solving linear systems by elimination assignment quizlet

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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• Book title: Elementary Algebra 2e
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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

You can use the Mathway widget below to practice solving systems of equations by addition/elimination (or skip the widget and proceed to the next page ). Try the entered exercise, or type in your own exercise. Then click on the arrow and select "Solve by Addition/Elimination" from the menu to compare your answer to Mathway's.

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4.2: Solving Linear Systems by Substitution

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• Page ID 18348

Learning Objectives

• Solve linear systems using the substitution method.

The Substitution Method

In this section, we will define a completely algebraic technique for solving systems. The idea is to solve one equation for one of the variables and substitute the result into the other equation. After performing this substitution step, we will be left with a single equation with one variable, which can be solved using algebra. This is called the substitution method, and the steps are outlined in the following example.

Example \(\PageIndex{1}\)

Solve by substitution:

Step 1 : Solve for either variable in either equation. If you choose the first equation, you can isolate \(y\) in one step.

\(\begin{aligned} 2x+y&=7\\2x+y\color{Cerulean}{-2x}&=7\color{Cerulean}{-2x} \\ y&=-2x+7 \end{aligned}\)

Step 2 : Substitute the expression \(−2x+7\) for the \(y\) variable in the other equation.

This leaves you with an equivalent equation with one variable, which can be solved using the techniques learned up to this point.

Step 3 : Solve for the remaining variable. To solve for \(x\), first distribute \(−2\):

Step 4 : Back substitute to find the value of the other coordinate. Substitute \(x= 1\) into either of the original equations or their equivalents. Typically, we use the equivalent equation that we found when isolating a variable in step 1.

\(\begin{aligned} y&=-2x+7\\&=-2(\color{OliveGreen}{1}\color{black}{)+7}\\&=-2+7\\&=5 \end{aligned}\)

The solution to the system is \((1, 5)\). Be sure to present the solution as an ordered pair.

Step 5 : Check. Verify that these coordinates solve both equations of the original system:

\(\color{Cerulean}{Check:}\:\:\color{black}{(1,5)}\)

\(\begin{array}{c|c} {Equation\:1:}&{Equation\:2:}\\{2x+y=7}&{3x-2y=-7}\\{2(\color{OliveGreen}{1}\color{black}{)+(}\color{OliveGreen}{5}\color{black}{)=7}}&{3(\color{OliveGreen}{1}\color{black}{)-2(}\color{OliveGreen}{5}\color{black}{)=-7}}\\{2+5=7}&{3-10=-7}\\{7=7\quad\color{Cerulean}{\checkmark}}&{-7=-7\quad\color{Cerulean}{\checkmark}} \end{array}\)

The graph of this linear system follows:

The substitution method for solving systems is a completely algebraic method. Thus graphing the lines is not required.

Example \(\PageIndex{2}\)

\(\left\{\begin{aligned}2x−y&=1\\2x−y&=3\end{aligned}\right.\).

In this example, we can see that \(x\) has a coefficient of \(1\) in the second equation. This indicates that it can be isolated in one step as follows:

\(\begin{aligned} x-y&=3 \\ x-y\color{Cerulean}{+y}&=3\color{Cerulean}{+y} \\ x&=3+y \end{aligned}\)

\(\left\{\begin{aligned} 2\color{Cerulean}{x}\color{black}{-y}&=12 \\ x-y&=3 \Rightarrow \color{Cerulean}{x}\color{black}{=3+y} \end{aligned} \right.\)

Substitute \(3+y\) for \(x\) in the first equation. Use parentheses and take care to distribute.

\(\begin{aligned} 2x-y&=12\\2(\color{OliveGreen}{3+y}\color{black}{)-y}&=12 \\6+2y-y&=12 \\6+y&=12 \\ 6+y\color{Cerulean}{-6}&=12\color{Cerulean}{-6}\\y&=6 \end{aligned}\)

Use \(x=3+y\) to find \(x\).

