# SOLVING LINEAR EQUATIONS BY ELIMINATION METHOD EXAMPLES

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.

## Simultaneous Equations Elimination Method - Examples

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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