Rajasthan Board RBSE Solutions for Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables Ex 3.3 Textbook Exercise Questions and Answers.

## RBSE Class 10 Maths Solutions Chapter 3 Pair of Linear Equations in Two Variables Exercise 3.3

Question 1.

Solve the following pair of linear equations by the substitution method.

(i) x + y = 14

x – y = 4

(ii) s – t = 3

\(\frac{s}{3}\) + \(\frac{t}{2}\) = 6

(iii) 3x – y = 3

9x – 3y = 9

(iv) 0.2x + 0.3y = 1.3

0.4x + 0.5y = 2.3

(v) \(\sqrt{2}\)x + \(\sqrt{3}\)y = 0

\(\sqrt{3}\)x – \(\sqrt{8}\)y = 0

(vi) \(\frac{3 x}{2}-\frac{5 y}{3}\) = -2

\(\frac{x}{3}-\frac{y}{2}\) = \(\frac{13}{6}\)

Solution:

(i) The given pair of linear equations is

x + y = 14 …..(1)

and x – y = 4 (2)

From (2) x = 4 + y …..(3)

Substituting this value of x in equation (1)

4 + y + y = 14

or 2y = 14 – 4

or 2y = 10

or y = \(\frac{10}{2}\) = 5

Substituting this value of y in equation (3),

x = 4 + 5 = 9

Hence, x = 9 and y = 5

(ii) The given pair of linear equations is

s – t = 3 ……(1)

and \(\frac{s}{3}+\frac{t}{2}\) = 6

or \(\frac{2 s+3 t}{6}\) = 6

or 2s + 3t = 36 …(2)

From (1), s = 3 + t …(3)

Substituting this value of 5 in equation (2),

2 (3 + t) + 3t = 36

or 6 + 2t + 3t = 36

or 6 + 5t = 36

or 5t = 36 – 6

or t = \(\frac{30}{5}\) = 6

Substituting this value of t in equation (3),

s = 3 + 6 = 9

Hence, s = 9 and t = 6

(iii) The given pair of linear equations is

3x – y = 3 …(1)

and 9x – 3y = 9 …(2)

From (1),

3x – 3 = y

or y = 3x – 3 …(3)

Substituting this value of y in equation (2),

9x – 3 (3x – 3) = 9

or 9x – 9x + 9 = 9

This statement is true for all values of x. So we cannot find any specific value of x in the form of solution. Therefore we cannot find out any value of y also. This situation has arisen because both the given equation are one and the same.

Hence there are infinitely many solutions of equations (1) and (2).

(iv) The given pair of linear equations is

0.2x + 0.3y = 1.3

Substituting this value of x in equation (2)

4\(\left[\frac{13-3 y}{2}\right]\) + 5y = 23

26 – 6y + 5y = 23

-y = 23 – 26 = -3

y = 3

Substituting this value of y in equation (3)

Hence, x = 2 and y = 3

Question 2.

Solve 2x + 3y = 11 and 2x – 4y = – 24 and hence find the value of ‘m’ for which y = mx +3.

Solution:

The given pair of linear

equations is 2x + 3y = 11 ….(1)

and 2x – Ay = – 24 ….(2)

From (2),

2x = 4y – 24

⇒ 2x = 2 (2y – 12)

⇒ x = 2y – 12 ….(3)

Substituting this value of x in equation (1)

2 (2y – 12) + 3y = 11

or 4y – 24 + 3y = 11

or 7y = 11 + 24

or 7y = 35

or y = \(\frac{35}{7}\) = 5

Substituting this value of y in equation (3)

x = 2 (5) – 12

= 10 – 12 = – 2

Now, take y = mx + 3

Substituting x = – 2, y = 5

5 = m (- 2) + 3

or 5 – 3 = -2 m

or 2 = -2m

or – 2m = 2

or m = – 1

Hence, x = – 2, y = 5 and m = – 1

Question 3.

Form the pair of linear equations for the following problems and find their solution by substitution method.

(i) The difference between two numbers is 26 and one number is three times the other. Find them.

(ii) The larger of two supplementary angles exceeds the smaller by 18 degrees. Find them.

(iii) The coach of a cricket team buys 7 bats and 6 balls for Rs 3800. Later, she buys 3 bats and 5 balls for Rs 1750. Find the cost of each bat and each ball.

(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered. For a distance of 10 km, the charge paid is Rs 105 and for a journey of 15 km, the charge paid is Rs 155. What are the fixed charges and the charge per km? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes \(\frac{9}{11}\), if 2 is added to both the numerator and the denominator. If, 3 is added to both the numerator and the denominator it becomes \(\frac{5}{6}\). Find the fraction.

