SOLVED DIFFERENT PROBLEMS USING ELIMINATION METHOD

Question 1 :

Solve the pair of linear equations using elimination method.

10/(x + y) + 2/(x –y)  =  4 and 15/(x + y) – 5/(x – y)  =  -2

Solution :

Let 1/(x + y) = a  and 1/(x – y) = b

      10 a + 2 b  =  4  --------(1)

      15 a – 5 b  =  -2  --------(2)

(1) ⋅ 5 =>     50 a + 10 b  =  20

(2) ⋅ 2 =>     30 a – 10 b  =  -4                 

                 -------------------------   

                   80 a  =  16

a  =  16/80

a  =  1/5

Substitute a = 1/5 in the first equation

10(1/5) + 2b  =  4

2 + 2b  =  4

2b  =  2

b  =  1

(3) + (4)

      x + y  =  5

      x – y  =  1

    ------------

       2x  =  6

          x  =  3

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

3 + y  =  5

y  =  2

Question 2 :

Solve the pair of linear equations using elimination method.

1/(3x + y) + 1/(3x – y)  =  3/4 and 1/2(3x + y) - 1/2(3x – y)  =  -1/8

Solution :

1/(3x + y)  =  a and  1/(3x – y)  =  b

a + b  =  3/4

4(a  + b)  =  3

 4 a + 4 b  =  3 ------(1)

(a/2) – (b/2)  =  -1/8

4a – 4b  =  -1------(2)

        4a + 4b  =  3

        4a – 4b  =  1

     ------------------

          8 a = 4  ==> a = 1/2

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

4(1/2) + 4b  =  3

2 + 4b  =  3

4 b  =  1

b  =  1/4

         3 x + y = 2

         3 x – y = 4

        --------------

          6 x  =  6

              x  =  1

By applying x = 1 in the (3) equation, we get

3(1) + y  =  2

y  =  2 – 3

y  =  -1

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