\(\begin{aligned} x&=3+y\\&=3+\color{OliveGreen}{+6}\\&=9 \end{aligned}\)

\((9, 6)\). The check is left to the reader.

Example \(\PageIndex{3}\)

\(\left\{\begin{aligned}3x−5y&=1\\7x&=−1\end{aligned}\right.\).

In this example, the variable \(x\) is already isolated. Hence we can substitute \(x=−1\) into the first equation.

\((−1, −4)\). It is a good exercise to graph this particular system to compare the substitution method to the graphing method for solving systems.

Exercise \(\PageIndex{1}\)

\(\left\{\begin{aligned}3x+y&=4\\8x+2y&=10\end{aligned}\right.\).

Solving systems algebraically frequently requires work with fractions.

Example \(\PageIndex{4}\)

\(\left\{\begin{aligned}2x+8y&=5\\24x−4y&=−15\end{aligned}\right.\).

Begin by solving for \(x\) in the first equation.

\(\begin{aligned} 2x+8y&=5\\2x+8y\color{Cerulean}{-8y}&=5\color{Cerulean}{-8y} \\ \frac{2x}{\color{Cerulean}{2}}&=\frac{-8y+5}{\color{Cerulean}{2}} \\ x&=\frac{-8y}{2}+\frac{5}{2} \\ x&=-4y+\frac{5}{2} \end{aligned}\)

\(\left\{\begin{aligned} 2x+8y&=5 \Rightarrow \color{Cerulean}{x}\color{black}{=-4y+\frac{5}{2}} \\ 24\color{Cerulean}{x}\color{black}{-4y}&=-15\end{aligned}\right.\)

Next, substitute into the second equation and solve for \(y\).

Back substitute into the equation used in the substitution step:

\(\begin{aligned} x&=-4y+\frac{5}{2} \\ &=-4\left(\color{OliveGreen}{\frac{3}{4}} \right)\color{black}{+\frac{5}{2}} \\ &=-3+\frac{5}{2} \\ &=-\frac{6}{2} + \frac{5}{2} \\ &=-\frac{1}{2} \end{aligned}\)

\((-\frac{1}{2},\frac{3}{4})\)

As we know, not all linear systems have only one ordered pair solution. Recall that some systems have infinitely many ordered pair solutions and some do not have any solutions. Next, we explore what happens when using the substitution method to solve a dependent system.

Example \(\PageIndex{5}\)

\(\left\{\begin{aligned}−5x+y&=−1\\10x−2y&=2\end{aligned}\right.\).

Since the first equation has a term with coefficient \(1\), we choose to solve for that first.

Next, substitute this expression in for \(y\) in the second equation.

\(\begin{aligned} 10x-2y&=2 \\ 10x-2(\color{OliveGreen}{5x-1}\color{black}{)}&=2 \\ 10x-10x+2&=2 \\ 2&=2 \quad\color{Cerulean}{True} \end{aligned}\)

This process led to a true statement; hence the equation is an identity and any real number is a solution. This indicates that the system is dependent. The simultaneous solutions take the form \((x, mx + b)\), or in this case, \((x, 5x − 1)\), where \(x\) is any real number.

\((x, 5x−1)\)

To have a better understanding of the previous example, rewrite both equations in slope-intercept form and graph them on the same set of axes.

We can see that both equations represent the same line, and thus the system is dependent. Now explore what happens when solving an inconsistent system using the substitution method.

Example \(\PageIndex{6}\)

\(\left\{\begin{aligned}−7x+3y&=3\\14x−6y&=−16\end{aligned}\right.\).

Solve for \(y\) in the first equation.

\(\begin{aligned} -7x+3y&=3 \\ -7x+3y\color{Cerulean}{+7x}&=3\color{Cerulean}{+7x} \\3y&=7x+3 \\ \frac{3y}{\color{Cerulean}{3}}&=\frac{7x+3}{\color{Cerulean}{3}}\\y&=\frac{7}{3}x+1 \end{aligned}\)

\(\left\{\begin{aligned} -7x+3y&=3 \Rightarrow \color{Cerulean}{y}\color{black}{=\frac{7}{3}x+1} \\ 14x-6\color{Cerulean}{y}&=-16\end{aligned}\right.\)

Substitute into the second equation and solve.

