3. Pair of Linear Equations in Two Variables Class 10 MATHEMATICS Solutions English Medium-Exercise 3.4

EXERCISE 3.4

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Q1. Solve the following pair of linear equations by the elimination method and the substitution method :

(i) x + y = 5 and 2x - 3y = 4      

(ii) 3x + 4y = 10 and 2x - 2y = 2

(iii) 3x - 5y - 4 = 0 and 9x = 2y + 7

Solution:

Q1. (i) x + y = 5 .......... (i)

        2x - 3y = 4 ......... (ii)

Equation (i) × 3 = 3x + 3y = 15  ...(iii)

Equation (ii) ×1 = 2x - 3y = 4  ... (iv)

[Note: Here the coefficient of y has been equal and the signs are unlike, therefore these will be added]

Now adding Equation (iii) and (iv) 

Solution: (ii) 3x + 4y = 10 ........ (i)

2x - 2y = 2   ........ (ii)

Equation (i) × 1 ⇒ 3x + 4y = 10 ........ (iii)

Equation (ii) ×2 ⇒ 4x -4y = 4........ (iv)

Adding equation (i) and (ii)   

Now putting the value of x = 2 in equation (i)

3x + 4y = 10

⇒ 3(2) + 4y = 10

⇒ 6 + 4y = 10

⇒4y = 10- 6

⇒ 4y = 4

⇒ y = 1

Therefore, solution of given pair of linear equation are x = 2 and y = 1

Solution:(iii) 3x - 5y - 4 = 0

Or       3x – 5y = 4 ......... (i)

         9x = 2y + 7

Or       9x - 2y = 7 ......... (ii)

Equation (i) × 3  ⇒ 9x - 15y = 12 ..... (iii)

Equation (ii) × 1 ⇒ 9x - 2y = 7     .....(iv)

Subtracting equation (iv) from equation (iii)  

Or      3x + 4y = - 6 .......... (i)

Or      3x - y = 9  ............. (ii)

Subtracting equation (ii) from equation (i)

Now, putting y = - 3 in equation (i)  

        3x + 4y = - 6 

⇒   3x + 4(-3) = - 6 

⇒   3x - 12 = - 6 

⇒   3x = 12 - 6 

⇒   3x = 6 

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⇒   x = 2

Therefore, solution of given pair of linear equation are x = 2 and y = - 3

Q2. Form the pair of linear equations in the following problems, and find their solutions (if they exist) by the elimination method :

Solution: Let the numerator be x and denominator of fraction be y.

(Note: Here in equation (i) and (ii) has the equal coefficients of y, therefore it need not to equalize them.)

Now, subtracting equation (ii) f rom equation (i) 

∴  x = 3

Now, putting the value of x = 3 in equation (i)

     x - y = - 2

⇒  3 - y = - 2 

⇒   y = 3 + 2 

⇒   y = 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?

Solution: Let the Nuri’s age be x years  

And Sonu’s age be y years

Situation I  

Five years ago,

Nuri’s age = x - 5 years

Sonu’s age = y - 5 years

According to question,

⇒ x - 5 = 5(y - 5)

⇒ x - 5 = 5y - 25

⇒ x - 5y = 5 - 25

⇒ x - 5y = - 20 ............ (i)

Situation II  

Ten years later,

Nuri’s age = x + 10 years

Sonu’s age = y + 10 years

According to question,

⇒ x + 10 = 2(y + 10)

⇒ x + 10 = 2y + 20

⇒ x - 2y = 20 - 10

⇒ x - 2y = 10  ............ (ii)

(Since the coefficients of x are automatically equal, the coefficients will not equal.)

Now, subtracting equation (ii) from equation (i)

Putting y = 10 in equation (i)

     x -  5y = -  20

Or   x -  5(10) = -  20

Or   x -  50 = -  20

Or   x = 50 -  20

Or x = 30

Therefore Nuri’s age is 30 years and Sonu’s age is 10 years.

(iii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.

Solution: Let the unit digit of the required number be x.

And ten’s digit be y.

Then the real number = 10y + x,

And reversed number = 10x + y 

Situation I

            x + y = 9 ........... (i)

Situation II

            9(number) = 2(reversed number)

⇒      9(10y + x) = 2(10x + y)

⇒       90y + 9x = 20x + 2y

⇒       20x - 9x + 2y - 90y = 0

⇒       11x - 88y = 0

⇒       x - 8y = 0

⇒       x = 8y ........... (ii)

Putting x = 8y in equation (i)  

      x + y = 9

Or   8y + y = 9

Or    9y = 9

Or     y =  = 1

Putting y = 1 in equation (ii)  

x = 8y = 8 × 1 = 8

Therefore, required number = 10y + x

                 = 10 × 1 + 8

                 = 18 Answer

(iv) Meena went to a bank to withdraw Rs 2000. She asked the cashier to give her Rs 50 and Rs 100 notes only. Meena got 25 notes in all. Find how many notes of Rs 50 and Rs 100 she received.

Solution: Let the numbers of 50 notes = x

And number of 100 rupees notes = y

Situation I

Total numbers of notes  = 25

∴ x + y = 25  ........... (i)

Situation II

     x notes of 50 + y notes of 100 = 2000  

Therefore,  50x + 100y = 2000

Or    x + 2y = 40   ........... (ii) (simplifying)

Subtracting equation(ii) from equation (i)  

∴  y = 15

Now putting y = 15 in equation (i)

     x + y = 25

Or   x + 15 = 25

Or   x = 25 - 10

Or   x = 10

∴ Number of 50 notes is 10 and 100 notes is 15.

(v) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid Rs 27 for a book kept for seven days, while Susy paid Rs 21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.

Putting y = 3 in equation (i)

            x + 7y = 27

Or        x + 7(3) = 27

Or    x + 21 = 27

Or    x = 27 - 21

Or    x = 6

Therefore, fixed charge = ₹ 6 and additional charge = ₹ 3