The document discusses methods for solving systems of linear equations in two variables, specifically the elimination method. It provides examples of using the elimination method to solve sample systems of linear equations. The key steps are: 1) rewriting the equations so coefficients of the variable being eliminated are opposite, 2) adding/subtracting the equations to eliminate one variable, 3) solving the resulting equation for the eliminated variable, and 4) substituting back into one of the original equations to solve for the other variable. Five sample systems are provided and the reader is prompted to try solving them using the elimination method.

1. .
 Teacher nilvett ferreira

3. LINEAR EQUATIONS IN TWO VARIABLES ARE OF THE
FORM
2x + 3y = 7
14x + 3y = -9
3x - 7y = 11
6x + y = 2

4. Since the topic deals with a pair of linear equations in two
variables , we are going to study two equations at a time.
Solving a pair of linear equations in two
variables means finding the value of x and
finding the value of y

5. Methods of solving a pair of linear equations in
two variables.
1.Elimination method
2.Substitution method
3.Cross multiplication method

6. Solve by elimination method
7x +5y =12
4x + 3y = 9
(1)
(2)
28x + 20y =48
28x + 21y = 63
(-) (-) (-)
-y =-15
y = 15
By using the value of y in equation 1)
The solution is x = -9, y= 15
x 4
x 7
7x + 5y = 12
28x + 20y =48 (3)
(4)
Eqn (3) –eqn (4)
(Taking the opposite coefficients
of x)
Solve by elimination method
7x +5y =12
+ 3y = 9
28x + 21y = 63
7x + 5(15) = 12
7 x + 75 = 12
7 x = 12 - 75
7 x = -63
x = -63
7 , x = -9
Eg 1

7. Solve by elimination method
7x +5y =12
4x + 3y = 9
(1)
(2)
21x + 15y =36
20x + 15y = 45
(-) (-) (-)
x = - 9
By using the value of x in equation 2)
The solution is x = -9, y= 15
x 3
x 5
4x + 3y = 9
21x + 15y =36 (3)
(4)
Eqn (3) –eqn (4)
(Taking the opposite coefficients
of y)
Solve by elimination method
7x +5y =12
4x + 3y 9
20x + 15y = 45
4(-9) + 3y = 9
-36 + 3y = 9
3y = 9 + 36
3 y = 45
y = 45 ,
3 y = 15

8. Solve by elimination method
7x +15y =20
x + 2y = 3
(1)
(2)
7x + 15y =20
7x + 14y = 21
(-) (-) (-)
y =-1
By using the value of y in equation 2)
The solution is x =5, y= -1
x 1
x 7
x + 2y = 3
7x + 15y =20 (3)
(4)
Eqn (3) –eqn (4)
(Taking the opposite coefficients
of x)
Solve by elimination method
7x +15y =20
x + 2y = 3
7x + 14y = 21
x + 2(-1) = 3
x -2 = 3
x = 3 + 2
x = 5
Eg. 2

9. Solve by elimination method
3x +10y = -14
8x - 3y = 22
(1)
(2)
89x = 178
x = 178
89
By using the value of x in equation 1)
The solution is x = 2, y= -2
x 3
x 10
3x + 10y = -14
9x + 30y = -42 (3)
(4)
(Taking the opposite coefficients
of y)
Solve by elimination method
3x +10y = -14
8x - 3y 22
Eg 3
x = 178
89 1
2
x = 2
80x - 30y = 220
3(2) + 10y = -14
6 + 10y = -14
10y = -14 -6
10y = -20
y = -20 ,
10 y = -2
(No need to change the sign since
y coefficients are already opposite)

10. Solve by elimination method ( Try to solve )
(i) 3x + 4y = 5
5x – 2y = -9
(ii) 4x + 3y = 17
5x -2y = 4
(iii) 5x + 7y = -1
2x - 3y = 17
(iv) 2x - 3y = 12
4x - 5y = 22
(v) 7x - 3y = 11
5x - 2y = 8

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