Solving leads to a false statement. This indicates that the equation is a contradiction. There is no solution for \(x\) and hence no solution to the system.

No solution, \(Ø\)

A false statement indicates that the system is inconsistent, or in geometric terms, that the lines are parallel and do not intersect. To illustrate this, determine the slope-intercept form of each line and graph them on the same set of axes.

In slope-intercept form, it is easy to see that the two lines have the same slope but different \(y\)-intercepts.

Exercise \(\PageIndex{2}\)

\(\left\{\begin{aligned}2x−5y&=3\\4x−10y&=6\end{aligned}\right.\).

\((x, \frac{2}{5}x−\frac{3}{5}) \)

Key Takeaways

• The substitution method is a completely algebraic method for solving a system of equations.
• The substitution method requires that we solve for one of the variables and then substitute the result into the other equation. After performing the substitution step, the resulting equation has one variable and can be solved using the techniques learned up to this point.
• When the value of one of the variables is determined, go back and substitute it into one of the original equations, or their equivalent equations, to determine the corresponding value of the other variable.
• Solutions to systems of two linear equations with two variables, if they exist, are ordered pairs \((x, y)\).
• If the process of solving a system of equations leads to a false statement, then the system is inconsistent and there is no solution, \(Ø\).
• If the process of solving a system of equations leads to a true statement, then the system is dependent and there are infinitely many solutions that can be expressed using the form \((x, mx + b)\).

Exercise \(\PageIndex{3}\) Substitution Method

Solve by substitution.

• \(\left\{\begin{aligned} y&=4x−1\\−3x+y&=1\end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=3x−8\\4x−y&=2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x&=2y−3\\x+3y&=−8 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x&=−4y+12\\x+3y&=12 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=3x−5\\x+2y&=2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=x\\2x+3y&=10 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=4x+1\\−4x+y&=2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−3x+5\\3x+y&=5 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=2x+3\\2x−y&=−3 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=5x−1\\x−2y&=5 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−7x+1\\3x−y&=4 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x&=6y+2\\5x−2y&=0 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−2−2x\\−y&=−6 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x&=−3x−4\\y&=−3 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−\frac{1}{5}x+\frac{3}{7}\\x−5y&=9 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=\frac{2}{3}x−\frac{1}{6}\\x−9y&=0 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=\frac{1}{2}x+\frac{1}{3}\\x−6y&=4 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=−\frac{3}{8}x+\frac{1}{2}\\2x+4y&=1 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x+y&=6\\2x+3y=\frac{1}{6} \end{aligned}\right.\)
• \(\left\{\begin{aligned}x−y&=3\\−2x+3y&=−2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}2x+y&=2\\3x−2y&=\frac{1}{7} \end{aligned}\right.\)
• \(\left\{\begin{aligned}x−3y&=−\frac{1}{13}\\x+5y&=−5 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x+2y&=−3\\3x−4y&=−2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}5x−y&=12\\9x−y&=10 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x+2y&=−6\\−4x−8y&=24 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x+3y&=−6\\−2x−6y&=−12 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−3x+y&=−4\\6x−2y&=−2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x−5y&=−10\\2x−10y&=−20 \end{aligned}\right.\)
• \(\left\{\begin{aligned}3x−y&=9\\4x+3y&=−1 \end{aligned}\right.\)
• \(\left\{\begin{aligned}2x−y&=5\\4x+2y&=−2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−x+4y&=0\\2x−5y&=−6 \end{aligned}\right.\)
• \(\left\{\begin{aligned}3y−x&=5\\5x+2y&=−8 \end{aligned}\right.\)
• \(\left\{\begin{aligned}2x−5y&=1\\4x+10y&=2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}3x−7y&=−3\\6x+14y&=0 \end{aligned}\right.\)
• \(\left\{\begin{aligned}10x−y&=3\\−5x+12y&=1 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−\frac{1}{3}x+\frac{1}{6}y&=\frac{2}{3}\\ \frac{1}{2}x−\frac{1}{3}y&=−\frac{3}{2} \end{aligned}\right.\)
• \(\left\{\begin{aligned}\frac{1}{3}x+\frac{2}{3}y&=1 \\ \frac{1}{4}x−\frac{1}{3}y&=−\frac{1}{12} \end{aligned}\right.\)
• \(\left\{\begin{aligned}\frac{1}{7}x−y&=\frac{1}{2}\\ \frac{1}{4}x+\frac{1}{2}y&=2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−\frac{3}{5}x+\frac{2}{5}y&=\frac{1}{2}\\ \frac{1}{3}x−\frac{1}{12}y&=−\frac{1}{3} \end{aligned}\right.\)
• \(\left\{\begin{aligned}\frac{1}{2}x&=\frac{2}{3}y\\x−\frac{2}{3}y&=2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−\frac{1}{2}x+\frac{1}{2}y&=\frac{5}{8} \\ \frac{1}{4}x+\frac{1}{2}y&=\frac{1}{4} \end{aligned}\right.\)
• \(\left\{\begin{aligned}x−y&=0\\−x+2y&=3 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=3x\\2x−3y&=0 \end{aligned}\right.\)
• \(\left\{\begin{aligned}2x+3y&=18\\−6x+3y&=−6 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−3x+4y&=20\\ 2x+8y&=8 \end{aligned}\right.\)
• \(\left\{\begin{aligned}5x−3y&=−1\\ 3x+2y&=7 \end{aligned}\right.\)
• \(\left\{\begin{aligned}−3x+7y&=2\\ 2x+7y&=1 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=3\\y&=−3 \end{aligned}\right.\)
• \(\left\{\begin{aligned}x&=5\\x&=−2 \end{aligned}\right.\)
• \(\left\{\begin{aligned}y&=4\\y&=4\end{aligned}\right.\)

