solving linear systems by elimination with fractions

Solving Systems Of Equations With Fractions Or Decimals

In these lessons, we help Algebra 1 students learn to solve systems of linear equations with fractions.

Related Pages Solving Systems of Equations or Simultaneous Equations using algebra Solving Simultaneous Equations Using Matrices More Lessons On Matrices Algebra Lessons

How To Solve Systems Of Equations With Fractions?

When a system includes an equation with fractions as coefficients: Step 1. Eliminate the fractions by multiplying each side of the equation by a common denominator. Step 2: Solve the resulting system using the addition method, elimination method, or the substitution method.

The following diagrams show how to solve systems of equations using the Substitution Method and the Elimination Method.

Systems of Equations with Fractions

Students learn to solve systems of linear equations that involve fractions. Students also learn to solve linear systems of equations by the method of their choice using the following rules: if one of the variables cancels out when the equations are added together, then use addition, and if a variable is already isolated in one of the equations, then use substitution.

Example: Solve the following systems of equations y = 1/3 x - 1 3x - 2y = 4

Solving a System Involving Fractions with Substitution

How to solve a system of 2 equations with 2 variables involving fractions using substitution?

Example: y = -3/4 x + 12 y = 1/3 x - 2

Solving a System Involving Fractions with Elimination

How to solve a linear system by the process of elimination where the equations contain fractions?

Example: x/2 + y/8 = 4 x/3 - y/2 = -2

Solving systems of linear equations involving decimals and fractions

• Scale the system so that integer coefficients exist.
• Use elimination to solve for the solutions, if they exist.

Examples: System 1 0.05x - 0.03y = 0.21 0.07x + 0.02y = 0.16

System 2 x/4 + y/6 = 1 x - y = 3

System 3 (x + 3)/4 + (y - 1)/3 = 1 2x - y = 1

Solving System Of Linear Equations by Adding (with decimals)

Examples: 3.5x + 2.5y = 17 -1.5x - 7.5y = -33

Solving Systems of Linear Equations with Fractions

Solve a system of linear equations using multiplication with addition or subtraction.

Examples: x/2 + y/3 = -4 x/5 + y/5 = -2

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5.3 Solve Systems of Equations by Elimination

Learning objectives.

By the end of this section, you will be able to:

• Solve a system of equations by elimination
• Solve applications of systems of equations by elimination
• Choose the most convenient method to solve a system of linear equations

Be Prepared 5.8

Before you get started, take this readiness quiz.

Simplify −5 ( 6 − 3 a ) −5 ( 6 − 3 a ) . If you missed this problem, review Example 1.136 .

Be Prepared 5.9

Solve the equation 1 3 x + 5 8 = 31 24 1 3 x + 5 8 = 31 24 . If you missed this problem, review Example 2.48 .

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.

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 ,

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:

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

Let’s try another one:

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:

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:

When we add these equations,

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.

Example 5.25

How to solve a system of equations by elimination.

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

Try It 5.49

Solve the system by elimination. { 3 x + y = 5 2 x − 3 y = 7 { 3 x + y = 5 2 x − 3 y = 7

Try It 5.50

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

The steps are listed below for easy reference.

How to solve a system of equations by elimination.

• Step 1. Write both equations in standard form. If any coefficients are fractions, clear them.
• Decide which variable you will eliminate.
• Multiply one or both equations so that the coefficients of that variable are opposites.
• Step 3. Add the equations resulting from Step 2 to eliminate one variable.
• Step 4. Solve for the remaining variable.
• Step 5. Substitute the solution from Step 4 into one of the original equations. Then solve for the other variable.
• Step 6. Write the solution as an ordered pair.
• Step 7. 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.

Example 5.26

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

Try It 5.51

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

Try It 5.52

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

In Example 5.27 , we will be able to make the coefficients of one variable opposites by multiplying one equation by a constant.

Example 5.27

Solve the system by elimination. { 3 x − 2 y = −2 5 x − 6 y = 10 { 3 x − 2 y = −2 5 x − 6 y = 10

Try It 5.53

Solve the system by elimination. { 4 x − 3 y = 1 5 x − 9 y = −4 { 4 x − 3 y = 1 5 x − 9 y = −4

Try It 5.54

Solve the system by elimination. { 3 x + 2 y = 2 6 x + 5 y = 8 { 3 x + 2 y = 2 6 x + 5 y = 8

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.

Example 5.28

Solve the system by elimination. { 4 x − 3 y = 9 7 x + 2 y = −6 { 4 x − 3 y = 9 7 x + 2 y = −6

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

Try It 5.55

Solve the system by elimination. { 3 x − 4 y = −9 5 x + 3 y = 14 { 3 x − 4 y = −9 5 x + 3 y = 14

Try It 5.56

Solve the system by elimination. { 7 x + 8 y = 4 3 x − 5 y = 27 { 7 x + 8 y = 4 3 x − 5 y = 27

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

Example 5.29

Solve the system by elimination. { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2 { x + 1 2 y = 6 3 2 x + 2 3 y = 17 2

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

Try It 5.57

Solve the system by elimination. { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2 { 1 3 x − 1 2 y = 1 3 4 x − y = 5 2

Try It 5.58

Solve the system by elimination. { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6 { x + 3 5 y = − 1 5 − 1 2 x − 2 3 y = 5 6

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.

Example 5.30

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

This is a true statement. The equations are consistent but dependent. Their graphs would be the same line. The system has infinitely many solutions.

After we cleared the fractions in the second equation, did you notice that the two equations were the same? That means we have coincident lines.

Try It 5.59

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

Try It 5.60

Solve the system by elimination. { x + 2 y = 6 y = − 1 2 x + 3 { x + 2 y = 6 y = − 1 2 x + 3

Example 5.31

Solve the system by elimination. { −6 x + 15 y = 10 2 x − 5 y = −5 { −6 x + 15 y = 10 2 x − 5 y = −5

This statement is false. The equations are inconsistent and so their graphs would be parallel lines.

The system does not have a solution.

Try It 5.61

Solve the system by elimination. { −3 x + 2 y = 8 9 x − 6 y = 13 { −3 x + 2 y = 8 9 x − 6 y = 13

Try It 5.62

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

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.

Example 5.32

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

Try It 5.63

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

Try It 5.64

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

Example 5.33

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?

Try It 5.65

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?

Try It 5.66

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?

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.

