## Elimination Calculator

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## Most Used Actions

Number line.

- elimination\:x+y+z=25,\:5x+3y+2z=0,\:y-z=6
- elimination\:x+2y=2x-5,\:x-y=3
- elimination\:5x+3y=7,\:3x-5y=-23
- elimination\:x+z=1,\:x+2z=4

elimination-system-of-equations-calculator

- High School Math Solutions – Systems of Equations Calculator, Elimination A system of equations is a collection of two or more equations with the same set of variables. In this blog post,... Read More

## 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.

- 5.3 Solve Systems of Equations by Elimination
- Introduction
- 1.1 Introduction to Whole Numbers
- 1.2 Use the Language of Algebra
- 1.3 Add and Subtract Integers
- 1.4 Multiply and Divide Integers
- 1.5 Visualize Fractions
- 1.6 Add and Subtract Fractions
- 1.7 Decimals
- 1.8 The Real Numbers
- 1.9 Properties of Real Numbers
- 1.10 Systems of Measurement
- Key Concepts
- Review Exercises
- Practice Test
- 2.1 Solve Equations Using the Subtraction and Addition Properties of Equality
- 2.2 Solve Equations using the Division and Multiplication Properties of Equality
- 2.3 Solve Equations with Variables and Constants on Both Sides
- 2.4 Use a General Strategy to Solve Linear Equations
- 2.5 Solve Equations with Fractions or Decimals
- 2.6 Solve a Formula for a Specific Variable
- 2.7 Solve Linear Inequalities
- 3.1 Use a Problem-Solving Strategy
- 3.2 Solve Percent Applications
- 3.3 Solve Mixture Applications
- 3.4 Solve Geometry Applications: Triangles, Rectangles, and the Pythagorean Theorem
- 3.5 Solve Uniform Motion Applications
- 3.6 Solve Applications with Linear Inequalities
- 4.1 Use the Rectangular Coordinate System
- 4.2 Graph Linear Equations in Two Variables
- 4.3 Graph with Intercepts
- 4.4 Understand Slope of a Line
- 4.5 Use the Slope-Intercept Form of an Equation of a Line
- 4.6 Find the Equation of a Line
- 4.7 Graphs of Linear Inequalities
- 5.1 Solve Systems of Equations by Graphing
- 5.2 Solving Systems of Equations by Substitution
- 5.4 Solve Applications with Systems of Equations
- 5.5 Solve Mixture Applications with Systems of Equations
- 5.6 Graphing Systems of Linear Inequalities
- 6.1 Add and Subtract Polynomials
- 6.2 Use Multiplication Properties of Exponents
- 6.3 Multiply Polynomials
- 6.4 Special Products
- 6.5 Divide Monomials
- 6.6 Divide Polynomials
- 6.7 Integer Exponents and Scientific Notation
- 7.1 Greatest Common Factor and Factor by Grouping
- 7.2 Factor Trinomials of the Form x2+bx+c
- 7.3 Factor Trinomials of the Form ax2+bx+c
- 7.4 Factor Special Products
- 7.5 General Strategy for Factoring Polynomials
- 7.6 Quadratic Equations
- 8.1 Simplify Rational Expressions
- 8.2 Multiply and Divide Rational Expressions
- 8.3 Add and Subtract Rational Expressions with a Common Denominator
- 8.4 Add and Subtract Rational Expressions with Unlike Denominators
- 8.5 Simplify Complex Rational Expressions
- 8.6 Solve Rational Equations
- 8.7 Solve Proportion and Similar Figure Applications
- 8.8 Solve Uniform Motion and Work Applications
- 8.9 Use Direct and Inverse Variation
- 9.1 Simplify and Use Square Roots
- 9.2 Simplify Square Roots
- 9.3 Add and Subtract Square Roots
- 9.4 Multiply Square Roots
- 9.5 Divide Square Roots
- 9.6 Solve Equations with Square Roots
- 9.7 Higher Roots
- 9.8 Rational Exponents
- 10.1 Solve Quadratic Equations Using the Square Root Property
- 10.2 Solve Quadratic Equations by Completing the Square
- 10.3 Solve Quadratic Equations Using the Quadratic Formula
- 10.4 Solve Applications Modeled by Quadratic Equations
- 10.5 Graphing Quadratic Equations in Two Variables

## 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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- Authors: Lynn Marecek, MaryAnne Anthony-Smith, Andrea Honeycutt Mathis
- Publisher/website: OpenStax
- Book title: Elementary Algebra 2e
- Publication date: Apr 22, 2020
- Location: Houston, Texas
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## Solving Systems of Linear Equations Using Elimination

Systems of linear equations:.

A system of linear equations is just a set of two or more linear equations.

In two variables ( x and y ) , the graph of a system of two equations is a pair of lines in the plane.

There are three possibilities:

- The lines intersect at zero points. (The lines are parallel.)
- The lines intersect at exactly one point. (Most cases.)
- The lines intersect at infinitely many points. (The two equations represent the same line.)

How to Solve a System of Linear Equations Using The Elimination Method (aka The Addition Method, aka The Linear Combination Method)

- Step 1 : Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x -terms or the y -terms cancel out.
- Step 2 : Then solve for x (or y , whichever's left) and substitute back to get the other coordinate.

Now, how do we know that a linear equation obtained by the addition of the first equation with a scalar multiplication of the second is equivalent to the first?

Let us take an example. Consider the system

3 x + 2 y = 3 x − y = − 4 .

Consider the equation obtained by multiplying the second equation by a constant m and then adding the resultant equation with the first one.

That is, ( 3 x + 2 y ) + m ( x − y ) = ( 3 ) + m ( − 4 ) .

What we need to prove is that this equation is equivalent to the equation 3 x + 2 y = 3 .

We have x − y = − 4 ⇒ m ( x − y ) = − 4 m .

Since m ( x − y ) = − 4 m , subtract m ( x − y ) from the left side and − 4 m from the right side of the equation ( 3 x + 2 y ) + m ( x − y ) = ( 3 ) − 4 m which will retain the balance.

