Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.4 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.4

Question 1.

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

(iv) \(\frac{x}{2}\) + \(\frac{2y}{3}\) = – 1 and x – \(\frac{y}{3}\) = 3

Solution:

(i) The given pair of linear equations is

x + y = 5 ….(1)

and 2x – 3y = 4 …….(2)

Elimination Method—

Multiplying (1) by 2

2x + 2y= 10 ….(3)

Subtracting (2) from equation (3)

or y = \(\frac{6}{5}\)

Substituting this value of y in equation (1)

x + \(\frac{6}{5}\) =5

or x = 5 – \(\frac{6}{5}\)

= \(\frac{25-6}{5}\) = \(\frac{19}{5}\)

Hence, x = \(\frac{19}{5}\) and y = \(\frac{6}{5}\)

Substitution Method—

From (2), 2x = 4 + 3y

or x = \(\frac{4+3y}{2}\) ….(4)

Substituting this value of x in (1)

\(\frac{4+3y}{2}\) + y = 5

or \(\frac{4+3y+2y}{2}\) = 5

or 4 + 5y = 10

or 5y = 10 – 4 = 6

or y = \(\frac{6}{5}\)

Substituting this value of y in (4)

Hence, x = \(\frac{19}{5}\) and y = \(\frac{6}{5}\)

(ii) The given pair of linear equations is

3x + 4y = 10 ….(1)

and 2x – 2y = 2 …..(2)

Elimination Method—

Multiplying equation (2) by 2

4x – 4y = 4 …(3)

Adding equations (3) and (1)

4x – 4y = 4

3x + 4y = 10

7x = 14

or x = \(\frac{14}{7}\) = 2

Substituting this value of x in (1)

3 (2) + 4y = 10

or 6 + 4y = 10

or 4y = 10 – 6

or 4y = 4

or y = \(\frac{4}{4}\) = 1

Hence, x = 2 and y = 1

Substitution Method—

From (2), 2x = 2 + 2y

or x = y + 1 ….(4)

Substituting this value of x in (1)

3 (y + 1) + 4y = 10

or 3y + 3 + 4y = 10

or 7y = 10 – 3

or 7y = 7

or y = 1

Substituting this value of y in (3)

x = 1 + 1 = 2

Hence, x = 2 and y = 1

(iii) The given pair of linear equations is

3x – 5y – 4 = 0 ….(1)

and 9x = 2y + 7

or 9x – 2y – 7 = 0 ….(2)

Elimination Method—

Multiplying equation (1) by (3)

9x – 15y – 12 = 0 ….(3)

Subtracting (2) from equation (3)

or 3x + 4y = -6 …(1)

and x – \(\frac{y}{3}\) = 3

or \(\frac{3 x-y}{3}\) = 3

or 3x – y = 9 …(3)

Elimination Method—

Substituting this value of y in (1)

3x + 4(-3) = -6

or 3x – 12 = -6

or 3x = -6 + 12

or 3x = 6

or x = \(\frac{6}{3}\) = 2

So, x = 2, y = -3

Substitution Method-

From (2), y = 3x – 9 …..(3)

Substituting this value of y in (1)

3x + 4(3x – 9) = -6

or 3x + 12x – 36 = -6

or 15x = -6 + 36

or 15x = 30

or x = \(\frac{30}{15}\) = 2

Substituting this value of x in (3)

y = 3(2) – 9

= 6 – 9 = -3

Hence, x = 2 and y = -3

Question 2.

Form the pair of linear equations in the following problems, and find their solutions (if they exist) 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 \(\frac{1}{2}\) if we only add 1 to the denominator. What is the fraction?

(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?

(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.

(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.

(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.

Solution:

(i) Let the numerator of the fraction = x

and denominator of the fraction = y

∴ Required fraction = \(\frac{x}{y}\)

According to the first condition,

\(\frac{x+1}{y+1}\) = 1

or x + 1 = y – 1

or x – y + 2=0 ……(1)

According to the second condition,

\(\frac{x}{y+1}\) = \(\frac{1}{2}\)

or 2x = y + 1

or 2x – y – 1 = 0 ….(2)

Subtracting equation (1) from equation (2)

2x – y – 1 = 0

x – y + 2 = 0

x – 3 = 0

or x = 3

Substituting this value of x in (2)

2 × 3 – y – 1 = 0

or 6 – y – 1 = 0

or 5 – y = 0

or y = 5

Hence, the required fraction is \(\frac{3}{5}\).

(ii) Let the present age of Nuri = x years

and the present age of Sonu = y years

Five years ago

Age of Nuri = (x – 5) years and

Age of Sonu = (y – 5) years

According to the first condition,

x – 5 = 3(y – 5)

⇒ x – 5 = 3y – 15

⇒ x – 3y + 10 = 0 …(1)

Ten years later

Age of Nuri = (x + 10) years

and Age of Sonu = (y + 10) years

According to the second condition,

x + 10 = 2 (y + 10)

or x + 10 = 2y + 20

or x – 2y – 10 = 0 …(2)

Substracting equation (2) from equation (1)

– y + 20 = 0

or – y = – 20

or y = 20

Subtituting this value of y in (2)

x – 2 (20) – 0 = 0

or x – 40 – 10 = 0

or x = 50

(iii) Let the unit’s digit = x

and ten’s digit = y

∴ Required number = 10y + x

According to the first condition,

x + y = 9 ..(1)

Reversing the order of the digits

unit’s digit = y

and ten’s digit = x

∴ Number = 10x + y

According to the second condition,

9 [10y + x] = 2 [10x + y]

or 90y + 9x = 20x + 2y

or 90y + 9x – 20x – 2y = 0

or – 11x + 88y = 0

or x – 8y = 0 ……(2)

Substituting this value of y in (2)

x – 8 × 1=0

or x = 8

Hence, Required number

= 10y + x

= 10 × 1 + 8 = 18

(iv) Let the number of notes of Rs. 50 Meena received = x

and, the number of notes of Rs. 100 Meena received = y

According to the first condition,

x + y = 25 …. (1)

According to the second condition,

50x + 100y = 2000

or x + 2y = 40 …. (2)

Subtracting equation (1) from equation (2)

Substituting this value of y in (1)

x + 15= 25

or x = 25 – 15 = 10

Hence, the number of notes of Rs. 50 and Rs. 100 Meena received is 10 and 15 respectively.

(v) Let the fixed charge for the first three days = Rs. x

and an additional charge for each day thereafter = Rs. y

In case of Saritha

x + 4y = 27 ….(1)

In case of Susy

x + 2y = 21 ….(2)

Subtracting equation (2) from equation (1)

Substituting this value of y in equation (2)

x + 2 (3)= 21

or x + 6 = 21

or x = 21 – 6 = 15

Hence the fixed charge for first three days and an additional charge for each day thereafter is Rs. 15 and Rs. 3 respectively.

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