Example 5.34

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

ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

• ⓐ { 3 x + 8 y = 40 7 x − 4 y = −32 { 3 x + 8 y = 40 7 x − 4 y = −32 Since both equations are in standard form, using elimination will be most convenient.
• ⓑ { 5 x + 6 y = 12 y = 2 3 x − 1 { 5 x + 6 y = 12 y = 2 3 x − 1

>Since one equation is already solved for y , using substitution will be most convenient.

Try It 5.67

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

ⓐ { 4 x − 5 y = −32 3 x + 2 y = −1 { 4 x − 5 y = −32 3 x + 2 y = −1 ⓑ { x = 2 y − 1 3 x − 5 y = − 7 { x = 2 y − 1 3 x − 5 y = − 7

Try It 5.68

ⓐ { y = 2 x − 1 3 x − 4 y = − 6 { y = 2 x − 1 3 x − 4 y = − 6 ⓑ { 6 x − 2 y = 12 3 x + 7 y = −13 { 6 x − 2 y = 12 3 x + 7 y = −13

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

Section 5.3 Exercises

Practice makes perfect.

In the following exercises, solve the systems of equations by elimination.

{ 5 x + 2 y = 2 −3 x − y = 0 { 5 x + 2 y = 2 −3 x − y = 0

{ −3 x + y = −9 x − 2 y = −12 { −3 x + y = −9 x − 2 y = −12

{ 6 x − 5 y = −1 2 x + y = 13 { 6 x − 5 y = −1 2 x + y = 13

{ 3 x − y = −7 4 x + 2 y = −6 { 3 x − y = −7 4 x + 2 y = −6

{ x + y = −1 x − y = −5 { x + y = −1 x − y = −5

{ x + y = −8 x − y = −6 { x + y = −8 x − y = −6

{ 3 x − 2 y = 1 − x + 2 y = 9 { 3 x − 2 y = 1 − x + 2 y = 9

{ −7 x + 6 y = −10 x − 6 y = 22 { −7 x + 6 y = −10 x − 6 y = 22

{ 3 x + 2 y = −3 − x − 2 y = −19 { 3 x + 2 y = −3 − x − 2 y = −19

{ 5 x + 2 y = 1 −5 x − 4 y = −7 { 5 x + 2 y = 1 −5 x − 4 y = −7

{ 6 x + 4 y = −4 −6 x − 5 y = 8 { 6 x + 4 y = −4 −6 x − 5 y = 8

{ 3 x − 4 y = −11 x − 2 y = −5 { 3 x − 4 y = −11 x − 2 y = −5

{ 5 x − 7 y = 29 x + 3 y = −3 { 5 x − 7 y = 29 x + 3 y = −3

{ 6 x − 5 y = −75 − x − 2 y = −13 { 6 x − 5 y = −75 − x − 2 y = −13

{ − x + 4 y = 8 3 x + 5 y = 10 { − x + 4 y = 8 3 x + 5 y = 10

{ 2 x − 5 y = 7 3 x − y = 17 { 2 x − 5 y = 7 3 x − y = 17

{ 5 x − 3 y = −1 2 x − y = 2 { 5 x − 3 y = −1 2 x − y = 2

{ 7 x + y = −4 13 x + 3 y = 4 { 7 x + y = −4 13 x + 3 y = 4

{ −3 x + 5 y = −13 2 x + y = −26 { −3 x + 5 y = −13 2 x + y = −26

{ 3 x − 5 y = −9 5 x + 2 y = 16 { 3 x − 5 y = −9 5 x + 2 y = 16

{ 4 x − 3 y = 3 2 x + 5 y = −31 { 4 x − 3 y = 3 2 x + 5 y = −31

{ 4 x + 7 y = 14 −2 x + 3 y = 32 { 4 x + 7 y = 14 −2 x + 3 y = 32

{ 5 x + 2 y = 21 7 x − 4 y = 9 { 5 x + 2 y = 21 7 x − 4 y = 9

{ 3 x + 8 y = −3 2 x + 5 y = −3 { 3 x + 8 y = −3 2 x + 5 y = −3

{ 11 x + 9 y = −5 7 x + 5 y = −1 { 11 x + 9 y = −5 7 x + 5 y = −1

{ 3 x + 8 y = 67 5 x + 3 y = 60 { 3 x + 8 y = 67 5 x + 3 y = 60

{ 2 x + 9 y = −4 3 x + 13 y = −7 { 2 x + 9 y = −4 3 x + 13 y = −7

{ 1 3 x − y = −3 x + 5 2 y = 2 { 1 3 x − y = −3 x + 5 2 y = 2

{ x + 1 2 y = 3 2 1 5 x − 1 5 y = 3 { x + 1 2 y = 3 2 1 5 x − 1 5 y = 3

{ x + 1 3 y = −1 1 2 x − 1 3 y = −2 { x + 1 3 y = −1 1 2 x − 1 3 y = −2

{ 1 3 x − y = −3 2 3 x + 5 2 y = 3 { 1 3 x − y = −3 2 3 x + 5 2 y = 3

{ 2 x + y = 3 6 x + 3 y = 9 { 2 x + y = 3 6 x + 3 y = 9

{ x − 4 y = −1 −3 x + 12 y = 3 { x − 4 y = −1 −3 x + 12 y = 3

{ −3 x − y = 8 6 x + 2 y = −16 { −3 x − y = 8 6 x + 2 y = −16

{ 4 x + 3 y = 2 20 x + 15 y = 10 { 4 x + 3 y = 2 20 x + 15 y = 10

{ 3 x + 2 y = 6 −6 x − 4 y = −12 { 3 x + 2 y = 6 −6 x − 4 y = −12

{ 5 x − 8 y = 12 10 x − 16 y = 20 { 5 x − 8 y = 12 10 x − 16 y = 20

{ −11 x + 12 y = 60 −22 x + 24 y = 90 { −11 x + 12 y = 60 −22 x + 24 y = 90

{ 7 x − 9 y = 16 −21 x + 27 y = −24 { 7 x − 9 y = 16 −21 x + 27 y = −24

{ 5 x − 3 y = 15 y = 5 3 x − 2 { 5 x − 3 y = 15 y = 5 3 x − 2

{ 2 x + 4 y = 7 y = − 1 2 x − 4 { 2 x + 4 y = 7 y = − 1 2 x − 4

In the following exercises, translate to a system of equations and solve.