( 3 x + 2 y ) + m ( x − y ) − m ( x − y ) = ( 3 ) − 4 m − ( − 4 m )

Cancelling common terms we get, 3 x + 2 y = 3 which is equivalent to the first equation.

Therefore, the systems of equations 3 x + 2 y = 3 x − y = − 4 and ( 3 x + 2 y ) + m ( x − y ) = 3 + m ( − 4 ) x − y = − 4 are equivalent.

In general, for any system of equations K = L and P = Q , it ca be shown that K + m P = L + m Q is equivalent to K = L .

Solve the system { 4 x + 3 y = − 2 8 x − 2 y = 12

− 8 x − 6 y = 4 8 x − 2 y = 12 _ − 8 y = 16

Solve for y .

y = − 2

Substitute for y in either of the original equations and solve for x .

4 x + 3 ( − 2 ) = − 2 4 x − 6 = − 2 4 x = 4 x = 1

The solution is ( 1 , − 2 ) .

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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}

- Properties of exponents
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- Monomials and polynomials
- Special products of polynomials
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- The quadratic formula
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- Simplify radical expressions
- Radical equations
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- Simplify rational expression
- Multiply rational expressions
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## Chapter 1: Real Numbers

- Real Numbers
- Euclid’s Division Lemma
- Fundamental Theorem of Arithmetic
- HCF and LCM
- Irrational Numbers
- Decimal Expansions of Rational Numbers
- Rational Numbers

## Chapter 2: Polynomials

- Algebraic Expressions
- Polynomial Formula
- Types of Polynomials
- Zeros of Polynomial
- Geometrical meaning of the Zeroes of a Polynomial
- Factorization of Polynomial
- Division Algorithm for Polynomials
- Algebraic Identities
- Relationship between Zeroes and Coefficients of a Polynomial
- Division Algorithm Problems and Solutions

## Chapter 3: Pair of Linear equations in two variables

- Linear Equation in Two Variables
- Pair of Linear Equations in Two Variables
- Graphical Methods of Solving Pair of Linear Equations in Two Variables
- Solve the Linear Equation using Substitution Method

## Solving Linear Equations Using the Elimination Method

Chapter 4: quadratic equations.

- Quadratic Equations
- Roots of Quadratic Equations
- Solving Quadratic Equations
- How to find the Discriminant of a Quadratic Equation?

## Chapter 5: Arithmetic Progressions

- Introduction to Arithmetic Progressions | Class 10 Maths
- Sequences and Series
- Arithmetic Progression
- Arithmetic Progression – Common difference and Nth term | Class 10 Maths
- How to find the nth term of an Arithmetic Sequence?
- Arithmetic Progression – Sum of First n Terms | Class 10 Maths
- Arithmetic Mean

## Chapter 6: Triangles

- Similar Triangles
- Criteria for Similarity of Triangles
- Basic Proportionality Theorem
- Pythagoras Theorem

## Chapter 7: Coordinate Geometry

- Coordinate Geometry
- Distance formula – Coordinate Geometry | Class 10 Maths
- Distance Between Two Points
- Section Formula
- How to find the ratio in which a point divides a line?
- How to find the Trisection Points of a line?
- How to find the Centroid of a Triangle?
- Area of a Triangle – Coordinate Geometry | Class 10 Maths

## Chapter 8: Introduction to Trigonometry

- Trigonometric Ratios
- How to use the Unit Circle in Trigonometry?
- Trigonometric ratios of some Specific Angles
- Trigonometric Identities

## Chapter 9: Some Applications of Trigonometry

- Height and Distance

## Chapter 10: Circles

- Theorem – The tangent at any point of a circle is perpendicular to the radius through the point of contact – Circles | Class 10 Maths
- Theorem – The lengths of tangents drawn from an external point to a circle are equal – Circles | Class 10 Maths
- Number of Tangents from a Point on a Circle

## Chapter 11: Constructions

- Division of Line Segment in Given Ratio – Constructions | Class 10 Maths
- Construction of Similar Triangles
- Constructions of Tangents to a Circle

## Chapter 12: Areas related to circles

- Area of a Circle
- Circumference of a Circle
- Sector of a Circle
- Arc Length Formula

## Chapter 13: Surface Areas and Volume

- Surface Area of Cuboid
- Volume of Cuboid
- Surface Area of Cube
- Volume of a Cube
- Surface Area of a Cylinder
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- Surface Area of Cone
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- Surface Area of a Hemisphere
- Volume of Hemisphere
- Volume of Combination of Solids
- Frustum of Cone
- Conversion of solids – Surface Areas and Volumes
- Surface Areas and Volumes

## Chapter 14: Statistics

- How to find Mean of grouped data by direct method?
- Shortcut Method for Arithmetic Mean
- How to calculate the mean using Step deviation method?
- Graphical determination of Median
- Ogive (Cumulative Frequency Curve) and its Types

## Chapter 15: Probability

- Types of Events in Probability
- Dependent and Independent Events
- Experimental Probability
- Probability Theory
- NCERT Solutions for Class 10 Maths
- RD Sharma Class 10 Solutions

If an equation is written in the form ax + by + c = 0, where a, b, and c are real integers and the coefficients of x and y, i.e. a and b, are not equal to zero, it is said to be a linear equation in two variables. For example, 3x + y = 4 is a linear equation in two variables- x and y. The numbers that come before these variables are called coefficients. Thus the coefficient of x is 3 and that of y is 1.

## Linear equation using 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 of different signs (negative in one equation, positive in the other), you add the equations to eliminate a variable and when the coefficients of one variable have the same sign (either negative in both equations or positive in both equations), you subtract the equations to eliminate a variable. It is to be noted that the variable to be eliminated needs to have the same coefficient in both equations.

The elimination method of solving a pair of linear equations is shown below, followed by the steps it entails:

Generalized example

Suppose we are given two equations of the form ax + by = c and mx + ny = q.

Given: ax + by = c ⇢ (1) mx + ny = q ⇢ (2) In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, we multiply equation (1) by m and equation (2) by a in order to equate the coefficients of x. amx + bmy = cm amx + any = aq Subtracting the two equations, we obtain: (amx – amx) + (bmy – any) = cm – aq ⇒ y(bm – an) = cm – aq ⇒ y = (cm – aq)/(bm – an) The equation (1) can be re- written as x = c – by/a. Now substitute y = (cm – aq)/(bm – an) in the equation, x = c – by/a.

