Solving Linear Equations Using the Elimination Method

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 Equations in Two Variables

Linear equations in two variables are equations written in the form:

ax + by = c

where 𝑥 and 𝑦 are the variables, and a, b, and c are constants. These equations represent straight lines when graphed on a coordinate plane

Linear equations using the Elimination Method

First of all, the Elimination Method cannot be applied to one single Linear Equation, It can be applied to a pair of Linear Equations also called a System of Linear Equations (i.e. a group of Linear Equations ). A system of Linear Equations is represented as:

Representation of System of Linear Equations (or Pair of Linear Equations)

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 (or System) of linear equations is shown below, followed by the steps it entails:

For example, consider the system of linear equations given in the above image:

• 2x – y = -1 ⇢ (1)
• x + y = 4 ⇢(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 1 and equation (2) by 2 in order to equate the coefficients of x. we get :

• 2x – y = -1 ⇢ (1)
• 2x + 2y = 8 ⇢ (2)

Now subtracting equation (1) from equation (2) , we get :

Subtracting Equation 1 from Equation 2

0 + 3y = 9

3y = 9 ….(dividing both sided by 3)

y = 3 …. (result 1)

Now substitute the result 1 (i.e. y=3) in any of the given Equations, here we will use Equation 2 :

Equation 2 :⇢ x + y = 4 ….(put y=3)

x + 3 = 4

x = 4 – 3

x = 1 ….(result 2)

Thus we got the Solution as : x = 1 and y = 3

Note : The result be same even if you substitute y=3 in Equation 1

Steps to Solve a Linear Equation using Elimination Method

Step 1 : Make sure that both the linear equations are of the form ax + by = c

Step 2 : 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 equation(1) with 1 and equation(2) with 2 in order to eliminate x by changing its coefficient in equation(2) from 1 to 2.

Step 3 : Add or subtract the equations to eliminate the variable with equal coefficients. In the above example, variable x was eliminated.

Step 4 : Solve for the value of the variable left after elimination of one variable is done. In the example above, after eliminating x, the value of y was calculated..

Step 5 : 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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Solution:

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.

Verification of Solution

If at any stage during solving the system of linear equations if you feel like you might have made a mistake and can`t help but feel doubtful OR YOU JUST ASKED YOUR DUMB FRIEND FOR ANSWERS DURING A TEST , and you must re-check your answer before submitting your answer but don`t have enough time to re-solve the entire question from the start. Here is simple trick :

Step 1 : Take the Coordinates(x, y) that you think might be the answer.

Step 2 : Put these Coordinates(x, y) in the given Equations one-by-by and solve.

Step 3 : If the Left-hand side (LHS) becomes equal to the Right-hand side (RHS) for all the equations in the system. So, we can confirm that the Coordinates taken are the correct Solution to the System of Linear Equation. (i.e. for condition LHS = RHS )

For example, Lets consider the following System of Linear Equations

• 2x – y = -1
• x + y = 4

Now lets suppose my friend told me that the answer is x = 1 and y = 3 but I doubt him/her, to confirm the answer I will simply take the Coordinates(x=1 and y=3) and put it in given equations, we get :

Equation 1 (i.e. 2x – y = -1) at x=1 and y=3 is,

2(1) – (3) = -1

2 – 3 = -1

-1 = -1 ….(LHS = RHS)

Equation 2 (i.e. x + y = 4) at x=1 and y=3 is,

(1) + (3) = 4

4 = 4 ….(LHS = RHS)

Now since both the equations satisfies the condition LHS = RHS

Thus we can say the values (i.e. x=1 and y=3) were indeed correct. Therefore the required Coordinates are (1, 3)

Also read:

• What is a System of linear equations
• Linear Equations in One Variable
• Difference between Linear and Non-Linear Equations
• Solve Linear Equations with Variable on both sides
• Introduction to Two-Variable Linear Equations in Straight Lines

Conclusion

The elimination method is a powerful and straightforward technique for solving systems of linear equations. It is particularly useful when the equations can be easily manipulated to eliminate one variable, making the solution process more efficient. Understanding the elimination method and its application is essential for solving linear systems effectively.