The sum of two numbers is 65. Their difference is 25. Find the numbers.

The sum of two numbers is 37. Their difference is 9. Find the numbers.

The sum of two numbers is −27. Their difference is −59. Find the numbers.

The sum of two numbers is −45. Their difference is −89. Find the numbers.

Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?

Peter is buying office supplies. He is able to buy 3 packages of paper and 4 staplers for $40 or he is able to buy 5 packages of paper and 6 staplers for $62. How much does a package of paper cost? How much does a stapler cost?

The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

The total number of calories in 2 hot dogs and 3 cups of cottage cheese is 960 calories. The total number of calories in 5 hot dogs and 2 cups of cottage cheese is 1190 calories. How many calories are in a hot dog? How many calories are in a cup of cottage cheese?

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

ⓐ { 8 x − 15 y = −32 6 x + 3 y = −5 { 8 x − 15 y = −32 6 x + 3 y = −5 ⓑ { x = 4 y − 3 4 x − 2 y = −6 { x = 4 y − 3 4 x − 2 y = −6

ⓐ { y = 7 x − 5 3 x − 2 y = 16 { y = 7 x − 5 3 x − 2 y = 16 ⓑ { 12 x − 5 y = −42 3 x + 7 y = −15 { 12 x − 5 y = −42 3 x + 7 y = −15

ⓐ { y = 4 x + 9 5 x − 2 y = −21 { y = 4 x + 9 5 x − 2 y = −21 ⓑ { 9 x − 4 y = 24 3 x + 5 y = −14 { 9 x − 4 y = 24 3 x + 5 y = −14

ⓐ { 14 x − 15 y = −30 7 x + 2 y = 10 { 14 x − 15 y = −30 7 x + 2 y = 10 ⓑ { x = 9 y − 11 2 x − 7 y = −27 { x = 9 y − 11 2 x − 7 y = −27

Everyday Math

Norris can row 3 miles upstream against the current in 1 hour, the same amount of time it takes him to row 5 miles downstream, with the current. Solve the system. { r − c = 3 r + c = 5 { r − c = 3 r + c = 5

• ⓐ for r r , his rowing speed in still water.
• ⓑ Then solve for c c , the speed of the river current.

Josie wants to make 10 pounds of trail mix using nuts and raisins, and she wants the total cost of the trail mix to be $54. Nuts cost $6 per pound and raisins cost $3 per pound. Solve the system { n + r = 10 6 n + 3 r = 54 { n + r = 10 6 n + 3 r = 54 to find n n , the number of pounds of nuts, and r r , the number of pounds of raisins she should use.

Writing Exercises

Solve the system { x + y = 10 5 x + 8 y = 56 { x + y = 10 5 x + 8 y = 56

ⓐ by substitution ⓑ by graphing ⓒ Which method do you prefer? Why?

Solve the system { x + y = −12 y = 4 − 1 2 x { x + y = −12 y = 4 − 1 2 x

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ What does this checklist tell you about your mastery of this section? What steps will you take to improve?

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Solving Linear Systems by Elimination

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4. Systems of Equations

4.3 Solve Systems of Equations by Elimination

Lynn Marecek and MaryAnne Anthony-Smith

Learning Objectives

By the end of this section it is expected that you will be able to:

• Solve a system of equations by elimination
• Solve applications of systems of equations by elimination
• Choose the most convenient method to solve a system of linear equations

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.

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 ,

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:

Let’s try another one:

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:

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:

When we add these equations,

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.

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.
• 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.

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

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.

In this example, we cannot multiply just one equation by any constant to get opposite coefficients. So we will strategically multiply both equations by a constant to get the opposites.

What other constants could we have chosen to eliminate one of the variables? Would the solution be the same?

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

In this example, both equations have fractions. Our first step will be to multiply each equation by its LCD to clear the fractions.

When we were solving systems of linear 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.

Solve the system by elimination:

no solution

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.

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

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

The numbers are 25 and 17.

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?

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?

The bag of diapers costs ?11 and the can of formula costs ?13.

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.

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

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

a) Since both equations are in standard form, using elimination will be most convenient.

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

4.3 exercise set.

In the following exercises, solve the systems of equations by elimination.

In the following exercises, translate to a system of equations and solve.

• The sum of two numbers is 65. Their difference is 25. Find the numbers .
• The sum of two numbers is −27. Their difference is −59. Find the numbers.
• Andrea is buying some new shirts and sweaters. She is able to buy 3 shirts and 2 sweaters for $114 or she is able to buy 2 shirts and 4 sweaters for $164. How much does a shirt cost? How much does a sweater cost?
• The total amount of sodium in 2 hot dogs and 3 cups of cottage cheese is 4720 mg. The total amount of sodium in 5 hot dogs and 2 cups of cottage cheese is 6300 mg. How much sodium is in a hot dog? How much sodium is in a cup of cottage cheese?

In the following exercises, decide whether it would be more convenient to solve the system of equations by substitution or elimination.

• infinitely many solutions
• inconsistent, no solution
• The numbers are 20 and 45.
• The numbers are 16 and −43.
• A shirt costs $16 and a sweater costs $33.
• There are 860 mg in a hot dog. There are 1,000 mg in a cup of cottage cheese.
• elimination
• substitution

Business/Technical Mathematics Copyright © 2021 by Lynn Marecek and MaryAnne Anthony-Smith is licensed under a Creative Commons Attribution 4.0 International License , except where otherwise noted.

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How to solve systems of equations by Elimination

Step by step tutorial for systems of linear equations (in 2 variables)

Video on Solving by Elimination

What is the Elimination Method?

It is one way to solve a system of equations.

The basic idea is if you have 2 equations, you can sometimes do a single operation and then add the 2 equations in a way that eleiminates 1 of the 2 variables as the example that follows shows.

Elimination Example 1

$$ y = x + 1 \\ y = -x $$

Practice Problems

Ultimate math solver (free) free algebra solver ... type anything in there, popular pages @ mathwarehouse.com.

Elimination Method

Elimination method (systems of linear equations).

The main concept behind the elimination method is to create terms with opposite coefficients because they cancel each other when added. In the end, we should deal with a simple linear equation to solve, like a one-step equation in [latex]x[/latex] or in [latex]y[/latex].

