**Solving Linear Equations by Elimination Method Examples :**

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

11x - 7y = xy ------ (1)

9x - 4y = 6xy ------ (2)

Divide the entire equation by (xy)

11/y – 7/x = 1 |
9/y – 4/x = 6 |

Let 1/x = a and 1/y = b

-7a + 11b = 1 -------- (3)

-4a + 9b = 6 --------- (4)

(3) ⋅ 4 - (4) ⋅ 7

44b - 63b = 4 - 42

-19b = -38

b = 2

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

-7a + 11(2) = 1

-7a + 22 = 1

-7a = -21

a = 3

x = 1/3 and y = 1/2

Hence the solution is (1/3, 1/2)

**Verification :**

9x - 4y = 6xy

9(1/3) - 4 (1/2) = 6 (1/3)(1/2)

3 - 2 = 1

1 = 1

**Question 2 :**

Solve the following system of linear equations by elimination method

(3/y) + (5/x) = 20/xy ------- (1)

(2/x) + (5/y) = 15/xy ------- (2)

Multiply the entire equations by xy

(3/y) xy + (5/x) xy = (20/xy) xy

(2/x) xy + (5/ y) xy = (15/xy) xy

3x + 5y = 20 ------- (3)

2x + 5y = 15 ------- (4)

(3) - (4)

3x - 2x = 20 - 15

x = 5

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

2(5) + 5y = 15

10 + 5y = 15

5y = 15 - 10

5y = 5

y = 1

Hence the solution is (5, 1)

**Verification :**

By applying the value of x and y in (3), we get

3x + 5y = 20

3(5) + 5(1) = 20

15 + 5 = 20

20 = 20

**Question 3 :**

Solve the following system of linear equations by elimination-method

8x – 3y = 5xy

6x – 5y = - 2xy

8x – 3y = 5xy ------ (1)

6x – 5y = - 2xy ------ (2)

First, we have to divide the first and second equations by xy

8/y – 3/x = 5

6/y – 5/x = -2

Let 1/x = a and 1/y = b

-3a + 8b = 5 ------- (3)

- 5a + 6b = -2 ------- (4)

(3) ⋅ 5 - (4) ⋅ 3

40b - 18b = 25 + 6

22b = 31

b = 31/22

By applying the value of b in (3), we get

-3a + 8(31/22) = 5

-3a + 4(31/11) = 5

-3a = 5 - 124/11

-3a = -69/11

a = 23/11

x = 11/23 and y = 22/31

Hence the solution is (11/23, 22/31).

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

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