2x−y=−1**Answer**

In Exercise 5.3.75.3.7, we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Exercise 5.3.75.3.7

Solve the system by elimination. {3x−2y=−25x−6y=10{3x−2y=−25x−6y=10**Answer**

Exercise 5.3.85.3.8

Solve the system by elimination.{4x−3y=15x−9y=−4{4x−3y=15x−9y=−4**Answer**

Exercise 5.3.95.3.9

Solve the system by elimination.{3x+2y=26x+5y=8{3x+2y=26x+5y=8**Answer**

Now we’ll do an example where we need to multiply both equations by constants in order to make the coefficients of one variable opposites.

Exercise 5.3.105.3.10

Solve the system by elimination. {4x−3y=97x+2y=−6{4x−3y=97x+2y=−6**Answer**

Exercise 5.3.115.3.11

Solve the system by elimination. {3x−4y=−95x+3y=14{3x−4y=−95x+3y=14**Answer**

Exercise 5.3.125.3.12

Solve the system by elimination. {7x+8y=43x−5y=27{7x+8y=43x−5y=27**Answer**

When the system of equations contains fractions, we will first clear the fractions by multiplying each equation by its LCD.

Exercise 5.3.135.3.13

Solve the system by elimination. {x+12y=632x+23y=172{x+12y=632x+23y=172**Answer**

Exercise 5.3.145.3.14

Solve the system by elimination. {13x−12y=134x−y=52{13x−12y=134x−y=52**Answer**

Exercise 5.3.155.3.15

Solve the system by elimination. {x+35y=−15−12x−23y=56{x+35y=−15−12x−23y=56**Answer**

In the Solving Systems of Equations by Graphing we saw that not all systems of linear equations have a single ordered pair as a solution. When the two equations were really the same line, there were infinitely many solutions. We called that a consistent system. When the two equations described parallel lines, there was no solution. We called that an inconsistent system.

Exercise 5.3.165.3.16

Solve the system by elimination.{3x+4y=12y=3−34x{3x+4y=12y=3−34x**Answer**

Exercise 5.3.175.3.17

Solve the system by elimination. {5x−3y=15y=−5+53x{5x−3y=15y=−5+53x**Answer**

Exercise 5.3.185.3.18

Solve the system by elimination. {x+2y=6y=−12x+3{x+2y=6y=−12x+3**Answer**

Exercise 5.3.195.3.19

Solve the system by elimination. {−6x+15y=102x−5y=−5{−6x+15y=102x−5y=−5**Answer**

Exercise 5.3.205.3.20

Solve the system by elimination. {−3x+2y=89x−6y=13{−3x+2y=89x−6y=13**Answer**

Exercise 5.3.215.3.21

Solve the system by elimination. {7x−3y=−2−14x+6y=8{7x−3y=−2−14x+6y=8**Answer**

# Solve Applications of Systems of Equations by Elimination

Some applications problems translate directly into equations in standard form, so we will use the elimination method to solve them. As before, we use our Problem Solving Strategy to help us stay focused and organized.

Exercise 5.3.225.3.22

The sum of two numbers is 39. Their difference is 9. Find the numbers.**Answer**

Exercise 5.3.235.3.23

The sum of two numbers is 42. Their difference is 8. Find the numbers.**Answer**

Exercise 5.3.245.3.24

The sum of two numbers is −15. Their difference is −35. Find the numbers.**Answer**

Exercise 5.3.255.3.25

Joe stops at a burger restaurant every day on his way to work. Monday he had one order of medium fries and two small sodas, which had a total of 620 calories. Tuesday he had two orders of medium fries and one small soda, for a total of 820 calories. How many calories are there in one order of medium fries? How many calories in one small soda?**Answer**

Exercise 5.3.265.3.26

Malik stops at the grocery store to buy a bag of diapers and 2 cans of formula. He spends a total of $37. The next week he stops and buys 2 bags of diapers and 5 cans of formula for a total of $87. How much does a bag of diapers cost? How much is one can of formula?**Answer**

Exercise 5.3.275.3.27

To get her daily intake of fruit for the day, Sasha eats a banana and 8 strawberries on Wednesday for a calorie count of 145. On the following Wednesday, she eats two bananas and 5 strawberries for a total of 235 calories for the fruit. How many calories are there in a banana? How many calories are in a strawberry?**Answer**

# Choose the Most Convenient Method to Solve a System of Linear Equations

When you will have to solve a system of linear equations in a later math class, you will usually not be told which method to use. You will need to make that decision yourself. So you’ll want to choose the method that is easiest to do and minimizes your chance of making mistakes.

Exercise 5.3.285.3.28

For each system of linear equations decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

- {3x+8y=407x−4y=−32{3x+8y=407x−4y=−32
- {5x+6y=12y=23x−1{5x+6y=12y=23x−1

**Answer**

Exercise 5.3.295.3.29

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

- {4x−5y=−323x+2y=−1{4x−5y=−323x+2y=−1
- {x=2y−13x−5y=−7{x=2y−13x−5y=−7

**Answer**

Exercise 5.3.305.3.30

For each system of linear equations, decide whether it would be more convenient to solve it by substitution or elimination. Explain your answer.

- {y=2x−13x−4y=−6{y=2x−13x−4y=−6
- {6x−2y=123x+7y=−13{6x−2y=123x+7y=−13

**Answer**

Note

Access these online resources for additional instruction and practice with solving systems of linear equations by elimination.

- Instructional Video-Solving Systems of Equations by Elimination
- Instructional Video-Solving by Elimination
- Instructional Video-Solving Systems by Elimination

# Key Concepts

**To Solve a System of Equations by Elimination**- Write both equations in standard form. If any coefficients are fractions, clear them.
- Make the coefficients of one variable opposites.
- Decide which variable you will eliminate.
- Multiply one or both equations so that the coefficients of that variable are opposites.

- Add the equations resulting from Step 2 to eliminate one variable.
- Solve for the remaining variable.
- Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
- Write the solution as an ordered pair.
- Check that the ordered pair is a solution to
**both**original equations.

## Conclusion

When the coefficients of one variable are opposites you add the equations to eliminate a variable and when the coefficients of one variable are equal you subtract the equations to eliminate a variable.

DD