Two Ideal Cases of the Elimination Method

I can summarize the “big” ideas about the elimination method when solving systems of linear equations using the illustrations below. Here I present two ideal cases that I want to achieve during the solving process. Take a look at them and hopefully, it makes sense. Otherwise, go directly to the six (6) worked examples to see how actual problems are being solved.

Case 1 : By Adding the Two Equations, the Variable “[latex]x[/latex]” is Eliminated

The coefficients of variable [latex]x[/latex] are opposites.

Case 2 : By Adding the Two Equations, the Variable “[latex]y[/latex]” is Eliminated

The coefficients of variable [latex]y[/latex] are opposites.

Examples of Solving Systems of Linear Equations by Elimination Method

Example 1: Solve the system of linear equations by elimination method.

I have observed that adding the [latex]x[/latex]-column will not eliminate the variable [latex]x[/latex]. However, if I add the [latex]y[/latex]-column the variable [latex]y[/latex] disappears. This happens because the coefficients of [latex]y[/latex] are opposite of each other in terms of signs. Now, I will proceed with the second option.

After doing so, I end up with an easy equation.

I divide both sides by the coefficient of [latex]x[/latex] which gives the answer of [latex]x = 4[/latex].

The next step is to find the corresponding value of [latex]y[/latex]. This is easy to find since I already know what [latex]x[/latex] is. I will pick any of the two original equations, which in this case, I chose the top equation. Then, I will plug in the value of [latex]x = 4[/latex] to get [latex]y[/latex]. The process or procedure of solving for [latex]y[/latex] should be similar below.

Here I get [latex]y = – \,4[/latex]. The final answer in point notation is shown below.

Graphically, the solution looks like this.

Example 2: Solve the system of linear equations by elimination method.

This is quite interesting because no variables will cancel when added. What I want is to introduce a multiplier to one of the equations, or both, and then observe if I arrive at some coefficients that only differ in signs.

There are a few ways to do just that. However, looking at the [latex]x[/latex]-column, I can easily make the [latex]− 3[/latex] into [latex]− 12[/latex] by multiplying the top equation by [latex]+ 4[/latex]. At this point, I can proceed with the addition of [latex]x[/latex]-column.

Multiplying the entire equation by any nonzero number does not change its original meaning. What will change is just its form. I call this process, equation “revision” or “modification”.

This is a one-step equation so I solve [latex]y[/latex] by dividing both sides by its coefficient.

Great! I obtained the value [latex]y = 2[/latex]. Next, I will solve [latex]x[/latex] using back substitution using either of the original equations. For this, I will utilize the top equation because it is less complicated.

I obtained the value [latex]x = – 1[/latex]. I can now write the final answer as the ordered pair written below.

The graph below verifies that our solution is correct.

Example 3: Use the method of elimination or linear combination to solve.

There’s some twist on this problem because the coefficients of [latex]x[/latex] variables are exactly the same, both [latex]- 2[/latex].  The only thing I need to fix here is to make one of them positive.  Now, I decided to multiply the top equation by [latex]− 1[/latex]. It should also work just fine if I multiply the bottom by [latex]− 1[/latex].

You should see that the plan works since adding the [latex]x[/latex]-column results to the cancellation of [latex]x[/latex].

I solved the value of [latex]y[/latex] by dividing both sides by [latex]− 17[/latex] which results to [latex]y = 3[/latex]. This time, I will back solve the value of [latex]x[/latex] using the bottom equation because I know what [latex]y[/latex] is.

After a few steps in solving the equation above, I arrive at [latex]x = 2[/latex]. The final answer as an ordered pair is shown below.

Indeed, the two lines intersect at the point that we found in our calculations.

Example 4: Use the method of elimination or linear combination to solve.

This example follows along the line of example 3 where we have exactly the same coefficients. I see that variable [latex]y[/latex] both have coefficients of [latex]8[/latex]. So, I will need to tweak it a bit in order to make their signs opposite. I now have two options on how to proceed. I can multiply the top equation by [latex]− 1[/latex] or the bottom by [latex]− 1[/latex] as well. For this exercise, I choose to do the latter.

Applying the [latex]− 1[/latex] multiplier on the bottom equation and adding them together results to [latex]y[/latex] going away.

Solve the simple equation that arises from it.

I got [latex]x = 4[/latex] by dividing both sides by [latex]− 9[/latex]. The next obvious step is to solve for the other variable [latex]y[/latex] using back substitution. Pick any of the original equations, plug [latex]x = 4[/latex], and you will get [latex]y[/latex] in no time.

The answer is [latex]y = – \,1[/latex]. The final answer in the ordered pair form is shown below.

The graphical solution looks like this.

Example 5: Use the method of elimination or linear combination to solve.

This type of problem requires us to simultaneously multiply both the top and bottom equations by some number in order to generate coefficients with opposite signs.

If I decide to eliminate [latex]x[/latex], I can multiply the top equation by [latex]− 2[/latex] and the bottom by [latex]9[/latex]. By doing so, I should end up with [latex]x[/latex] terms, [latex]18x[/latex] and [latex] – 18x[/latex], respectively, which would cancel when added together.

For this exercise, I want to eliminate [latex]y[/latex]. Therefore, I will multiply the top by [latex]5[/latex] and the bottom by [latex]3[/latex].

As predicted, I was able to get rid of [latex]y[/latex] which leaves us with a simple equation to deal with.

You should arrive at [latex]x = 1[/latex]. Proceed on solving the other variable which is [latex]y[/latex]. Plug in [latex]x = 1[/latex] then solve for [latex]y[/latex].

I got [latex]y = – \,2[/latex]. Putting it together, our final answer is the ordered pair below .

The graphical representation of the two lines intersecting at the solved point is…

Example 6: Use the method of elimination or linear combination to solve.

This last example is very similar to the previous one. As it stands, no variables will be eliminated after adding the columns of [latex]x[/latex] and [latex]y[/latex]. However, I can eliminate the [latex]x[/latex] variables by multiplying the first equation by [latex]5[/latex] and the second by [latex]− 4[/latex], and then add them together. The rest is history!

I will end up solving a simple equation as shown.

I now have [latex]y = 5[/latex] after dividing both sides by [latex]8[/latex]. I will then substitute this value of [latex]y[/latex] to any of the original equations to solve for the corresponding [latex]x[/latex]-value.

This yields an answer of [latex]x = – \,6[/latex]. The final answer should be [latex]\left( {x,y} \right) = \left( { – \,6,5} \right)[/latex] .