Steps to solve a linear equation using elimination method

- Make sure that the linear equations are of the form ax + by = m and cx + dy = n.
- In order to solve the given equations by elimination, the coefficients of one of the variables in both equations must be equal. Look for the numbers, which when multiplied by the coefficients of the given equations, would equate them. Just like we multiplied the above equations with m and a in order to eliminate x by making changing its coefficient in equation one from a to am and from m to am in the second equation.
- Add or subtract the equations to eliminate the variable with equal coefficients. In the above example, variable x was eliminated.
- Solve for the value of the other variable. In the example above, after eliminating x, the value of y was calculated.
- Substitute the value of the variable into any of the given equations and solve for the variable that was eliminated earlier.

## Sample Problems

Question 1: Solve 2x + y = 3 and 6x − y = 9 using elimination method.

Given: 2x + y = 3 ⇢ (1) 6x − y = 9 ⇢ (2) In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, multiply the first equation by -3,and multiply the second equation by 1. −6x − 6y = −9 6x − 3y = 9 Add these equations to eliminate x: −9y = 0 ⇒ y = 0 Substitute y = 0 in equation (1): 2x + 0 = 3 ⇒ x = 3/2 Thus, by elimination method, x = 3/2 and y = 0.

Question 2: Solve 4x + 2y = 10, 5x − y = 4 using elimination method.

Given: 4x + 2y =10 ⇢ (1) 5x − y = 4 ⇢ (2) In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, multiply the first equation by 1,and multiply the second equation by 2. 4x + 2y = 10 10x – 2y = 8 Add these equations to eliminate y: 14x = 18 ⇒ x = 18/14 ⇒ x = 9/7 Substitute x = 9/7 in equation (1): 2x + 0 = 3 ⇒ 4(9/7) + 2y = 10 ⇒ y = 17/7 Thus, by elimination method, x = 9/7 and y = 17/7.

Question 3: Solve 9a + 2b = 6, 4a − 7b = 2 using elimination method.

Given: 9a + 2b = 6 ⇢ (1) 4a − 7b = 2 ⇢ (2) In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Therefore, multiply the first equation by 7,and multiply the second equation by 2. 63a + 14b = 42 8a − 14b = 4 Add these equations to eliminate b: 71a = 46 ⇒ a = 46/71 Substitute a = 46/71 in equation (1): 9(46/71) + 2b = 6 ⇒ b = 6/71 Thus, by elimination method, a = 46/71 and b = 6/71.

Question 4: Solve: 3u + 2t = 8; 5u + 9t = 2 using elimination method.

Given: 3u + 2t = 8 ⇢ (1) 5u + 9t = 2 ⇢ (2) In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Multiply the first equation by 5,and multiply the second equation by −3. 10t + 15u = 40 −27t − 15u = −6 Add these equations to eliminate u: −17t = 34 ⇒ t = −2 Substituting t = −2 in equation (1), we have: 3u + 2(−2) = 8 ⇒ 3u = 12 ⇒ u = 4 Thus, by elimination method, t = −2 and u = 4.

Question 5: Solve 7p + 4q = 7, 8p + 5q = 5 using elimination method.

Given: 3u + 2t = 8 ⇢ (1) 5u + 9t = 2 ⇢ (2) In order to solve the given equations by elimination, the coefficients of one of the variables must be equal. Multiply the first equation by 5, and multiply the second equation by −4. 35p + 20q = 35 −32p − 20q = −20 Add these equations to eliminate q: 3p = 15 ⇒ p = 15/3 ⇒ p = 5 Substituting p = 5 in equation (1), we have: 7(5) + 4q = 7 ⇒ 4q = -28 ⇒ q = -7 Thus, by elimination method, p = 5 and q = -7.

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## Elimination Method Calculator

What are systems of linear equations, what is the elimination method in math, elimination method system of equations steps, how to use this elimination method calculator, how to solve with the elimination method, how to use the elimination method in special situations, elimination method system of equations examples.

Welcome to Omni's elimination method calculator ! It's here to help whenever you need to use the elimination method to solve a system of equations . The other name for this method is the linear combination method . So, what is the elimination method? How to solve a system with the elimination method? Scroll down!

In the article below, we give the definition of the elimination method, explain a little bit of the math behind it, and go step-by-step through several examples of systems solved with the elimination method so that you can grasp all the details. We will also teach you how to do the elimination method when a system has infinitely many solutions or no solutions at all.

We say that an equation is linear if all the variables that appear in that equation are in the first power. In particular, variables cannot be squared or cubed, nor put under a root or in the denominator of a fraction. The only thing you can do to variables is to multiply them by numbers and add such expressions together . If we have several linear equations and want to find numbers that solve all of these equations simultaneously, then we say we want to solve a system of linear equations .

Our elimination method calculator works for systems of two linear equations in two variables . In general, such a system takes the form:

a 1 x + b 1 y = c 1

a 2 x + b 2 y = c 2

- x and y are the variables;
- a 1 , b 1 , c 1 are the coefficients of the first equation; and
- a 2 , b 2 , c 2 are the coefficients of the second equation.

The elimination method is one methods used to solve systems of linear equations . The main idea behind this method is to get rid of one of the variables so that we can focus on a simpler equation . In particular, when we have a system of two linear equations in two variables and eliminate one variable, we are left with a single equation in just one variable!

How do we eliminate variables? We multiply one or both equations by numbers that make the coefficients of a variable become opposite numbers on each equation (e.g., so we get 2x and -2x ). Then we add the equations together - creating a resulting equation that doesn't contain that variable! We can now easily solve this equation using standard methods for solving equations with one variable.

Once we have found the value of this variable, we substitute it into one of the original equations. This way, we obtain another equation with one variable . We solve it and that's it! This is how we use the elimination method to solve the system of equations. Go to the next section to learn more about the elimination method steps.