1. \((2, 7)\)

3. \((−5, −1)\)

5. \((2, 6)\)

7. \(∅\)

9. \((x, 2x+3)\)

11. \((\frac{1}{2}, −\frac{5}{2})\)

13. \((4, −2)\)

15. \((3, \frac{12}{5})\)

17. \((−3, −\frac{7}{6})\)

19. \((2, 4)\)

21. \((3, −4)\)

23. \((−\frac{8}{5}, −\frac{7}{10})\)

25. \((x, −\frac{1}{2}x−3)\)

27. \(∅\)

29. \((2, −3)\)

31. \((−8, −2)\)

33. \((\frac{1}{2}, 0)\)

35. \(∅\)

37. \((1, 1)\)

39. \((−\frac{11}{10}, −\frac{2}{5})\)

41. \((−\frac{1}{2}, \frac{3}{4})\)

43. \((0, 0)\)

45. \((−4, 2)\)

47. \((−\frac{1}{5}, \frac{1}{5})\)

49. \(∅\)

Exercise \(\PageIndex{4}\) Substitution Method

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

• The sum of two numbers is \(19\). The larger number is \(1\) less than three times the smaller.
• The sum of two numbers is \(15\). The larger is \(3\) more than twice the smaller.
• The difference of two numbers is \(7\) and their sum is \(1\).
• The difference of two numbers is \(3\) and their sum is \(−7\).
• Where on the graph of \(−5x+3y=30\) does the \(x\)-coordinate equal the \(y\)-coordinate?
• Where on the graph of \(\frac{1}{2}x−\frac{1}{3}y=1\) does the \(x\)-coordinate equal the \(y\)-coordinate?

1. The two numbers are \(5\) and \(14\).

3. The two numbers are \(4\) and \(−3\).

5. \((−\frac{1}{5}, −\frac{1}{5})\)

Exercise \(\PageIndex{5}\) Discussion Board Topics

• Describe what drives the choice of variable to solve for when beginning the process of solving by substitution.
• Discuss the merits and drawbacks of the substitution method.