This point is where the two lines intersect, as shown below.

You may also be interested in these related math lessons or tutorials:

Substitution Method (Systems of Linear Equations)

Systems of Non-Linear Equations

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Elimination Method

The elimination method of solving systems of equations is also called the addition method. To solve a system of equations by elimination we transform the system such that one variable "cancels out".

Example 1: Solve the system of equations by elimination

In this example we will "cancel out" the y term. To do so, we can add the equations together.

Now we can find: $x = 2$

In order to solve for y, take the value for x and substitute it back into either one of the original equations.

The solution is $(x, y) = (2, 1)$.

Example 2: Solve the system using elimination

Look at the x - coefficients. Multiply the first equation by -4, to set up the x-coefficients to cancel.

Now we can find: $y = -2$

Take the value for y and substitute it back into either one of the original equations.

The solution is $(x, y) = (1, -2)$.

Example 3: Solve the system using elimination method

In this example, we will multiply the first row by -3 and the second row by 2 ; then we will add down as before.

Now we can find: y = -1

Substitute y = -1 back into first equation:

The solution is $(x, y) = (3, -1)$.

Exercise: Solve the following systems using elimination method

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4.3 Solving Linear Systems by Elimination

Learning objectives.

• Solve linear systems using the elimination method.
• Solve linear systems with fractions and decimals.
• Identify the weaknesses and strengths of each method for solving linear systems.

The Elimination Method

In this section, the goal is to develop another completely algebraic method for solving a system of linear equations. We begin by defining what it means to add equations together. In the following example, notice that if we add the expressions on both sides of the equal sign, we obtain another true statement.

This is true in general: if A , B , C , and D are algebraic expressions, then we have the following addition property of equations If A , B , C , and D are algebraic expressions, where A = B and C = D , then A + C = B + D . :

For the system

we add the two equations together:

The sum of y and − y is zero and that term is eliminated. This leaves us with a linear equation with one variable that can be easily solved:

At this point, we have the x coordinate of the simultaneous solution, so all that is left to do is back substitute to find the corresponding y -value.

Hence the solution to the system is (3, 2). This process describes the elimination (or addition) method A means of solving a system by adding equivalent equations in such a way as to eliminate a variable. for solving linear systems. Of course, the variable is not always so easily eliminated. Typically, we have to find an equivalent system by applying the multiplication property of equality to one or both of the equations as a means to line up one of the variables to eliminate. The goal is to arrange that either the x terms or the y terms are opposites, so that when the equations are added, the terms eliminate. The steps for the elimination method are outlined in the following example.

Example 1: Solve by elimination: { 2 x + y = 7 3 x − 2 y = − 7 .

Step 1: Multiply one, or both, of the equations to set up the elimination of one of the variables. In this example, we will eliminate the variable y by multiplying both sides of the first equation by 2. Take care to distribute.

This leaves us with an equivalent system where the variable y is lined up to eliminate.

Step 2: Add the equations together to eliminate one of the variables.

Step 3: Solve for the remaining variable.

Step 3: Back substitute into either equation or its equivalent equation.

Step 4: Check. Remember that the solution must solve both of the original equations.

Answer: (1, 5)

Occasionally, we will have to multiply both equations to line up one of the variables to eliminate. We want the resulting equivalent equations to have terms with opposite coefficients.

Example 2: Solve by elimination: { 5 x − 3 y = − 1 3 x + 2 y = 7 .

Solution: We choose to eliminate the terms with variable y because the coefficients have different signs. To do this, we first determine the least common multiple of the coefficients; in this case, the LCM(3, 2) is 6. Therefore, multiply both sides of both equations by the appropriate values to obtain coefficients of −6 and 6.

This results in the following equivalent system:

The y terms are now lined up to eliminate.

Back substitute.

Answer: (1, 2)

Sometimes linear systems are not given in standard form. When this is the case, it is best to first rearrange the equations before beginning the steps to solve by elimination.

Example 3: Solve by elimination: { 5 x + 12 y = 11 3 y = 4 x + 1 .

Solution: First, rewrite the second equation in standard form.

This results in the following equivalent system where like terms are aligned in columns:

We can eliminate the term with variable y if we multiply the second equation by −4.

Next, we add the equations together,

Answer: (1/3, 7/9)

Try this! Solve by elimination: { 2 x + y = − 3 − 3 x − 2 y = 4 .

Answer: (−2, 1)

Video Solution

At this point, we explore what happens when solving dependent and inconsistent systems using the elimination method.

Example 4: Solve by elimination: { 3 x − y = 7 6 x − 2 y = 14 .

Solution: To eliminate the variable x , we could multiply the first equation by −2.

Now adding the equations we have

A true statement indicates that this is a dependent system. The lines coincide, and we need y in terms of x to present the solution set in the form ( x ,   m x + b ) . Choose one of the original equations and solve for y . Since the equations are equivalent, it does not matter which one we choose.

Answer: ( x ,   3 x − 7 )

Example 5: Solve by elimination: { − x + 3 y = 9   2 x − 6 y = 12 .

Solution: We can eliminate x by multiplying the first equation by 2.

A false statement indicates that the system is inconsistent. The lines are parallel and do not intersect.

Answer: No solution, ∅

Try this! Solve by elimination: { 3 x + 15 y = − 15 2 x + 10 y = 30 .

Clearing Fractions and Decimals

Given a linear system where the equations have fractional coefficients, it is usually best to clear the fractions before beginning the elimination method.

Example 6: Solve: { − 1 10 x + 1 2 y = 4 5   1 7 x + 1 3 y = − 2 21 .

Solution: Recall that we can clear fractions by multiplying both sides of an equation by the least common denominator (LCD). Take care to distribute and then simplify.

This results in an equivalent system where the equations have integer coefficients,

Solve using the elimination method.

Answer: (−3, 1)

We can use a similar technique to clear decimals before solving.

Example 7: Solve: { 3 x − 0.6 y = − 0.9 − 0.5 x + 0.12 y = 0.16 .

Solution: Multiply each equation by the lowest power of 10 necessary to result in integer coefficients. In this case, multiply the first equation by 10 and the second equation by 100.

This results in an equivalent system where the equations have integer coefficients:

Answer: (−0.2, 0.5)

Try this! Solve using elimination: { 1 3 x − 2 3 y = 3 1 3 x − 1 2 y = 8 3 .