Let us not forget about other methods for solving systems of linear equations ! Once you have learned how to do the elimination method, you make sure to visit the following Omni tools:

- Substitution method calculator ;
- Reduced row echelon form calculator ; and
- Cramer's rule calculator .

You already know what the elimination method is all about, so let's discuss how to do the elimination method when given a specific system of linear equations in more detail.

If needed, rearrange the equations so that the variables appear in the same order.

If needed, multiply the equations so that one variable can be eliminated by addition.

Add the equations together to eliminate this variable. This is essence of the solving by elimination method!

You get a one-variable equation - solve for this variable.

Substitute the value for this variable into one of the original equations.

Solve for the other variable.

Just to be sure, you may want to test your solution . Substitute it into the system and see if everything is OK.

The most critical step (and the only one which can cause problems) is to transform the system in a way that allows for the elimination of a variables , i.e., Step 2. In the next section, we will explain it in more detail and show you a bit of the math behind the elimination method. After that we will move onto discussing several examples of elimination method.

The use of the elimination method calculator is straightforward:

- Enter the coefficients in their respective fields.
- The complete solution appears below the elimination method calculator.
- If you just need the pair of numbers that satisfy the system, they are near the end of the calculator's output.
- All the elimination method steps , along with explanations, are here as well in case you need them.
- If you need the solution computed with a higher precision (number of sig figs), go to the advanced mode of the elimination method calculator to set the desired precision. By default, we display six sig figs.

The best situation you can encounter is when the coefficients of one variable are opposite numbers . When you add the equations together, this variable vanishes!

More often, however, there are no opposite coefficients. It is your task to create them by multiplying both sides of one or both equations by suitably chosen multipliers . Only then will you be able to use the elimination method to solve the system of equations. The main task is to guess the multipliers. A simple example is when the coefficients of a variable are equal - in such case it is enough to multiply one of the equations by -1 . This creates opposite coefficients. Then you only need to add the equations to eliminate this variable.

In general, for the system of equations:

we resort to the notion of the least common multiple of two numbers. Namely, we define the multipliers m 1 and m 2 as follows:

m 1 := LCM(a 1 , a 2 ) / a 1

m 2 := LCM(a 1 , a 2 ) / a 2

and multiply the first equation by m 1 and the second equation by -m 2 . As a result, we get the following system:

LCM(a 1 , a 2 )x + [LCM(a 1 , a 2 )b 1 /a 1 ]y = LCM(a 1 ,a 2 )c 1 /a 1

-LCM(a 1 , a 2 )x - [LCM(a 1 , a 2 )b 2 /a 2 ]y = -LCM(a 1 , a 2 )c 2 /a 2

As you can see, we have created opposite coefficients for the variable x (they are equal to LCM(a 1 , a 2 ) and -LCM(a 1 , a 2 ) ). By adding these equations together, we eliminate x and end up with an equation containing only one variable: y . Solving such single-variable equations is very easy, as you'll see in the examples below.

When trying to eliminate a variable, you may sometimes eliminate both variables ! What to do in such a situation? If you have eliminated both variables, you will end up with a statement concerning numbers. This statement is either true or false. Examples of true statements are:

4 = 4 or 0 = 0 ,

and examples of false statements are:

4 = 5 or 0 = 1 .

It's not difficult, right?

Depending on whether the statement you got is true or false, you can make conclusions about the system :

If you eliminated both variables and the final statement is false , then your system of equations has no solution .

If you eliminated both variables and the final statement is true , then your system has infinitely many solutions .

In this section, we will look at several examples to get a better idea of how to use the elimination method in math to solve systems of equations.

Use the elimination method to solve the system of equations:

3x - 4y = 6

-x + 4y = 2

Eliminate y by adding the two equations together:

Solve for x :

Substitute x = 4 into the second equation:

-4 + 4y = 2

Solve for y :

Solution: x = 4, y = 1.5

We test the solution:

3 ⋅ 4 - 4 ⋅ 1.5 = 12 - 6 = 6

So the first equation is OK.

-4 + 4 ⋅ 1.5 = -4 + 6 = 2

And the second equation is OK as well.

Solve using elimination method :

2x + 3y = 5

2x + 7y = -3

We want to eliminate x this time. To this end, we first multiply the first equation by -1 :

-2x - 3y = -5

Add the equations, which results in eliminating x :

Substitute y = -2 into the first equation:

2x + 3 ⋅ (-2) = 5

Solution: x = 5.5, y = -2

Now, we will see how to solve with the elimination method the following system of linear equations :

3x - 3y = 0

We see that neither variable has equal or opposite coefficients. We will have to create them using multipliers, as we explained above. Let's eliminate x . First, calculate the least common multiplicity of 2 and 3 :

LCM(2, 3) = 6 .

The multipliers are:

m 1 := 6 / 3 = 2 and m 2 := -6 / 2 = -3 .

Hence, we multiply the first equation by 2 and the second equation by -3 :

6x - 6y = 0

-6x - 3y = -9

Add the equations:

Substitute y = 1 into the first equation:

3x - 3 ⋅ 1 = 0

Solution: x = 1, y = 1

Next, let's see how to use the elimination method in case of the system :

6x - 3y = 12

To eliminate y , multiply the second equation by -3 so that the coefficients of y are opposite numbers:

-6x + 3y = -12

We eliminated both variables and arrived at a true statement. Therefore, there are infinitely many solutions for this system of equations!

Finally, let's solve using the elimination method:

-4x + 8y = 5

3x - 6y = -1

To eliminate x , we multiply the first equation by 3 and the second equation by 4 :

-12x + 24y = 15

12x - 24y = -4

We eliminated both variables and arrived at a blatantly false statement. We conclude that our system of equations has no solution.

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Home / United States / Math Classes / 8th Grade Math / Solving System of Linear Equations by Elimination Method

## Solving System of Linear Equations by Elimination Method

A system of linear equations is a collection of two or more linear equations that relates multiple variables. Here we wi ll learn how to solve a system of linear equations equations using the elimination method. We will also look at some solved examples to understand the steps involved in the elimination method. ...Read More Read Less

## About System of Linear Equations Using Elimination Method

- What is the Expression?
- What are Equations?

## Linear Equation

Solving linear equations by elimination method, steps in the elimination method.

