10th CBSE maths solution for exercise 3.4 part 2

This page 10th cbse maths solution for exercise 3.4 part 2 is going to provide you solution for every problems that you find in the exercise no 3.4

10th CBSE maths solution for Exercise 3.4 part 2

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

(3 x + 4 y)/6 = -1

3 x + 4 y = -6  ---------(1)

x – (y/3) = 3

(3 x – y)/3 = 3

3 x – y = 9  ---------(2)

Elimination method:

  Now we are going to subtract these two equations to eliminate y


                   5 y = 15

                    y = 15/5

                      y = 3

Apply y = 3 in first or second equation to get the value of y

                 3 x – y = 9

                 3 x - 3 = 9

                 3 x = 9 + 3

                  3 x = 12

                    x = 12/3

                    x = 4

Therefore solution is x = 4 and y = 3

Substitution method:

3 x + 4 y = -6  ---------(1)

 3 x – y = 9  ---------(2)

Step 1:

Find the value of one variable in terms of another variable

-y = 9 – 3 x

 y = 3x - 9

Step 2:

Substitute this value of y in the other equation, and reduce it to an equation in one variable.

          3 x + 4 (3x-9) = -6

          3 x + 12 x – 36 = - 6

              15 x – 36 = -6

                  15 x = -6 + 36

                    15 x = 30  

                        x = 30/15

                        x = 2

Step 3:

Apply x = 2 in the equation y = 3x - 9

                y = 3 (2) – 9

                y = 6 - 9

                y = -3

Therefore solution is x = 2 and y = -3


(2) Form the pair of linear equations in the following problems, and find their solutions (if they exists) by the elimination method:

(i) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1, it becomes 1/2 if we only add 1 to the denominator. What is the fraction?

Solution:

Let “x/y” be the required fraction

x – Numerator

y – Denominator

(x + 1)/(y – 1) = 1

 (x + 1) =  (y - 1)

 x – y = -1- 1

x – y = -2 ---------(1)

x/(y  + 1) = ½

 2 x = (y + 1)

 2 x – y = 1---------(2)

Subtract the second equation from the first equation

        x = 3

Substitute x = 3 in the second equation

 2 (3) – y = 1

 6 – y = 1

-y = 1 – 6

 -y = -5

  y = 5

 Therefore the required fraction is 3/5


(ii) Five years ago, Nuri was thrice as old as sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?

Let “x” be the present age of Nuri

Let “y” be the present age of Sonu

Five years ago:

Nuri’s age = x – 5

Sonu’s age = y – 5

(x – 5) = 3 (y – 5)

x – 5 = 3 y – 15

x – 3y = -15 + 5

 x – 3 y = -10 -----(1)

Ten years later:

Nuri’s age = x + 10

Sonu’s age = y + 10

(x + 10) = 2 (y + 10)

x + 10 = 2 y + 20

x – 2 y = 20 – 10

 x – 2 y = 10 ------(2)

Subtract the second equation from the first equation

     y = 20

Substitute y = 20 in the first or second equation to get the value of x

 x - 2 (20) = 10

 x – 40 = 10

   x = 10 + 40

   x = 50




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