Answer: (5, −2)

Summary of the Methods for Solving Linear Systems

We have developed three methods for solving linear systems of two equations with two variables. In this section, we summarize the strengths and weaknesses of each method.

The graphing method is useful for understanding what a system of equations is and what the solutions must look like. When the equations of a system are graphed on the same set of axes, we can see that the solution is the point where the graphs intersect. The graphing is made easy when the equations are in slope-intercept form. For example,

The simultaneous solution (−1, 10) corresponds to the point of intersection. One drawback of this method is that it is very inaccurate. When the coordinates of the solution are not integers, the method is practically unusable. If we have a choice, we typically avoid this method in favor of the more accurate algebraic techniques.

The substitution method, on the other hand, is a completely algebraic method. It requires you to solve for one of the variables and substitute the result into the other equation. The resulting equation has one variable for which you can solve. This method is particularly useful when there is a variable within the system with coefficient of 1. For example,

In this case, it is easy to solve for y in the first equation and then substitute the result into the other equation. One drawback of this method is that it often leads to equivalent equations with fractional coefficients, which are tedious to work with. If there is not a coefficient of 1, then it usually is best to choose the elimination method.

The elimination method is a completely algebraic method that makes use of the addition property of equations. We multiply one or both of the equations to obtain equivalent equations where one of the variables is eliminated if we add them together. For example,

Here we multiply both sides of the first equation by 5 and both sides of the second equation by −2. This results in an equivalent system where the variable x is eliminated when we add the equations together. Of course, there are other combinations of numbers that achieve the same result. We could even choose to eliminate the variable y . No matter which variable is eliminated first, the solution will be the same. Note that the substitution method, in this case, would require tedious calculations with fractional coefficients. One weakness of the elimination method, as we will see later in our study of algebra, is that it does not always work for nonlinear systems.

Key Takeaways

• The elimination method is a completely algebraic method for solving a system of equations.
• Multiply one or both of the equations in a system by certain numbers to obtain an equivalent system consisting of like terms with opposite coefficients. Adding these equivalent equations together eliminates a variable, and the resulting equation has one variable for which you can solve.
• It is a good practice to first rewrite the equations in standard form before beginning the elimination method.
• When the value of one of the variables is determined, back substitute into one of the original equations, or their equivalent equations, and determine the corresponding value of the other variable.

Topic Exercises

Part A: Elimination Method

Solve by elimination.

1. { x + y = 3 2 x − y = 9

2. { x − y = − 6 5 x + y = − 18

3. { x + 3 y = 5 − x − 2 y = 0

4. { − x + 4 y = 4 x − y = − 7

5. { − x + y = 2 x − y = − 3

6. { 3 x − y = − 2 6 x + 4 y = 2

7. { 5 x + 2 y = − 3 10 x − y = 4

8. { − 2 x + 14 y = 28 x − 7 y = 21

9. { − 2 x + y = 4 12 x − 6 y = − 24

10. { x + 8 y = 3 3 x + 12 y = 6

11. { 2 x − 3 y = 15 4 x + 10 y = 14

12. { 4 x + 3 y = − 10 3 x − 9 y = 15

13. { − 4 x − 5 y = − 3 8 x + 3 y = − 15

14. { − 2 x + 7 y = 56 4 x − 2 y = − 112

15. { − 9 x − 15 y = − 15 3 x + 5 y = − 10

16. { 6 x − 7 y = 4 2 x + 6 y = − 7

17. { 4 x + 2 y = 4 − 5 x − 3 y = − 7

18. { 5 x − 3 y = − 1 3 x + 2 y = 7

19. { 7 x + 3 y = 9 2 x + 5 y = − 14

20. { 9 x − 3 y = 3 7 x + 2 y = − 15

21. { 5 x − 3 y = − 7 − 7 x + 6 y = 11

22. { 2 x + 9 y = 8 3 x + 7 y = − 1

23. { 2 x + 2 y = 5 3 x + 3 y = − 5

24. { − 3 x + 6 y = − 12 2 x − 4 y = 8

25. { 25 x + 15 y = − 1 15 x + 10 y = − 1

26. { 2 x − 3 y = 2 18 x − 12 y = 5

27. { y = − 2 x − 3 − 3 x − 2 y = 4

28. { 28 x + 6 y = 9 6 y = 4 x − 15

29. { y = 5 x + 15 y = − 5 x + 5

30. { 2 x − 3 y = 9 5 x − 8 y = − 16  

31. { 1 2 x − 1 3 y = 1 6 5 2 x + y = 7 2

32. { 1 4 x − 1 9 y = 1 x + y = 3 4

33. { 1 2 x − 1 4 y = 1 3 1 4 x + 1 2 y = − 19 6

34. { − 14 3 x + 2 y = 4 − 1 3 x + 1 7 y = 4 21

35. { 0.025 x + 0.1 y = 0.5 0.11 x + 0.04 y = − 0.2

36. { 1.3 x + 0.1 y = 0.35 0.5 x + y = − 2.75

37. { x + y = 5 0.02 x + 0.03 y = 0.125

38. { x + y = 30 0.05 x + 0.1 y = 2.4

Set up a linear system and solve it using the elimination method.

39. The sum of two numbers is 14. The larger number is 1 less than two times the smaller.

40. The sum of two numbers is 30. The larger is 2 more than three times the smaller.

41. The difference of two numbers is 13 and their sum is 11.

42. The difference of two numbers is 2 and their sum is −12.

Part B: Mixed Exercises

Solve using any method.

