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Last updated on October 17, 2025

Elimination Method

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To solve linear equations, we have different methods; one of the common methods used is the elimination method. It is applicable when there are two equations. In this article, we will learn about the elimination method, its steps, and how to solve equations using it.

Elimination Method for US Students
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What is the Elimination Method?

The elimination method is a technique used to solve linear equations by eliminating a variable, either by multiplying or dividing both equations to make the coefficients of one variable equal. Then, using basic arithmetic operations to eliminate a variable; hence, the process is also called the addition-subtraction method.

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What are the Steps to Use Elimination Method?

When solving linear equations using the elimination method, we remove one variable and then solve the equation using addition or subtraction. Follow these steps to use the elimination method:


 

Step 1: Write the equations in standard form

 

Before solving the equations, check if they are in standard form. If not, arrange them in the form ax + by = c.


 
Step 2: Multiplying or dividing the equation


First, we multiply or divide one or both equations by a non-zero number to make the coefficients of a variable equal. 



Step 3: Adding or subtracting the equation to eliminate a variable 

 

Since the coefficients of a variable in both equations are the same, we can add or subtract the equations to eliminate that variable.


 

Step 4: Simplifying the equation


Now we have an equation with one variable, and by solving the equation, we can find the value of this variable. 


 

Step 5: Substitute to find the other variable


To find the value of the other variable, we substitute the value we found in the given equation and solve the equation. 


 

For example, solving 2x + 3y = 12 and 4x - 3y = 6

 

  • Here, both equations are of the form ax + by = c and have one variable that has a common coefficient.
  • That is the coefficients of y are 3 and -3, as they have opposite signs, we add the equations.

 

  1. Adding the equations:

    \((2x + 3y) + (4x - 3y) = 12 + 6 \\ 6x + 0y = 18 → 6x = 18\\ x = 18/6 \\ x = 3\\\)
    So, the value of x = 3
     
  2. Finding the value of y by substituting the value of x in 2x + 3y = 12

    \(2(3) + 3y = 12 \\ 6 + 3y = 12\\ 3y = 12 - 6 \\ 3y = 6\\ y = 6/3 = 2\\\)
    Therefore, x = 3 and y = 2
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Elimination Method: Infinitely Many Solutions

When the equations are coincident lines, which means both equations are on the same line. Then there will be more than one point of intersection, that is, every point on the line satisfies both equations.
 

 

 

For these equations, if we use the elimination method, the answer will be 0 = 0. That means there are infinitely many solutions, as x and y cancel out.

 

Let take an example of linear equation in one variable.
 

  • Equation 1 → 2(x + 3) - 5 = 0
  • Equation 2 → 2x + 1 = 0

 

Since, coefficients of x are equal, we can directly subtract the equation.

 

 

Let's take another example. This time, we will solve linear equations in two variables.

Solving x + 2y = 3 and 2x + 4y - 6 = 0

 

  1. Arranging the equations in the form ax + by = c

    \(x + 2y = 3 ⇒ equation \ 1 \\ 2x + 4y = 6 ⇒ equation \ 2\)
     
  2. Multiplying equation 1 by 2:

    \((x + 2y) × 2 = 3 × 2 \\ 2x + 4y = 6\)
     
  3. Subtract equation 1 from equation 2

    \((2x + 4y) - (2x + 4y) = 6 - 6 \\ 0 = 0\)
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Elimination Method: No Solutions

When two equations of lines are parallel, there are no possible solutions. Solving such equations gives two unequal numbers on both sides of the not equal sign. This eliminates both the variables. 



 

Let's understand it with an example of a linear equation in one variable
 

Equation 1 → 3(x - 2) + 4 = 0

Equation 2 → 3x - 5 = 0


Since, coefficients of x are equal, we can directly subtract the equation.

 

 

Now, let's take another example. This is a more advanced problem since, both the equations are of two variables. 


Suppose there are two lines of equation: x + 2y = 5 and 3x + 6y = 4.

 

  • Step 1: Arranging equation:

    \(x + 2y = 5 ⇒ equation \ 1 \\ 3x + 6y = 4 ⇒ equation \ 2\)

 

  • Step 2: Making the coefficients of x equal.  

    \(3 × (x + 2y = 5) \\ = 3x + 6y = 15 → equation \ 3\)

 

  • Step 3: Subtracting equation 3 from equation 2

    \(⇒ 3x + 6y - (3x + 6y) = 15 - 4 \\ ⇒ 4 = 15\)

 

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Solving a System of 3 Equations Using the Elimination Method

An equation with 3 variables is in the form Ax + By + Cz = D. We can use the elimination method to solve a system of 3 equations. We can learn it with an example. 