- Solved Examples
- Frequently Asked Questions

## What is an expression?

Expressions are mathematical statements that contain at least two terms containing numbers or variables , or both, and are connected by an operator. Addition, subtraction, multiplication, and division are some examples of mathematical operators.

For example, 3x + 5 is an expression in which 3x and 5 are terms separated by an addition operator.

## What are equations?

Equations are mathematical statements that have algebraic expressions on both sides of an equal sign (=). It depicts the equality relationship between the expressions written on the left-hand side and the expressions written on the right-hand side. L.H.S = R.H.S (left hand side = right hand side) appears in every math equation.

Check whether these are expressions or equations:

Linear equations are mathematical equations with 1 as the highest degree of the variables. In other words, the highest exponent of the terms in such equations is 1. These can be further divided into one-variable linear equations, two-variable linear equations, three-variable linear equations, and so on.

A linear equation with variables x and y has the standard form ax + by – c = 0 , where a and b are the coefficients of x and y, and c is the constant.

One of the methods of solving a system of linear equations is the elimination method . To get the equation in one variable using this method, we can either add or subtract the equations. We can add the equations to eliminate a variable if the coefficients of one of the variables are the same, and the sign of the coefficients is the opposite.

We can also subtract the equations to get the equation in one variable, if the coefficients of one of the variables are the same, and the sign of the coefficients is the same.

If we don’t have the equations in a form to directly add or subtract the equations to eliminate the variable, we can start by multiplying one or both equations by a constant value on both sides of the equation.

We do this to get an equivalent linear system of equations, after which, we can eliminate the variable by simply adding or subtracting the equations.

Step 1: To make the coefficients of any one of the variables (either x or y) numerically equal, multiply both the given equations by some suitable non-zero constants. TIP: Correlate this with the Least Common Multiple (LCM) .

Step 2: After that, add or subtract one equation from the other so that one variable is eliminated.

Step 3: To find the value of one variable (x or y), solve the equation in one variable (x or y).

Step 4: To get the value of another variable, substitute this value into any of the given equations.

## Solved System of Linear Equation Examples by Elimination Method

Solve the system by elimination.

x + 5y = 2 Equation 1.

x – 5y = 16 Equation 2.

Step 1: The coefficients of the y-terms are opposite in terms of the signs, as you can see. As a result, you can add the equations to get a single variable equation for x.

x + 4y = 2 Equation 1.

2x – 4y = 16 Equation 2.

3x = 18 Add the equations.

Step 2: Solve for x .

3x = 18 Equation from step 1.

x = 6 Divide each side by 3.

Step 3: Substitute 6 for x in one of the original equations and solve for y .

x + 4y = 2 Equation 1

6 + 4y = 2 Substitute 6 for x .

4y = -4 Subtract 4 from each side.

y = -1 Divide each side by 4.

Hence, the required solution is (6,-1).

-8x + 5y = 25 Equation 1

-2x – 4y = 14 Equation 2

Step 1: It’s important to note that no two similar terms have the same or opposite coefficients. Multiply equation 2 by 4, so that the x terms have a coefficient of -8.

-8x + 5y = 23 -8x + 5y = 23 Equation 1

-2x-3y=10 Multiply by 4 . -8x – 12y = 40 Revised equation 2

Step 2: Subtract the equations to obtain an equation in one variable, y .

-8x + 5y = 23 Equation 1.

-8x – 12y = 40 Revised equation 2.

17y = -17 Subtract the equations.

Step 3: Solve for y .

17y = -17 Equation from step 2

y = -1 Divide each side by 17.

Step 4: Substitute -1 for y in one of the original equations and solve for x .

-2x – 3y = 10 Equation 2.

-2x – 3 – 1 = 10 Substitute -1 for y .

-2x + 3 = 10 Multiply.

-2x = 7 Subtract 3 from each side.

x = – 7 2 = -3.5 Divide each side by -2 .

Hence, the required solution is (-3.5,-1).

## What is meant by the elimination method?

The elimination method is a method for removing one of the variables from a system of linear equations by using addition or subtraction in conjunction with multiplication or division of the coefficients of the variables.

## Mention the benefits of applying the elimination method.

The following are some of the benefits of using the elimination method:

1. There are fewer steps in the elimination method than in other methods.

2. When compared to other methods, it reduces the risk of making a mistake while solving a problem.

## What are the various approaches to solving a system of linear equations?

The following are the various methods for solving a system of linear equations:

1. Substitution Method

2. Graphical Method

3. Elimination Method

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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.

Content Continues Below

## MathHelp.com

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.

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## Systems of Linear Equations

A linear equation is not always in the form y = 3.5 − 0.5x ,

It can also be like y = 0.5(7 − x)

Or like y + 0.5x = 3.5

Or like y + 0.5x − 3.5 = 0 and more.

(Note: those are all the same linear equation!)

A System of Linear Equations is when we have two or more linear equations working together.

## Example: Here are two linear equations:

Together they are a system of linear equations.

Can you discover the values of x and y yourself? (Just have a go, play with them a bit.)

Let's try to build and solve a real world example:

## Example: You versus Horse

It's a race!

You can run 0.2 km every minute.

The Horse can run 0.5 km every minute. But it takes 6 minutes to saddle the horse.

How far can you get before the horse catches you?

We can make two equations ( d =distance in km, t =time in minutes)

- You run at 0.2km every minute, so d = 0.2t
- The horse runs at 0.5 km per minute, but we take 6 off its time: d = 0.5(t−6)

So we have a system of equations (that are linear ):

- d = 0.5(t−6)

We can solve it on a graph:

Do you see how the horse starts at 6 minutes, but then runs faster?

It seems you get caught after 10 minutes ... you only got 2 km away.

Run faster next time.

So now you know what a System of Linear Equations is.

Let us continue to find out more about them ....

There can be many ways to solve linear equations!

Let us see another example:

## Example: Solve these two equations:

- −3x + y = 2

The two equations are shown on this graph:

Our task is to find where the two lines cross.

Well, we can see where they cross, so it is already solved graphically.

But now let's solve it using Algebra!