43. { y = 2 x − 3 3 x + y = 12

44. { x + 3 y = − 5 y = 1 3 x + 5

45. { x = − 1 y = 3

46. { y = 1 2 x + 9 = 0

47. { y = x − x + y = 1

48. { y = 5 x y = − 10

49. { 3 y = 2 x − 24 3 x + 4 y = 2

50. { y = − 3 2 x + 1 − 2 y + 2 = 3 x

51. { 7 y = − 2 x − 1 7 x = 2 y + 23

52. { 5 x + 9 y − 14 = 0 3 x + 2 y − 5 = 0

53. { y = − 5 16 x + 10 y = 5 16 x − 10

54. { y = − 6 5 x + 12 x = 6

55. { 2 ( x − 3 ) + y = 0 3 ( 2 x + y − 1 ) = 15

56. { 3 − 2 ( x − y ) = − 3 4 x − 3 ( y + 1 ) = 8

57. { 2 ( x + 1 ) = 3 ( 2 y − 1 ) − 21 3 ( x + 2 ) = 1 − ( 3 y − 2 )

58. { x 2 − y 3 = − 7 x 3 − y 2 = − 8

59. { x 4 − y 2 = 3 4 x 3 + y 6 = 1 6

60. { 1 3 x − 2 3 y = 3 1 3 x − 1 2 y = 8 3

61. { − 1 10 x + 1 2 y = 4 5 1 7 x + 1 3 y = − 2 21

62. { y = − 5 3 x + 1 2 1 3 x + 1 5 y = 1 10

63. { − 1 7 x + y = − 2 3 − 1 14 x + 1 2 y = 1 3

64. { 1 15 x − 1 12 y = 1 3 − 3 10 x + 3 8 y = − 3 2

65. { x + y = 4,200 0.03 x + 0.0525 y = 193.5

66. { x + y = 350 0.2 x + 0.1 y = 52.5

67. { 0.2 x − 0.05 y = 0.43 0.3 x + 0.1 y = − 0.3

68. { 0.1 x + 0.3 y = 0.3 0.05 x − 0.5 y = − 0.63

69. { 0.15 x − 0.25 y = − 0.3 − 0.75 x + 1.25 y = − 4

70. { − 0.15 x + 1.25 y = 0.4 − 0.03 x + 0.25 y = 0.08

Part C: Discussion Board Topics

71. How do we choose the best method for solving a linear system?

72. What does it mean for a system to be dependent? How can we tell if a given system is dependent?

3: (−10, 5)

7: (1/5, −2)

9: ( x ,   2 x + 4 )

11: (6, −1)

13: (−3, 3)

17: (−1, 4)

19: (3, −4)

21: (−1, 2/3)

25: (1/5, −2/5)

27: (−2, 1)

29: (−1, 10)

33: (−2, −16/3)

35: (−4, 6)

37: (2.5, 2.5)

39: The two numbers are 5 and 9.

41: The two numbers are 12 and −1.

45: (−1, 3)

49: (6, −4)

51: (3, −1)

53: (32, 0)

55: ( x ,   − 2 x + 6 )

57: (−4, 3)

59: (1, −1)

61: (−3, 1)

65: (1,200, 3,000)

67: (0.8, −5.4)

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Systems of Linear Equations: Solving by Addition / Elimination

Definitions Graphing Special Cases Substitution Elimination/Addition Gaussian Elimination More Examples

The "addition" method of solving systems of linear equations is also called the "elimination" method. Under either name, this method is similar to the method you probably used when you were first learning how to solve one-variable linear equations .

Suppose, back in the day, they'd given you the equation " x  + 6 = 11 ". To solve this, you would probably have subtracted the six to the other side of the "equals" sign by putting a " −6 " under either side of the equation. Then you'd have drawn a horizontal line underneath (representing an "equals" line) and "added down" to get " x  = 5 " as the solution.

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Solving Systems by Addition

Your work would probably have looked something like this:

x + 6 = 11     −6    −6 x       =   5

You'll do something very similar when you solve systems of linear equations using the addition method. I'll demonstrate with some examples.

• Solve the following system using addition.

2 x + y = 9 3 x − y = 16

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When I was solving one-variable linear equations, back in the day, I would "cancel out" an unwanted number by adding its opposite. (In the example above, this would have been the −6 that was added in the second line, in order to cancel out the +6 .) Then I'd draw a horizontal "equals" line under what I'd added to both sides of the original equation, and I'd add down. This would get the variable by itself on one side of the "equals" sign.

I want to do something similar here. I know how to solve linear equations with one variable. Here I've got two. Can I get rid of one of these variables in the system, just as I'd have gotten rid of the −6 in the equation?

Looking at the system of equations they've given me, I see that I've got a + y in the first line, and a − y in the second line. If I added these, they'd cancel out, leaving me with just the variable x . In other words, if I add down, I should end up with a linear equation with just one variable, and I know how to solve those. So let's do that!

I write down the two equations, draw an "equals" bar under them, and add down:

2 x + y = 9 3 x − y = 16 5 x      = 25

Now I have a one-variable linear equation that I already know how to solve. I divide through on both sides by 5 to get x = 5 . This is half of the solution to this system.

(By the way, this adding of the two equations, or two "rows", is called a "row operation".)

To find the other half (that is, to find the y -value), I can plug this x -value back into either one of the original equations, and simplify for the value of y . (This process — of taking a partial solution and plugging it back in to some portion of the original exercise to find the rest of the solution — is called "back-solving".)

I can use either of the original equations to back-solve and find the value of y . The first equation has smaller numbers (and I'm lazy), so I'll back-solve in that one:

2(5) + y = 9   10 + y = 9            y = −1

This gives me the other half of the solution, so my answer is:

( x , y ) = (5, −1)

In case you're wondering how I knew which was the "right" equation to use for the backsolving, I didn't. Because it doesn't matter. Solutions to systems are intersection points; intersection points will, by definition, be on both of the lines; so either equation will work just fine. You'll get the same answer either way.

Check it out: if I'd have used the other equation for the back-solving, here would be my working:

3(5) − y = 16   15 − y = 16         − y = 1            y = −1

...which is the same result as before.

x − 2 y = −9 x + 3 y = 16

Note that the x -terms would cancel out if only they'd had opposite signs. But I can create this opposite-sign cancellation by multiplying either one of the equations by −1 , and then adding down as usual. It doesn't matter which equation I choose, as long as I am careful to multiply the −1 through the entire equation. (That means both sides of the "equals" sign!)

I flipped a coin; I'll multiply the second equation.

(The " −1 R 2 " notation over the arrow in the above image indicates that I multiplied row 2 by −1 . This " R n " notation, indicating that you're doing something with the n -th row, is standard. And this multiplying of a row by a numerical value is another "row operation".)

By setting up the x -terms to cancel out when the equations are added together, I have eliminated that variable. Now I can solve the resulting one-variable equation " −5 y = −25 " to get y = 5 .

To find the corresponding value of x , I plug this y -value back into either of the original equations. Back-solving in the first equation, I get:

x − 2(5) = −9 x − 10 = −9 x = 1

This gives me the other coordinate of the solution point, so my answer is:

( x , y ) = (1, 5)

A very common temptation is to write the solution to a system of equations in the form "(first number I found, second number I found)". Sometimes, though, as in this case, you find the y -value first and then the x -value second, and of course in points the x -value comes first. So just be careful to write the coordinates for your solutions correctly.

