10th cbse maths solution for exercise 3.6 part 3

This page 10th cbse maths solution for exercise 3.6 part 3 is going to provide you solution for every problems that you find in the exercise no 3.6

10th CBSE maths solution for Exercise 3.6 part 3

(vii) 10/(x + y) + 2/(x –y) = 4

         15/(x + y) – 5/(x – y) = -2

Solution:

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

      10 a + 2 b = 4  --------(1)

      15 a – 5 b = -2  --------(2)

(1) x 5 => 50 a + 10 b = 20

(2) x 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) + 2 b = 4

2 + 2 b = 4

      2 b = 4 – 2

    2 b = 2

       b = 2/2

        b = 1

1/(x + y) = 1/5

5 = x + y

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

1/(x – y) = 1

x – y = 1 -------(4)

       x + y = 5

      x – y = 1

    ---------------

       2 x = 6

          x = 3

Substitute x = 3 in the third equation

3 + y = 5

      y = 5- 3

     y = 2


 (viii) 1/(3 x + y) + 1/(3 x – y) = 3/4

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

Solution:

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

a + b = 3/4

4(a  + b) = 3

 4 a + 4 b = 3 ------(1)

(a/2) –(b/2) = -1/8

4 a – 4 b = -1------(2)

        4a + 4 b = 3

       4 a – 4 b = 1

     ------------------

          8 a = 4

  a = 1/2

Substitute a = 1/2 in the first equation

 4(1/2) + 4 b = 3

2 + 4 b = 3

     4 b = 3 - 2

     4 b = 1

       b = 1/4

1/(3 x + y) = 1/2

2 = 3 x + y

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

1/(3 x – y) = 3/4

4 = 3 x – y

3 x – y = 4 --------(4)

         3 x + y = 2

         3 x – y = 4

        --------------

          6 x  = 6

              x = 1

Substitute x = 1 in the third equation

3 (1) + y = 2

        y = 2 – 3

        y = -1

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