1. Answers may vary

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Solving Systems of Linear Equations by E...

Mathematics.

Solving Systems of Linear Equations by Elimination

20 questions

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• 1. Multiple Choice Edit 1.5 minutes 1 pt Solve by elimination: 4x+9y=28 -4x-y=-28 (-7,0) (6,0) (-6,0) (7,0)
• 2. Multiple Choice Edit 5 minutes 1 pt Solve by elimination: -9x-4y=-20 5x+4y=4 (-4,4) (4,4) (4,-4) (-4,-4)
• 3. Multiple Choice Edit 45 seconds 1 pt Solve by elimination: -x+2y=17 2x+2y=-10 ARN (-9,4) (-9,-4) (9,-4)
• 4. Multiple Choice Edit 3 minutes 1 pt Solve by elimination  -6x+y= -2 -3x-6y= 12 (2,0) no solution  (0,-2) infinite solutions 
• 5. Multiple Choice Edit 45 seconds 1 pt Solve the system using Multiplication.  2x + 3y = 12 5x - y = 13 x = -3, y = -2 x = 1.5, y = 2 x = 6, y = 0 x = 3, y = 2
• 6. Multiple Choice Edit 45 seconds 1 pt Solve the system by multiplication. 3x + 2y = 16 7x + y = 19 (-2,5) (-2,-5) (2,-5) (2,5)

what does the top equation need to be multiplied by to create a zero pair?

2x - 6y = 20

2x + 5y = -11

• 8. Multiple Choice Edit 45 seconds 1 pt What is the first step in solving a system of equations by elimination? Get both equations in standard form and line up the like terms. Get both equations in slope-intercept form. Get both equations equal to zero. Put the x terms first.
• 9. Multiple Choice Edit 5 minutes 1 pt Solve the system by elimination.  −4x − 4y = 0 4x + 4y = 0  (−6, −4)  Infinite number of solutions   (−6, 10)  (6, 4) 
• 10. Multiple Choice Edit 5 minutes 1 pt Solve the system: x + 6y = 17 x - 3y = 8 (5, 2) (11, 1) (1, 11) no solution
• 11. Multiple Choice Edit 5 minutes 1 pt Solve by elimination: 3x+7y=23 -3x-7y=-17 No solution ARN (-3,3) (3,3)
• 12. Multiple Choice Edit 5 minutes 1 pt Solve : x - y = - 6 x + y = 8 (2, 3) no solution (1, 7) (8, 7)
• 13. Multiple Choice Edit 3 minutes 1 pt Solve  the system of equations.      x + 5y = -27 -2x - 5y = 24 (3, 6) (-3, -6) (-3, -4) (3, -6)

For the system below, what method would likely be BEST to use to solve?

4x - 2y = 7

Substitution

Elimination

Slope formula

• 15. Multiple Choice Edit 5 minutes 1 pt Solve  2 x + 4 y = 7  2 x – 4 y = -3  (2,3) (5⁄4,1) (1,5 ⁄4) (3,2)
• 16. Multiple Choice Edit 15 minutes 1 pt Solve the system of Equation (x,y) x + 4y = –3  x + 7y = –12  (-3,9) (9,-3) (3,9) (9,3)
• 17. Multiple Choice Edit 5 minutes 1 pt Solve the system by elimination.   12x + 36y = −3  −12x − 36y = 0  (10, 1)  (−10, −1)   (10, −1)   No solution 
• 18. Multiple Choice Edit 5 minutes 1 pt Find the solution to the system below.  3x + y = 19 2x - y = 6 No solution (4, 5) (5, 4) (14, 5)

Solve by elimination by multiplying bottom by negative 2 if you want.

3x + 4y = 10

6x - 4y = 8

Help -2(3x + 4y = 10)

10x - 7y = -18

Step 1. Identify the coefficient in front of x on top

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