Solve the system of equations: x + y + z = 6,


\(2x - y + z = 3, \\ 3x + 2y - z = 4. \)

 

  • Here, the equations are in the standard form: \(Ax + By + Cz = D\)


    \(x + y + z = 6        ⇒ 1\\ 2x - y + z = 3       ⇒ 2 \\ 3x + 2y - z = 4      ⇒3\)

 

  • Eliminating z between equations 1 and 2:

    Subtract equation 1 from 2:

    \((2x - y + z) - (x + y + z) = 3 - 6 \\ x - 2y = -3 \\ x - 2y = -3    ⇒ 4\)

 

  • Eliminating z in 1 and 3

    Adding 1 and 3 to eliminate z

    \((x + y + z) + (3x + 2y - z) = 6 + 4 \\ 4x + 3y = 10   ⇒5\)

 

  • Now we have 2 equations with two variables

    \(x - 2y = -3           ⇒ 4 \\ 4x + 3y = 10       ⇒ 5\)

 

  • Solving 4 and 5: 
    Multiplying equation 4 by 4:

    \(​​​​​​​(x - 2y = -3) × 4 \\ \quad 4x - 8y = -12 \)

    Subtract equation 5 to eliminate x

    \((4x - 8y) - (4x + 3y) = -12 - 10 \\ 0 - 11y = -22 \\ y = {-22 \over -11} = 2\)

     
  • Substituting y in equation 4:

    \(x - 2(2) = -3 \\ x - 4 = -3 \\ x = -3 + 4 \\ x = 1\)

     
  • Substituting x and y in equation 1 

    \(1 + 2 + z = 6  \\ 3 + z = 6\\ z = 6 - 3 \\ z = 3\)

 

Here, x = 1, y = 2, z = 3

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How to Solve Linear Equations Through Elimination Method?

The elimination method is used to solve two or more linear equations; it is not applicable to solve a single equation. A system of linear equations consists of two or more linear equations. The system of linear equations is enclosed in the symbol ‘{’
 

To solve a linear equation using the elimination method, we eliminate one variable to find its value.

 

  • If the equations have the same coefficient of one variable, we simply add or subtract the equations.
     
  • If there are no same coefficients, we multiply or divide the equations and then add or subtract them. 
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Tips and Tricks to Master Elimination Method

To make calculations easy, here are a few tips and tricks for children. 

 

  1. Always multiply or divide on both sides of the equations. 
  2. You can only add or subtract coefficients of the same variables. For example, 2x + 3y can't be added. 
  3. For multiplication, you can use multiplication tables
  4. To check if the given equation of lines are parallel, use the condition: \(\frac{a_1}{a_2} = \frac{b_1}{b_2} \ne \frac{c_1}{c_2}\).
  5. Verify the calculated values of variables by substituting it in the given equation.  

 

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Common Mistakes and How to Avoid Them in Elimination Method

When solving linear equations using the elimination method, errors are common among students, and they tend to repeat the same errors often. Here are some common mistakes and the ways to avoid them in the elimination method. 
 

Mistake 1

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Not aligning the terms
 

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Students sometimes forget to align the terms in the standard form, and without proper order, it is hard to add or subtract from the equation.

For example, if the equation is given in the form 2x - 3 = y, you first need to rearrange the equation in the form ax + by = C, that is, 2x - y = 3. 
 

Mistake 2

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Multiply only one side of the equation
 

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When multiplying the equation to eliminate the value, students regularly multiply only one side of the equation.

For instance, when multiplying 3x + 2y = 10 by 2, they multiply 3x + 2y by 2, but they won't multiply 10 by 2, which is incorrect.

So when multiplying an equation by a number, we should multiply both sides of the equation, that is (3x + 2y = 10) × 2 = 6x + 4y = 20.
 

Mistake 3

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Adding the equations when they should be subtracted 
 

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Confusing whether to add or subtract from the equation is common among students.

For example, to eliminate the value of x in (x + y = 6) and (x - y = 2), students add the equations instead of subtracting, which is wrong.

Always check whether to add or subtract the equations based on their signs.

Mistake 4

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Errors while substitution
 

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Substitution errors are common. For example, when y = 3 in x + 3y = 9, students sometimes substitute into the wrong equation.

So when substituting, make sure you are placing the correct variable and always try to double-check the value. 

Mistake 5

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Multiplying the equation when it is not required
 

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If the coefficients of a variable are already the same, we can directly add or subtract the equations without multiplying. But students sometimes multiply or divide the equations even if there is a common coefficient, so always check if there are any common coefficients or not. 

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Real-world applications of Elimination Method

In real life, we use the elimination method in fields like economics, budgeting, physics, and so on. Here are some real-world applications of the elimination method:

 

 

  1. To analyze the profit and cost, companies set up a system of equations, and by using the elimination method, we can determine the profit and cost.

     
  2. In chemistry, we use elimination methods to determine the proportions of different compounds needed to achieve the desired mixture. 

     
  3. In physics, to solve the problems related to force, acceleration, and motion, we use the elimination method. It is also applied in equations for electrical circuits.

     
  4. In geometry, elimination method are used to find the common point between two lines. In computer graphics and robotics, such methods are used to create path intersection.

     
  5. In machine learning and AI models, system of equations are used in algorithms such as in linear regression and neural training.
     