Hmmm ... how to solve this? There can be many ways! In this case both equations have "y" so let's try subtracting the whole second equation from the first:

Now let us simplify it:

So now we know the lines cross at x=1 .

And we can find the matching value of y using either of the two original equations (because we know they have the same value at x=1). Let's use the first one (you can try the second one yourself):

And the solution is:

x = 1 and y = 5

And the graph shows us we are right!

## Linear Equations

Only simple variables are allowed in linear equations. No x 2 , y 3 , √x, etc :

## Common Variables

For the equations to "work together" they share one or more variables:

A System of Equations has two or more equations in one or more variables

## Many Variables

So a System of Equations could have many equations and many variables.

## Example: 3 equations in 3 variables

There can be any combination:

- 2 equations in 3 variables,
- 6 equations in 4 variables,
- 9,000 equations in 567 variables,

When the number of equations is the same as the number of variables there is likely to be a solution. Not guaranteed, but likely.

In fact there are only three possible cases:

- No solution
- One solution
- Infinitely many solutions

When there is no solution the equations are called "inconsistent" .

One or infinitely many solutions are called "consistent"

Here is a diagram for 2 equations in 2 variables :

## Independent

"Independent" means that each equation gives new information. Otherwise they are "Dependent" .

Also called "Linear Independence" and "Linear Dependence"

- 2x + 2y = 6

Those equations are "Dependent" , because they are really the same equation , just multiplied by 2.

So the second equation gave no new information .

## Where the Equations are True

The trick is to find where all equations are true at the same time .

True? What does that mean?

The "you" line is true all along its length (but nowhere else).

Anywhere on that line d is equal to 0.2t

- at t=5 and d=1, the equation is true (Is d = 0.2t? Yes, as 1 = 0.2×5 is true)
- at t=5 and d=3, the equation is not true (Is d = 0.2t? No, as 3 = 0.2×5 is not true )

Likewise the "horse" line is also true all along its length (but nowhere else).

But only at the point where they cross (at t=10, d=2) are they both true .

So they have to be true simultaneously ...

... that is why some people call them "Simultaneous Linear Equations"

## Solve Using Algebra

It is common to use Algebra to solve them.

Here is the "Horse" example solved using Algebra:

The system of equations is:

In this case it seems easiest to set them equal to each other:

d = 0.2t = 0.5(t−6)

Now we know when you get caught!

And our solution is:

t = 10 minutes and d = 2 km

## Algebra vs Graphs

Why use Algebra when graphs are so easy? Because:

More than 2 variables can't be solved by a simple graph.

So Algebra comes to the rescue with two popular methods:

## Solving By Substitution

Solving by elimination.

We will see each one, with examples in 2 variables, and in 3 variables. Here goes ...

These are the steps:

- Write one of the equations so it is in the style "variable = ..."
- Replace (i.e. substitute) that variable in the other equation(s).
- Solve the other equation(s)
- (Repeat as necessary)

Here is an example with 2 equations in 2 variables :

- 3x + 2y = 19

We can start with any equation and any variable .

Let's use the second equation and the variable "y" (it looks the simplest equation).

Write one of the equations so it is in the style "variable = ...":

We can subtract x from both sides of x + y = 8 to get y = 8 − x . Now our equations look like this:

- y = 8 − x

Now replace "y" with "8 − x" in the other equation:

- 3x + 2 (8 − x) = 19

Solve using the usual algebra methods:

Expand 2(8−x) :

- 3x + 16 − 2x = 19

Then 3x−2x = x :

- x + 16 = 19

And lastly 19−16=3

Now we know what x is, we can put it in the y = 8 − x equation:

- y = 8 − 3 = 5

And the answer is:

x = 3 y = 5

Note: because there is a solution the equations are "consistent"

Check: why don't you check to see if x = 3 and y = 5 works in both equations?

## Solving By Substitution: 3 equations in 3 variables

OK! Let's move to a longer example: 3 equations in 3 variables .

This is not hard to do... it just takes a long time !

- z − 3y = 7
- 2x + y + 3z = 15

We should line up the variables neatly, or we may lose track of what we are doing:

WeI can start with any equation and any variable. Let's use the first equation and the variable "x".

Now replace "x" with "6 − z" in the other equations:

(Luckily there is only one other equation with x in it)

2(6−z) + y + 3z = 15 simplifies to y + z = 3 :

Good. We have made some progress, but not there yet.

Now repeat the process , but just for the last 2 equations.

Let's choose the last equation and the variable z:

Now replace "z" with "3 − y" in the other equation:

−3y + (3−y) = 7 simplifies to −4y = 4 , or in other words y = −1

Almost Done!

Knowing that y = −1 we can calculate that z = 3−y = 4 :

And knowing that z = 4 we can calculate that x = 6−z = 2 :

x = 2 y = −1 z = 4

Check: please check this yourself.

We can use this method for 4 or more equations and variables... just do the same steps again and again until it is solved.

Conclusion: Substitution works nicely, but does take a long time to do.

Elimination can be faster ... but needs to be kept neat.

"Eliminate" means to remove : this method works by removing variables until there is just one left.

The idea is that we can safely :

- multiply an equation by a constant (except zero),
- add (or subtract) an equation on to another equation

Like in these examples:

## WHY can we add equations to each other?

Imagine two really simple equations:

x − 5 = 3 5 = 5

We can add the "5 = 5" to "x − 5 = 3":

x − 5 + 5 = 3 + 5 x = 8

Try that yourself but use 5 = 3+2 as the 2nd equation

It will still work just fine, because both sides are equal (that is what the = is for!)

We can also swap equations around, so the 1st could become the 2nd, etc, if that helps.

OK, time for a full example. Let's use the 2 equations in 2 variables example from before:

Very important to keep things neat:

Now ... our aim is to eliminate a variable from an equation.

First we see there is a "2y" and a "y", so let's work on that.

Multiply the second equation by 2:

Subtract the second equation from the first equation:

Yay! Now we know what x is!

Next we see the 2nd equation has "2x", so let's halve it, and then subtract "x":

Multiply the second equation by ½ (i.e. divide by 2):

Subtract the first equation from the second equation:

x = 3 and y = 5

And here is the graph:

The blue line is where 3x + 2y = 19 is true

The red line is where x + y = 8 is true

At x=3, y=5 (where the lines cross) they are both true. That is the answer.

