**Example Problems of Elimination Method :**

In this section, we will see some example problems using the concept elimination method.

General form of linear equation in two variables is ax + by + c = 0

**Procedure for elimination method :**

- Multiply one or both of the equations by a suitable number(s) so that either the coefficients of first variable or the coefficients of second variable in both the become numerically equal.
- Add both the equations or subtract one equation from the other, as obtained in step 1, so that the terms with equal numerical coefficients cancel mutually.
- Solve the resulting equation to find the value of one of the unknowns.
- Substitute this value in any one of the two equations to find the value of the other unknown.

**Question 1 :**

Solve the following system of linear equations by elimination method

x + 2y = 7, x – 2y = 1

**Solution :**

x + 2y = 7 --------- (1)

x – 2y = 1 --------- (2)

The coefficients of x and y are equal in both the equations.

(1) + (2)

2x = 8

x = 8/2

x = 4

By applying the value of x = 4 in (1), we get

4 + 2y = 7

2y = 7 – 4

2 y = 3

y = 3/2

Hence the solution is (4, 3/2)

**Verification :**

Applying the value of x and y in any one of the equations, we get

x + 2y = 7

x = 4 and y = 3/2

4 + 2(3/2) = 7

4 + 3 = 7

7 = 7

**Question 2 :**

Solve the following system of linear equations by elimination method

3x + y = 8 , 5x + y = 10

**Solution :**

3x + y = 8 --------- (1)

5x + y = 10 --------- (2)

The coefficients of y are same.

(1) - (2)

-2x = -2

x = 1

By applying the value of x in (1), we get

3(1) + y = 8

y = 8 – 3

y = 5

**Verification :**

By applying the value of x and y in (1), we get

3x + y = 8

3(1) + 5 = 8

3 + 5 = 8

8 = 8

**Question 3 :**

Solve the following system of linear equations by elimination method

x + (y/2) = 4 and (x/3) + 2 y = 5

**Solution :**

2 x + y = 4 ⋅ 2 2 x + y = 8 |
x + 6 y = 5 ⋅ 3 x + 6 y = 15 |

2x + y = 8 --------- (1)

x + 6y = 15 --------- (2)

In order to make the coefficient of x as 2, let us multiply the 2^{nd} equation by 2.

(1) - 2 ⋅ (2)

y - 12y = 8 - 30

-11y = -22

y = 2

By applying the value of y in (2), we get

3 + 6y = 15

6y = 15 - 3

6y = 12

y = 2

Hence the solution is (3, 2).

**Verification : **

2x + y = 8

2(3) + 2 = 8

6 + 2 = 8

8 = 8

After having gone through the stuff and examples, we hope that the students would have understood, how to solve pair of linear equations by using elimination method.

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