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Solved Examples of Elimination Method

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Problem 1

Solve x + y = 10 and x - y = 2

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x = 6 and y = 4
 

Explanation

\(x + y =10 \\ x - y = 2\)

 

  1. To eliminate the y, add the equations:

    \((x + y) + (x - y) = 10 + 2 \\ 2x = 12\\ x = 12/2 \\ x = 6\)
     
  2. Substituting the value of x in x + y = 10

    \(6 +  y = 10  \\ y = 10 - 6\\ y = 4\)

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Problem 2

Solve 2x + 3y = 13 and 4x + 3y = 17

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x = 2 and y = 3
 

Explanation

\(2x + 3y = 13 \\ 4x + 3y = 17\)

 

  1. Subtracting the equations to eliminate y 

    \((4x + 3y) - (2x + 3y) = 17 - 13 \\ 2x = 4\\ x = 4/2 \\ x = 2\)
     
  2. Substituting the value of x in 2x + 3y = 13 to find the value of y
    \(​​​​​​\\2(2) + 3y = 13 \\ 4 + 3y = 13\\ 3y = 13 - 4 \\ 3y = 9\\ y = 9/3\\ y = 3\)

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Problem 3

Find two numbers whose sum is 9 and whose difference is 5

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The two numbers are: 7 and 2

Explanation

To find the numbers, let’s consider the numbers as x and y


\(x + y = 9         ⇒ 1\\ x - y = 5           ⇒ 2\)

 

 

  1. Adding 1 and 2 to eliminate y

    \((x + y) + (x - y) = 9 + 5 \\ 2x = 14\\ x = 14/2 \\ x = 7\)
     
  2. Substituting the value of x in 1 to find y

    \(x + y = 9 ⇒ 7 + y = 9\\ y = 9 - 7\\ y = 2\)

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Problem 4

Find the value of m and n in 9m - 4n = 18 and 5m + 3n = 27?

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Approximately, m = 3.44 and n = 3.24
 

Explanation

Let’s consider, 


\(9m - 4n = 18           ⇒1 \\ 5m + 3n = 27       ⇒2\)


 

 

  1. We multiply equations 1 by 3 and equation 2 by 4 to eliminate n

    \((9m - 4n = 18) × 3 ⇒ 27m - 12n = 54                       ⇒3 \\ (5m + 3n = 27) × 4 ⇒ 20m + 12n = 108                 ⇒4\)

     
  2. Adding 3 and 4

    \((27m - 12n) + (20m + 12n) = 54 + 108\\ 47m = 162\\ m = 162 ÷ 47 ≈ 3.44\\ m ≈ 3.44\)

     
  3. Substituting the value of m in 1

    \(9m - 4n = 18 \\ 9 × 3.44 - 4n = 18 \\ 30.96 - 4n = 18\\ -4n = 18 - 30.96 \\ -4n = -12.96\\ n = -12.96 ÷ -4 = 3.24\)

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Problem 5

The sum of two numbers is 30, and their difference is 10. Find the numbers.

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Here, the numbers are 20 and 10

Explanation

Let the numbers be x and y

\(x + y = 30 \\ x - y = 10\)


 

 

  1. To eliminate the value of y, we add both equations 

    \((x + y) + (x - y) = 30 + 10 \\ 2x = 40\\ x = 40/2\\ x = 20\)

     
  2. Substituting the value of x in x + y = 30

    \(20 + y = 30\\ y = 30 - 20 \\ y = 10 \)

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FAQs on Elimination Method

1.Why is the elimination method necessary for my child?

The elimination method helps children solve complex equations in advanced mathematics and real life problems. 

 

It also helps in the development of the brain, enhancing critical thinking skills and analytical reasoning.


 

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2.What are some common mistakes children make in the elimination method?

Incorrectly multiplying or subtracting, not verifying the answers, not multiplying on both sides are a few mistakes children make.

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3.How can I help my child in studying elimination methods?

Encourage your child to practice problems daily. Remind them to break the problems into small steps. Use real-life examples for better explanation.

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4.What are some real life applications of elimination methods?

Here is an example you can use to explain elimination method to your child

 

  1. Give your child 5 candies from two different candies, (let's say chocolate and mango).
  2. Now take a total of 12 candies, such that you have twice the number of chocolate candies and thrice the number of mango candies.
  3. Now, let's assume your child has x chocolate candies and y mango candies.
  4. Then it can be said that x + y = 5 and 2x + 3y = 12
  5. Now, the number of candies you and your child has can be calculated by solving x + y = 5 and 2x + 3y = 12
  6. Multiplying x+y = 5 by 2 and subtracting both equations, we get,2x + 2y - 10 - 2x -3y + 12 = -y + 2 ⇒ y = 2. 
  7. Substituting this value in x + y = 5, we get x = 3. 
  8. From here we get, your child had 3 chocolate candies and 2 mango candies. And you have 6 chocolate candy and 6 mango candies

 

Now, ask your child to count the candies, you both have to verify the answer. 

 

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5.What are some tips to make the learning elimination method fun and engaging for my child?

Apart from using real life examples, you can create games and fun quizzes, using cards and celebrate each victory along the way for making learning fun. 

 


 

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Jaskaran Singh Saluja

About the Author

Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.

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Fun Fact

: He loves to play the quiz with kids through algebra to make kids love it.

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