Here is another example:

- 2x − y = 4
- 6x − 3y = 3

Lay it out neatly:

Multiply the first equation by 3:

0 − 0 = 9 ???

What is going on here?

Quite simply, there is no solution.

And lastly:

- 6x − 3y = 12

0 − 0 = 0

Well, that is actually TRUE! Zero does equal zero ...

... that is because they are really the same equation ...

... so there are an Infinite Number of Solutions

And so now we have seen an example of each of the three possible cases:

## Solving By Elimination: 3 equations in 3 variables

Before we start on the next example, let's look at an improved way to do things.

Follow this method and we are less likely to make a mistake.

First of all, eliminate the variables in order :

- Eliminate x s first (from equation 2 and 3, in order)
- then eliminate y (from equation 3)

So this is how we eliminate them:

We then have this "triangle shape":

Now start at the bottom and work back up (called "Back-Substitution") (put in z to find y , then z and y to find x ):

And we are solved:

ALSO, we will find it is easier to do some of the calculations in our head, or on scratch paper, rather than always working within the set of equations:

- x + y + z = 6
- 2y + 5z = −4
- 2x + 5y − z = 27

Written neatly:

First, eliminate x from 2nd and 3rd equation.

There is no x in the 2nd equation ... move on to the 3rd equation:

Subtract 2 times the 1st equation from the 3rd equation (just do this in your head or on scratch paper):

And we get:

Next, eliminate y from 3rd equation.

We could subtract 1½ times the 2nd equation from the 3rd equation (because 1½ times 2 is 3) ...

... but we can avoid fractions if we:

- multiply the 3rd equation by 2 and
- multiply the 2nd equation by 3

and then do the subtraction ... like this:

And we end up with:

We now have that "triangle shape"!

Now go back up again "back-substituting":

We know z , so 2y+5z=−4 becomes 2y−10=−4 , then 2y=6 , so y=3 :

Then x+y+z=6 becomes x+3−2=6 , so x=6−3+2=5

x = 5 y = 3 z = −2

Check: please check for yourself.

## General Advice

Once you get used to the Elimination Method it becomes easier than Substitution, because you just follow the steps and the answers appear.

But sometimes Substitution can give a quicker result.

- Substitution is often easier for small cases (like 2 equations, or sometimes 3 equations)
- Elimination is easier for larger cases

And it always pays to look over the equations first, to see if there is an easy shortcut ... so experience helps.

## Solving Systems of Equations by Elimination: Explanation, Review, and Examples

- The Albert Team
- Last Updated On: March 25, 2022

Elimination is a method of simplification. Trying on clothes at a store helps us eliminate options and decide what to purchase. We eliminate items on a to-do list throughout the day to simplify our schedules. In mathematics, we use the elimination method for solving systems of equations to eliminate one variable and solve for the other.

This post will explain the process for solving systems of equations by elimination.

## Solving Systems of Linear Equations by Elimination

Substitution and elimination are two ways to solve systems of linear equations algebraically. In general, substitution is the best choice when one equation has a variable isolated. Thus, all things being equal, we would probably use substitution if our system of equations were:

So, when is it best to use the elimination method? We might use the elimination method to solve the system of equations if rewriting the equations to isolate a variable is more complicated.

Consider the following system:

It’s clear that if we try to solve one of these equations for either x or y , we will end up with fractions. Fractions can complicate the process of substitution. So, with this system of equations, the better option would be to solve the system of equations by elimination.

## What is the Elimination Method?

So, how do you solve systems of equations by elimination? Let’s start our overview of how to solve a system of equations by elimination with a list of steps.

- Determine which variable will be eliminated. If necessary, rewrite one or both equations to make the coefficients of that variable additive inverses.
- Combine the equations using addition. The new equation should have only one remaining variable.
- Solve the new equation for the remaining variable.
- Substitute the value for that variable into one of the original equations and solve.
- Check your work by graphing or substitution.

Check out this video for a full walkthrough of the elimination method in action.

Next, let’s put these steps to work with some examples of the elimination method.

## Solving Systems of Equations by Elimination Examples

Let’s take a closer look at the system below.

First, notice that the coefficients for the y terms are 5 and -5 . These two numbers are additive inverses , which means they have a sum of 0 . Therefore, if we were to add these equations, the y terms will cancel each other out.

But is adding equations a legitimate means to solve a system of equations? Consider the Addition Property of Equality , which states:

We’re going to use this property to solve systems of linear equations by elimination.

To start, let’s think about the second equation in two parts. We could separate each part and rewrite them as follows:

We know these statements are true because 6x-5y=3.5 . Therefore if one side of this equation equals z , then the other equals z as well.

Now, we can use the Addition Property of Equality to rewrite the first equation in our system, 2x+5y=12.5 , as:

…and then substitute in our values of z to get:

We’ve just added our two equations! When we simplify our new equation, we’ll eliminate the y term so that we can solve for x .

Next, we can substitute 2 for x in one of our original equations and solve for y .

So, the solution to our system is (2,1.7) . We can check our work by graphing, as shown below.

## Solving Systems of Equations by Elimination with Multiplication

What happens if the coefficients of a variable aren’t additive inverses in a system of equations? For example, consider the system below.

If we add these equations as they are now, the y terms would become -2y+(-2y)=-4y and the x terms would become 4x+7x=11x . We won’t have eliminated any variables.

However, if we multiply one of the equations by the factor -1 , we end up with coefficients that are additive inverses. Let’s try it:

Now, we can add our equations to eliminate the y variable and solve for x .

To solve other systems of equations by elimination, we’ll need to rewrite both equations to eliminate a variable. For example, consider the system below.

We could multiply the first equation by -2 and the second equation by 3 to allow us to eliminate x when we add the equations. Let’s use this process to solve the system by elimination.

## More Complex Systems of Equations with Multiplication

Finally, we’ll encounter systems where we need to rewrite equations more extensively to eliminate a variable. For example, consider the system below.

