NCERT Solutions Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3.3

Understanding the concepts of NCERT Solutions Class 10, Chapter 3, Exercise 3.3 is essential for students to grasp the fundamental linear equations in two variables. This study guide provides a detailed look at the explanation of each question in the NCERT Solutions Class 10 Maths Chapter 3 Exercise 3.3 PDF, along with step-by-step solutions.

1.0Download NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3 : Free PDF

2.0What is a Linear Equation in Two Variables?

A linear equation in two variables is an algebraic equation that is represented in the form of ax+by+c = 0. Here, a, b, and c are all real numbers, and the coefficients of a and b are not equal to zero. These equations are a fundamental part of mathematics and are used extensively in fields like physics, economics, and engineering. The graph of a linear equation in two variables is a straight line on the Cartesian plane. 

3.0Exercise Overview: Key Concepts 

A linear equation in two variables is an important part of algebra and real-life applications. The NCERT Solutions Class 10 Maths - Chapter 3 Exercise 3.3 focuses on solving simultaneous linear equations using the substitution method. This helps students understand how to simplify and solve equations efficiently. 

Here are the Key Concepts Covered in this Chapter 3 Exercise 3.3

• Solving linear equations using the substitution method
• Real-world applications of linear equations

Also Read: CBSE Notes for Class 10 Maths Chapter 3

4.0NCERT Class 10 Maths Chapter 3 Exercise 3.3 : Detailed Solutions

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

(i) x + y = 14, x - y = 4

(ii) s - t = 3, s/3 + t/2 = 6

(iii) 3x - y = 3, 9x - 3y = 9

(iv) 0.2x + 0.3y = 1.3, 0.4x + 0.5y = 2.3

(v) √2x + √3y = 0, √3x - √8y = 0

(vi) (3x/2) - (5y/3) = -2, (x/3) + (y/2) = 13/6

Sol. (i) x + y = 14  ...(1)

         x - y = 4   ...(2)

From (2), y = x - 4  ...(3)

Substituting y from (3) in (1), we get:

x + x - 4 = 14

2x = 18

x = 9

Substituting x = 9 in (3), we get:

y = 9 - 4 = 5

Therefore, x = 9, y = 5.

(ii) s - t = 3  ...(1)

      s/3 + t/2 = 6  ...(2)

From (1), s = t + 3  ...(3)

Substituting s from (3) in (2), we get:

(t + 3)/3 + t/2 = 6

2(t + 3) + 3t = 36

5t + 6 = 36

5t = 30

t = 6

From (3), s = 6 + 3 = 9.

Therefore, s = 9, t = 6.

(iii) 3x - y = 3  ...(1)

       9x - 3y = 9  ...(2)

From (1), y = 3x - 3  ...(3)

Substituting y from (3) in (2), we get:

9x - 3(3x - 3) = 9

9x - 9x + 9 = 9

9 = 9

This means the equations have infinite solutions.

(iv) 0.2x + 0.3y = 1.3  ...(1)

      0.4x + 0.5y = 2.3  ...(2)

From (1), y = (1.3 - 0.2x) / 0.3  ...(3)

Substituting y from (3) in (2), we get:

0.4x + 0.5((1.3 - 0.2x) / 0.3) = 2.3

0.4x + 13/6 - x/3 = 2.3

2x/5 - x/3 = 2.3 - 13/6

x/15 = 4/30

x = 2

Substituting x = 2 in (3), we get:

y = (1.3 - 0.2 * 2) / 0.3

y = (1.3 - 0.4) / 0.3

y = 0.9 / 0.3

y = 3

Therefore, x = 2, y = 3.

(v) √2x + √3y = 0  ...(1)

     √3x - √8y = 0  ...(2)

From (2), y = (√3x) / √8  ...(3)

Substituting y from (3) in (1), we get:

√2x + √3 * (√3x / √8) = 0

√2x + 3x / √8 = 0

(4x + 3x) / √8 = 0

7x = 0

x = 0

Substituting x = 0 in (3), we get:

y = 0

Therefore, x = 0, y = 0.

