NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variables

Question

Our NCERT Solutions Class 10 Science Chapter 3 Pair of Linear Equations in Two Variables will help you understand and clear all the NCERT question-answer doubts. We have covered all the topics and provided solutions to all in-text and exercise questions. One must refer to our solution for one-stop solutions for all concepts Linear Equations in Two Variables. Our NCERT solutions are designed by an expert in the subject matter who has a very good understanding of all the concepts in this field. Experts have used different kinds of methods to make these solutions easy to understand for the users using methods like diagrams, different equations, graphs, Shortcuts, tips, and tricks.

In this chapter of Linear Equations in Two Variable, we have provided NCERT Solutions Class 10 with all the related topics which deal with the queries of the students. Important topics mentioned in the chapter are

• Linear Equations in Two Variables
• Converting statements into Equations and drawing graphs of linear equations.
• Types of Graphs made by Pair of Linear Equations in Two Variables- Two lines that are Intersecting, Two lines that are parallel, Coincident Lines.
• Consistency of equations by finding the ratio of a₁/a₂, b₁/b₂, and c1/c₂ to check whether lines Intersecting Lines (unique solution), Coincident Lines (infinitely many solutions), and Parallel Lines (no solutions).
• Solutions of Equations from graphs.
• Solving pair of Linear Equations by Substitution Method, Elimination Method, and Cross Multiplication method.
• Solve statement questions with the help of the given word problem.
• Solve complicated equations with the help of substituting variable method.

Our NCERT Solutions Class 10 Maths Chapter 3 Pair of Linear Equations in Two Variable will help students comprehend the concepts with ease. Our NCERT solution is best for last-minute preparation for their examination. To score good marks one must completely go through all the solutions provided here.

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Answer 1

24. A train covered a certain distance at a uniform speed. If the train would have been 10 km/h faster, it would have taken 2 hours less than the scheduled time. And if the train were slower by 10 km/h; it would have taken 3 hours more than the scheduled time. Find the distance covered by the train.

Answer: 

Let the speed of the train be x km/h and the time taken by train to travel the given distance be t hours and the distance to travel was d km. We know that,

'

Or, d = xt ………………………………… (i) 

Using the information given in the question, we obtain

By using equation (i), we obtain 

− 2x + 10t = 20 ………………………… (ii)

By using equation (i), we obtain 

3x − 10t = 30   ....................................... (iii) 

Adding equations (ii) and (iii), we obtain 

x = 50 

Using equation (ii), we obtain 

(−2) × (50) + 10t = 20 

−100 + 10t = 20 10t = 120 t = 12 hours 

From equation (i), we obtain 

Distance to travel = d = xt 

= 50 × 12 

= 600 km 

Hence, the distance covered by the train is 600 km.

25. The students of a class are made to stand in rows. If 3 students are extra in a row, there would be 1 row less. If 3 students are less in a row, there would be 2 rows more. Find the number of students in the class.

Answer: 

Let the number of rows be x and number of students in a row be y. 

Total students of the class = Number of rows × Number of students in a row 

= xy 

Using the information given in the question,

Condition 1 

Total number of students = (x − 1) (y + 3) 

Or xy = (x − 1) (y + 3) = xy − y + 3x − 3 

Or 3x − y − 3 = 0 

Or 3x − y = 3 ……………………………………. (i)

Condition 2 

Total number of students = (x + 2) (y − 3) 

Or xy = xy + 2y − 3x − 6

Or 3x − 2y = −6 ………………………………(ii)

Subtracting equation (ii) from (i), 

(3x − y) − (3x − 2y) = 3 − (−6) 

− y + 2y = 3 + 6 y = 9

By using equation (i), we obtain 

3x − 9 = 3 

3x = 9 + 3 = 12 

x = 4 

Number of rows = x = 4 

Number of students in a row = y = 9 

Number of total students in a class = xy = 4 × 9 = 36

26. In a ∆ABC, ∠C = 3 ∠B = 2 (∠A + ∠B). Find the three angles.

Answer:

Given that, 

∠C = 3∠B = 2(∠A + ∠B) 

3∠B = 2(∠A + ∠B) 

3∠B = 2∠A + 2∠B 

∠B = 2∠A 

2 ∠A − ∠B = 0 ………………………………… (i) 

We know that the sum of the measures of all angles of a triangle is 180°. 

Therefore, 

∠A + ∠B + ∠C = 180° 

∠A + ∠B + 3 ∠B = 180° 

∠A + 4 ∠B = 180° …………………………….. (ii) 

Multiplying equation (i) by 4, we obtain 

8 ∠A − 4 

∠B = 0 …………………………….… (iii) 

Adding equations (ii) and (iii), we obtain 

9 ∠A = 180° 

∠A = 20° 

From equation (ii), we obtain 

20° + 4 ∠B = 180° 

4 ∠B = 160° 

∠B = 40° 

∠C = 3 ∠B 

= 3 × 40° = 120° 

Therefore, ∠A, ∠B, ∠C are 20°, 40°, and 120° respectively.