2 x − y = 9 3 x + 4 y = −14

Nothing cancels here, but I can multiply to create a cancellation. (As long as I multiply both sides of the equation by the same value, I won't have changed anything in mathematical terms. But I may be able to change things in practical terms, to create a cancellation.) If I multiply the first equation by 4 , this will set up the y -terms to cancel.

Solving this, I get that x = 2 . I'll use the first equation for backsolving, because the coefficients are smaller (and I'm lazy).

2(2) − y = 9 4 − y = 9 − y = 5 y = −5

Now I have the two coordinates of the solution point:

( x , y ) = (2, −5)

4 x − 3 y = 25 −3 x + 8 y = 10

Hmm... As the system stands, nothing cancels. But I know that I can multiply to create a cancellation.

In this case, neither variable is an obvious choice for cancellation, so I'll consider the least common multiples of the coefficients. I can multiply the equations (by 3 and 4 , respectively) to convert the x -terms to 12 x 's, or I can multiply them (by 8 and 3 , respectively) to convert the y -terms to 24 y 's. Since I'm lazy and 12 is smaller than 24 , I'll multiply to cancel the x -terms.

(I would get the same answer in the end if I set up the y -terms to cancel. It's not that how I'm doing it is "the right way"; it was just my choice. You could make a different choice, and your choice would be just as correct as mine.)

I will multiply the first row by 3 and the second row by 4 ; then I'll add down and solve.

Solving, I get that y = 5 . Neither equation looks particularly better than the other for back-solving, so I'll flip a coin and use the first equation.

4 x − 3(5) = 25 4 x − 15 = 25 4 x = 40 x = 10

Remembering to put the x -coordinate first in the solution, I get:

( x , y ) = (10, 5)

Usually when you are solving "by addition", you will need to create the cancellation. Warning: The most common mistake is to forget to multiply all the way through the equation, multiplying on both sides of the "equals" sign. Be careful of this; always multiply through the entire equation.

• Solve the following using addition.

12 x −  13 y =   2 −6 x + 6.5 y = −2

I think I'll multiply the second equation by 2 ; this will at least get rid of the decimal place.

Oops! This result isn't true! Zero is never equal to −2 !

All of my steps were correct, but I ended up with garbage. This tells me that my original assumption (being that the system had a solution) must have been wrong. So this is an inconsistent system (that i s, one that graphs as two parallel lines) with no solution (that is, having no intersection point).

no solution: inconsistent system

12 x − 3 y = 6   4 x −    y = 2

I think it'll be simplest to cancel off the y -terms, so I'll multiply the second row by −3 .

Well, yes, zero does equal zero, but...?

I already knew that zero equals zero. This information doesn't add anything to my store of knowledge. In particular, it doesn't help me narrow down my answer to one solution point. All my math was correct, so the issue lies elsewhere.

Then I remember: If the two equations are really the same one equation, then this "zero equals zero" result is the sort of thing I should expect. In fact, this result tells me that this system is a dependent system (that is, one that graphs as just one line) and, solving either of the original equations for " y = ", I find that the solution is the equation of the whole line, namely:

y = 4 x − 2

(Your text may format the answer as " ( s , 4 s − 2) ", or something like that.)

Remember the difference: a nonsense answer (like " 0 = −2 " in the exercise before the last one above) means that you have an inconsistent system with no solution; a useless-but-true answer (like " 0 = 0 " in the last exercise above) means that you have a dependent system where the set of all the points on the whole line is the solution.

Note: Some books use only " x " and " y " for their variables in systems of two equations, but many will also use additional variables. When you write the solution for an x , y -point, you know that the x -coordinate goes first and the y -coordinate goes second. When you are dealing with other variables, assume (unless explicitly told otherwise) that those variables, when used as coordinates of a point, are written in alphabetical order. For instance, if the variables in a given system are a and b , the solution point would be ( a ,  b ) ; it would not be ( b ,  a ) . Remember: Unless otherwise specified, the variables are always written in alphabetical order.

URL: https://www.purplemath.com/modules/systlin5.htm

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The elimination method for solving linear systems

• Solve the system IV
• Solve the system V

Another way of solving a linear system is to use the elimination method. In the elimination method you either add or subtract the equations to get an equation in one variable.

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.

\begin{cases} 3y+2x=6\\ 5y-2x=10 \end{cases}

We can eliminate the \(x\)-variable by addition of the two equations.

\begin{cases} 3y+2x=6 \\ \underline{+\: 5y-2x=10} \end{cases} 

$$=8y\: \: \: \: \; \; \; \; =16$$

$$\begin{matrix} \: \: \: y\: \: \: \: \: \; \; \; \; \; =2 \end{matrix}$$

The value of \(y\) can now be substituted into either of the original equations to find the value of \(x\)

$$3y+2x=6$$

$$3\cdot {\color{green} 2}+2x=6$$

The solution of the linear system is \((0, 2)\).

To avoid errors make sure that all like terms and equal signs are in the same columns before beginning the elimination.

If you don't have equations where you can eliminate a variable by addition or subtraction you directly you can begin by multiplying one or both of the equations with a constant to obtain an equivalent linear system where you can eliminate one of the variables by addition or subtraction.

\begin{cases} 3x+y=9\\ 5x+4y=22 \end{cases}

Begin by multiplying the first equation by \(-4\) so that the coefficients of \(y\) are opposites

\begin{cases} \color{green}{-4} \cdot (3x + y) = \color{green}{-4} \cdot 9\\ 5x + 4y = 22 \end{cases}

$$\Rightarrow$$

\begin{cases}-12x-4y=-36 \\ \underline{+5x+4y=22 }\end{cases}

$$=-7x\: \: \: \: \: \: \: \: \: \: =-14$$

$$\begin{matrix} \: \:\; \:\: x\: \: \: \: \: \: \: \: \: \: \:=2 \end{matrix}$$

Substitute \(x\) in either of the original equations to get the value of \(y\)

$$3\cdot {\color{green} 2}+y=9$$

The solution of the linear system is \((2, 3)\)

Video lesson

Solve the following linear system using the elimination method

\begin{cases} 2y - 4x = 2 \\ y = -x + 4 \end{cases}

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