(vi) (3x/2) - (5y/3) = -2  ...(1)

      (x/3) + (y/2) = 13/6  ...(2)

From (1), y = ((3x/2) + 2) / (5/3) = (9x + 12) / 10  ...(3)

Substituting y from (3) in (2), we get:

x/3 + ((9x + 12) / 10) / 2 = 13/6

x/3 + (9x + 12) / 20 = 13/6

x/3 + 9x/20 + 3/5 = 13/6

47x/60 = 47/30

x = 2

Substituting x = 2 in (3), we get:

y = (9 x 2 + 12) / 10

y = (18 + 12) / 10

y = 30 / 10

y = 3

Therefore, x = 2, y = 3.

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

Sol. 2x + 3y = 11  ...(1)

      2x - 4y = -24  ...(2)

Subtract equation (2) from (1), we get:

2x + 3y - 2x + 4y = 11 + 24

7y = 35

y = 5

Substituting the value of y in equation (1), we get:

2x + 3 * 5 = 11

2x = 11 - 15

2x = -4

x = -2

Now, x = -2, y = 5.

Putting the values of x and y in y = mx + 3, we get:

5 = -2m + 3

2 = -2m

m = -1

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 ₹ 3800. Later, she buys 3 bats and 5 balls for ₹ 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 ₹ 105 and for a journey of 15 km, the charge paid is ₹ 155. What are the fixed charges and the charge per kilometre? How much does a person have to pay for travelling a distance of 25 km?

(v) A fraction becomes 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 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?

Solutions:

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

x - y = 26  ...(1)

x = 3y      ...(2)

Substituting value of x from (2) in (1):

3y - y = 26

2y = 26

y = 13

Substituting value of y in (2):

Thus, the two numbers are 13 and 39.

(ii) Let the supplementary angles be x and y (x > y). Then,

x + y = 180   ...(1)

x - y = 18    ...(2)

From (2):

x = 18 + y   ...(3)

Substituting value of x from (3) in (1):

18 + y + y = 180

 2y = 180 - 18

 y = 162 / 2 = 81

 From (3):

 x = 18 + 81 = 99

 Thus, the angles are 99° and 81°.

(iii) Let the cost price of 1 bat be ₹ x and the cost price of 1 ball be ₹ y.

7x + 6y = 3800   ...(1)

3x + 5y = 1750   ...(2)

From (1):

7x = 3800 - 6y

x = (3800 - 6y) / 7   ...(3)

Substituting value of x from (3) in (2):

3((3800 - 6y) / 7) + 5y = 1750

11400 - 18y + 35y = 12250

17y = 850

y = 50

From (3):

x = (3800 - 300) / 7 = 500

Thus, the cost price of 1 bat is ₹ 500 and 1 ball is ₹ 50.

(iv) Let the fixed charge be ₹ x and the charge per km be ₹ y. Then,

x + 10y = 105   ...(1)

x + 15y = 155   ...(2)

From equation (1):

x = 105 - 10y   ...(3)

Substituting value of x from (3) in (2):

105 - 10y + 15y = 155

105 + 5y = 155

5y = 50

y = 10

From (3):

x = 105 - 10 × 10 = 5

Thus, the fixed charge is ₹ 5 and the charge per km is ₹ 10.

For 25 km, the charge is 5 + (25 * 10) = 255.

(v) Let x/y be the fraction where x and y are positive integers.

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

11x + 22 = 9y + 18

11x - 9y = -4   ...(1)

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

6x + 18 = 5y + 15

6x - 5y = -3   ...(2)

From (1):

11x = 9y - 4

x = (9y - 4) / 11   ...(3)

Substituting value of x from (3) in (2):

6((9y - 4) / 11) - 5y = -3

54y - 24 - 55y = -33

-y = -9

y = 9

From (3):

x = (9 × 9 - 4) / 11 = 7

Thus, the fraction is 7/9.

(vi) Let x (in years) be the present age of Jacob's son and y (in years) be the present age of Jacob. Then,

y + 5 = 3(x + 5)

3x - y = -10   ...(1)

y - 5 = 7(x - 5)

7x - y = 30    ...(2)

From (1):

y = 3x + 10   ...(3)

Substituting value of y from (3) in (2):

7x - (3x + 10) = 30

4x = 40

x = 10

From (3):

y = 40

Thus, Jacob's present age is 40 years and his son's age is 10 years.

5.0Benefits of Solving NCERT Solutions for Class 10 Maths Chapter 3 Exercise 3.3

• Strengthened foundational concepts
• Improved problem solving skills
• Exam preparation and confidence
• Enhanced understanding of key concepts
• Easy access and free download.