27. Draw the graphs of the equations 5x − y = 5 and 3x − y = 3. Determine the coordinates of the vertices of the triangle formed by these lines and the y axis.

Answer:

5x − y = 5 

Or, y = 5x − 5 

The solution table will be as follows.

3x − y = 3 

Or, y = 3x − 3 

The solution table will be as follows.

The graphical representation of these lines will be as follows.

It can be observed that the required triangle is ∆ABC formed by these lines and y-axis. The coordinates of vertices are A (1, 0), B (0, − 3), C (0, − 5).

Answer 2

28. Solve the following pair of linear equations. 

(i). px + qy = p − q , qx − py = p + q 

(ii). ax + by = c, bx + ay = 1 + c

(iii). x/a – y/b = 0, ax + by = a² + b² 

(iv). (a − b) x + (a + b) y = a²− 2ab − b²

(a + b) (x + y) = a² + b² 

(v) 152x − 378y = − 74 

− 378x + 152y = − 604

Answer:

(i) px + qy = p − q ……………………..… (1) 

qx − py = p + q ……………………..… (2) 

Multiplying equation (1) by p and equation (2) by q, we obtain 

p²x + pqy = p² − pq …………………………… (3) 

q²x − pqy = pq + q² ………………………..… (4) 

Adding equations (3) and (4), we obtain

p²x + q² x = p² + q² 

(p² + q²) x = p² + q²

From equation (1), we obtain 

p (1) + qy = p − q qy = − q y = − 1

(ii)ax + by = c ……………… (1) 

bx + ay = 1 + c ……….. (2) 

Multiplying equation (1) by a and equation (2) by b, we obtain

a²x + aby = ac ………………… (3) 

b²x + aby = b + bc …………… (4) 

Subtracting equation (4) from equation (3),

(a² − b²) x = ac − bc − b

From equation (1), we obtain

ax + by = c

(iii). x/a – y/b = 0......................(1)

ax + by = a² + b² ......................(2)

Multiplying equation (1) and (2) by b and a respectively, we obtain

b²x − aby = 0 …………….…… (3) 

a²x + aby = a³ + ab² ……………… (4) 

Adding equations (3) and (4), we obtain

b²x + a²x = a³ + ab² 

x (b² + a²) = a (a² + b²) x = a

By using (1), we obtain 

b (a) − ay = 0 

ab − ay = 0 

ay = ab 

y = b

(iv) (a − b) x + (a + b) y = a²− 2ab − b² …………………. (1) 

(a + b) (x + y) = a² + b² 

(a + b) x + (a + b) y = a² + b² ………………….… (2)

Subtracting equation (2) from (1), we obtain 

(a − b) x − (a + b) x = (a² − 2ab − b²) − (a² + b²)

(a − b − a − b) x = − 2ab − 2b² 

− 2bx = − 2b (a + b) 

x = a + b

Using equation (1), we obtain 

(a − b) (a + b) + (a + b) y = a² − 2ab − b²

a² − b² + (a + b) y = a²− 2ab − b²

(a + b) y = − 2ab

\(y = \frac{-2ab}{a + b}\)

(v) 152x − 378y = − 74

 76x − 189y = − 37

− 378x + 152y = − 604 

− 189x + 76y = − 302 … (2) 

Substituting the value of x in equation (2), we obtain

− (189)² y + 189 × 37 + (76)² y = − 302 × 76 

189 × 37 + 302 × 76 = (189)² y − (76)² y 

6993 + 22952 = (189 − 76) (189 + 76) y 

29945 = (113) (265) y 

y = 1 

From equation (1), we obtain

29. ABCD is a cyclic quadrilateral finds the angles of the cyclic quadrilateral.

Answer:

We know that the sum of the measures of opposite angles in a cyclic quadrilateral is 180°. 

Therefore, 

∠A + ∠C = 180 

4y + 20 − 4x = 180 

− 4x + 4y = 160 

x − y = − 40 ………………………(i)

Also, ∠B + ∠D = 180 

3y − 5 − 7x + 5 = 180 

− 7x + 3y = 180 …………………..(ii)

Multiplying equation (i) by 3, we obtain 

3x − 3y = − 120 ……………………(iii) 

Adding equations (ii) and (iii), we obtain 

− 7x + 3x = 180 − 120 

− 4x = 60 

x = −15

By using equation (i), we obtain 

x − y = − 40 

−15 − y = − 40 

y = −15 + 40 = 25 

∠A = 4y + 20 = 4(25) + 20 = 120° 

∠B = 3y − 5 = 3(25) − 5 = 70° 

∠C = − 4x = − 4(− 15) = 60° 

∠D = − 7x + 5 = − 7(−15) + 5 = 110°