SOLVING LINEAR EQUATIONS BY ELIMINATION METHOD

Solving Linear Equations by Elimination Method :

Here we are going to see some example problems of solving linear equations in two variables using elimination method.

Solving Linear Equations by Elimination Method - Practice questions

Question 1 :

Solve by the method of elimination

(i) 2x – y = 3; 3x + y = 7

Solution :

2x – y = 3  ----(1)

3x + y = 7 ------(2)

The coefficient of y in the 1st and 2nd equation are same.

(1) + (2)

         2x – y = 3

        3x + y = 7

       -------------

         5x   =  10

          x  =  10/5  =  2

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

2(2) - y  =  3

4 - y  =  3

y = 4 - 3 

y  =  1

Hence the solution is (2, 1).

(ii) x – y = 5; 3x + 2y = 25

Solution :

x – y = 5 -------(1)

3x + 2y = 25 ------(2)

The coefficient of y in the first equation is 1, the coefficient of y in the second equation is 2. So, we have to multiply the first equation by 2.

       2x - 2y  =  10

       3x + 2y  =  25

     -----------------

       5x  =  35

       x  =  35/5

      x  =  7

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

7 - y  =  5

y  =  7 - 5

y  =  2

Hence the solution is (7, 2).

(iii)  (x/10) + (y/5)  =  14 ; (x/8) + (y/6)  =  15

Solution :

(1) x 3 - (2)

(1) x 3  ==>  3x + 6y  =  420

                  3x + 4y  =  360

                (-)   (-)      (-)

                -----------------

                        2y  =  60

                        y  =  30

By applying the value of y in (1), w get

3x + 6(30)  =  420

3x + 180  =  420

3x  =  420 - 180

3x  =  240

x  =  240/3  =  80

Hence the solution is (80, 30).

After having gone through the stuff given above, we hope that the students would have understood how to solve system of linear equations using elimination method. 

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