Before we add these equations, we need to move the y term to the opposite side of the equals sign in the first equation and multiply the second equation by -2 .

Now, we can eliminate y terms and solve for x .

To summarize, the key to effectively rewriting the equations is selecting the correct factor to multiply one or both equations to create additive inverses. Additive inverses will cancel each other out and thus eliminate that variable when the equations are added.

Use the link here to access more solving systems of linear equations by elimination practice problems.

## Solving Systems of Equations by Elimination: Keys to Remember

- We use elimination to eliminate one variable in a two-variable system.
- One or both equations may need to be manipulated to make the coefficients of the same variable additive inverses.
- Since additive inverses sum to zero, adding the equations will eliminate one variable and allow us to solve for the other.
- Once the value of one variable is known, we substitute that value into one of the original equations and solve for the other variable.
- Finally, solutions can be confirmed algebraically or by graphing.

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The elimination method is a technique for solving systems of linear equations. Let's walk through a couple of examples. Example 1 We're asked to solve this system of equations: \begin {aligned} 2y+7x &= -5\\\\ 5y-7x &= 12 \end {aligned} 2y +7x 5y −7x = −5 = 12

Elimination Calculator gives you step-by-step help on solving systems by elimination. What do you want to calculate? Calculate it! Example (Click to try) x+y=5;x+2y=7 Try it now Enter your equations separated by a comma in the box, and press Calculate! Or click the example. About Elimination

Solve by elimination: { 2x + y = 7 3x − 2y = − 7. Solution: 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. 2(2x + y) = 2(7) 4x + 2y = 14

An old video where Sal introduces the elimination method for systems of linear equations. Created by Sal Khan. Questions Tips & Thanks Want to join the conversation? Sort by: Top Voted Rizelle 10 years ago for the first problem... the 4y= -8........ where did the -8 came from? • 5 comments ( 42 votes) Flag

HOW TO SOLVE A SYSTEM OF EQUATIONS BY ELIMINATION. Exercise 5.3. 4 Exercise 5.3. 5 Exercise 5.3. 6 Exercise 5.3. 7 Exercise 5.3. 8 Exercise 5.3. 9 Exercise 5.3. 10 Exercise 5.3. 11 Exercise 5.3. 12 Exercise 5.3. 13 Exercise 5.3. 14 Exercise 5.3. 15 Exercise 5.3. 16 Exercise 5.3. 17 Exercise 5.3. 18 Exercise 5.3. 19 Exercise 5.3. 20 Exercise 5.3. 21

High School Math Solutions - Systems of Equations Calculator, Nonlinear. In a previous post, we learned about how to solve a system of linear equations. In this post, we will learn how... Save to Notebook!

Step 1 is to.. Home Algebra Linear Equations Systems How to Solve by Elimination How to solve systems of equations by Elimination Step by step tutorial for systems of linear equations (in 2 variables) Table of contents top Video What is Elimination? Practice Problems Systems of Linear Equations Worksheets What is a System?

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.

How to Solve a System of Linear Equations Using The Elimination Method (aka The Addition Method, aka The Linear Combination Method) Step 1 1 : Add (or subtract) a multiple of one equation to (or from) the other equation, in such a way that either the x x -terms or the y y -terms cancel out.

Step 1: Firstly, multiply both the given equations by some suitable non-zero constants to make the coefficients of any one of the variables (either x or y) numerically equal. Step 2: After that, add or subtract one equation from the other in such a way that one variable gets eliminated. Now, if you get an equation in one variable, go to Step 3.

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The elimination method for solving linear systems Do excercises Show all 2 exercises 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.

Solve the simple equation that arises from it. I got x = 4 x = 4 by dividing both sides by − 9 −9. The next obvious step is to solve for the other variable y y using back substitution. Pick any of the original equations, plug x = 4 x = 4, and you will get y y in no time. The answer is y = - \,1 y = -1.

Let's solve a few more systems of equations using elimination, but in these it won't be kind of a one-step elimination. We're going to have to massage the equations a little bit in order to prepare them for elimination. So let's say that we have an equation, 5x minus 10y is equal to 15. And we have another equation, 3x minus 2y is equal to 3.

We can solve a system of linear equations using the method of elimination. The elimination method has us eliminate one of the variables from the system in order to solve for the value of the other variable, and then turn around and use that value to find the value of the eliminated variable.

Steps to solve a linear equation using elimination method Make sure that the linear equations are of the form ax + by = m and cx + dy = n. In order to solve the given equations by elimination, the coefficients of one of the variables in both equations must be equal.

Our elimination method calculator works for systems of two linear equations in two variables. In general, such a system takes the form: a1x + b1y = c1 a2x + b2y = c2 where: x and y are the variables; a1, b1, c1 are the coefficients of the first equation; and a2, b2, c2 are the coefficients of the second equation.

9 years ago. A rectangular matrix is in echelon form if it has the following three properties: 1. All nonzero rows are above any rows of all zeros. 2. Each leading entry of a row is in a column to the right of the leading entry of the row above it. 3. All entries in a column below a leading entry are zeros.

Schedule a free class About System of Linear Equations Using Elimination Method What is the Expression? What are Equations? Linear Equation Solving Linear Equations by Elimination Method Steps in the Elimination Method Solved Examples Frequently Asked Questions What is an expression?

2 x + y = 9 3 x − y = 16 Advertisement 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 .)

There can be many ways to solve linear equations! Let us see another example: Example: Solve these two equations: x + y = 6; −3x + y = 2; ... Solving By Elimination: 3 equations in 3 variables. Before we start on the next example, let's look at an improved way to do things.

Example of solving by addition: 2 x + y = 9 and 5 x − y = 5. Add the two equations together to eliminate y. Solve the new equation for x. Substitute x = 2 into one of the equations and solve for y. Check the solution by plugging it into both equations. Both equations are true, so ( 2, 5) is the solution.

We might use the elimination method to solve the system of equations if rewriting the equations to isolate a variable is more complicated. Consider the following system: 2x+5y=12.5 2x+5y = 12.5. 6x-5y=3.5 6x−5y = 3.5. It's clear that if we try to solve one of these equations for either x x or y y, we will end up with fractions.