(vi) Five years hence, the age of Jacob will be three times that of his son. Five years ago, Jacob’s age was seven times that of his son. What are their present ages?

Solution:

(i) Let the two numbers be x and y.

According to the first condition,

x – y = 26 ….(1)

According to the second condition,

x = 3y …(2)

Substituting this value of x in equation (1)

3y – y= 26

or 2y = 26

or y = \(\frac{26}{2}\) = 13

Substituting this value of y in equation (2)

x = 3 × 13 = 39

Hence the two numbers are 39, 13.

(ii) Let the two supplementary angles be

x°, y° and x° > y°.

According to the first condition,

x° + y° = 180° ….(1)

According to the second condition,

x° = y° + 18° ……(2)

Substituting this value of x° in equation (1)

y° + 18° + y° = 180°

or 2y° = 180° – 18°

or 2y° = 162°

or y° = \(\frac{162^{\circ}}{2}\) = 81°

Substituting this value of y° in equation (2)

x° = 81° + 18° = 99°

Hence the required angles are 99°, 81°.

(iii) Let the cost of a bat = ₹ x

and the cost of a ball = ₹ y

According to the first condition,

7x + 6y= 3 800 ….(1)

According to the second condition,

3x + 5y = 1750 ….(2)

From (1), 7x = 3800 – 6y

or x = \(\frac{3800-6 y}{7}\) …(3)

Substituting this value of x in equation (2)

3\(\left[\frac{3800-6 y}{7}\right]\) + 5y = 1750

or \(\frac{11400-18 y+35 y}{7}\) = 1750

or 11400+ 17y = 1750 × 7

or 11400+ 17y = 12250

or 17y = 12250 – 11400

or 17y = 850

or y = \(\frac{850}{17}\) = 50

Substituting this value of y in equation (3)

Hence,cost of one bat = ₹ 500

and cost of one ball = ₹ 50

(iv) Let the fixed charges of the taxi be ₹x and the charge of one km journey = ₹y

According to the first condition,

x + 10y = 105 ….(1)

According to the second condition,

x + 15y = 155 ….(2)

From (1),

x = 105 – 10y ….(3)

Substituting this value of x in equation (2)

105 – 10y + 15y = 155

or 5y = 155 – 105

or 5y = 50

or y = \(\frac{50}{5}\) = 10

Substituting this value of y in equation (3)

x = 105 – 10 × 10

= 105 – 100 = 5

Hence,the fixed charges of the taxi = ₹ 5

and the charge of 1 km journey = ₹ 10

Also, charges of 25 km joumery

= ₹ (10 × 25) + ₹ 5

= ₹ (250 + 5)

= ₹ 255

(v) Let the numerator of the fraction be x and the denominator be y.

∴ Required fraction = \(\frac{x}{y}\)

According to the first condition,

\(\frac{x+2}{y+2}\) = \(\frac{9}{11}\)

or 11 (x + 2) = 9 (y + 2)

or 11x + 22 = 9y + 18

or 11x = 9y + 18 – 22

or 11x = 9y – 4

or x = \(\frac{9 y-4}{11}\) …(1)

According to the second condition,

\(\frac{x+3}{y+3}\) = \(\frac{5}{6}\)

or 6 (x + 3) = 5(y + 3)

or 6x + 18 = 5y + 15

or 6x – 5y = 15 – 18

or 6x – 5y = – 3 ….(2)

Substituting this value of x in equation (1)

or -y – 24 = -3 × 11

or -y = -33 + 24

or -y = -9

or y = 9

Substituting this value of y in equation (1)

x = \(\frac{9 \times 9-4}{11}\) = \(\frac{81-4}{11}\)

= \(\frac{77}{11}\) = 7

Hence, the required fraction is\(\frac{7}{9}\).

(vi) Let the present age of Jacob = x years

and the present age of Jacob’s sun = y years

Five years hence,

Age of Jacob = (x + 5) years

Age of his son= (y + 5) years

According to the first condition,

x + 5 = 3(y + 5)

or x + 5 = 3y + 15

or x = 3y + 15 – 5

or x = 3y + 10 ….(1)

Five years ago

Age of Jacob = (x – 5) years

Age of his son= (y – 5) years

According to the second condition,

x – 5 = 7(y – 5)

or x – 5 = 7y – 35

or x – 7y = – 35 + 5

or x – 7y = – 30 ….(2)

Substituting this value of x in equation (1)

3y + 10 – 7y= – 30

or – 4y = – 30 – 10

or -4y = – 40

or y = 10

Substituting this value of y in equation (1)

x = 3 (10) + 10

= 30 + 10 = 40

Hence,

present age of Jacob = 40 years

and present age of his son = 10 years

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