RD Sharma Solutions Class 10 Maths Chapter 3 – Free PDF Download

RD Sharma Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables are provided here to help students prepare effectively for their board exams. Mathematics becomes easier to understand when students use the right methods and tools for studying. 

For students who find it challenging to grasp concepts or solve problems, RD Sharma Solutions are the perfect resource. Created by experts at Infinity Learn, these solutions are designed to match the students' understanding level, ensuring they meet their needs and excel in exams. The detailed answers will guide students towards securing high marks in the subject.

RD Sharma Class 10 Solutions for Chapter 3 are available here. Students can conveniently download the PDF of these solutions using the provided links. Class 10 introduces several new topics in Mathematics, and our experts have formulated these solutions to assist students in their exam preparation, ensuring they achieve excellent marks in Maths. The solutions are stepwise and detailed, making learning easier for students.

Students aiming to excel in their exams should practice RD Sharma Class 10 Solutions for Chapter 3. This solution guide not only supports exam preparation but also helps build a strong foundation in the subject. Chapter 3 is an extension of what students learned in middle school about linear equations in one variable. Here are some of the key concepts covered in this chapter:

• Systems of linear equations in two variables
• The solution of a system of linear equations in two variables
• Graphical and algebraic methods for solving a system of linear equations in two variables, including substitution, elimination, and cross-multiplication methods
• Consistent and inconsistent systems of equations
• Applications of linear equations in two variables to solve real-world problems from various areas.

Download RD Sharma Class 10 chapter 3 PDF with Solutions

For easy access, students can download the RD Sharma Class 10 Chapter 3 Solutions in PDF format, ensuring they have all the study material they need for effective learning.

Access the RD Sharma Solutions for Class 10 Maths Chapter 3 – Pair of Linear Equations in Two Variables

1. Solve the following system of linear equations:

2x + 3y = 12

4x - y = 10

Solution:

We will use the substitution method to solve the system.

From the second equation:

4x - y = 10

y = 4x - 10

Substitute y = 4x - 10 in the first equation:

2x + 3(4x - 10) = 12

2x + 12x - 30 = 12

14x = 42

x = 3

Now, substitute x = 3 in y = 4x - 10:

y = 4(3) - 10 = 12 - 10 = 2

Thus, the solution is x = 3 and y = 2.

2. Solve the following system of linear equations using the elimination method:

3x + 4y = 18

5x - 2y = 16

Solution:

We will multiply both equations to make the coefficients of y the same.

Multiply the first equation by 2 and the second equation by 4:

2(3x + 4y) = 2(18) ⟶ 6x + 8y = 36

4(5x - 2y) = 4(16) ⟶ 20x - 8y = 64

Now, add the two equations:

(6x + 8y) + (20x - 8y) = 36 + 64

26x = 100

x = 100 / 26 = 50 / 13

Substitute x = 50 / 13 into the first equation:

3(50 / 13) + 4y = 18

150 / 13 + 4y = 18

Multiply the whole equation by 13 to eliminate the fraction:

150 + 52y = 234

52y = 234 - 150 = 84

y = 84 / 52 = 21 / 13

Thus, the solution is x = 50 / 13 and y = 21 / 13.

3. Solve the system of linear equations using the substitution method:

x + 2y = 9

3x - 4y = 10

Solution:

From the first equation, solve for x:

x = 9 - 2y

Substitute x = 9 - 2y in the second equation:

3(9 - 2y) - 4y = 10

27 - 6y - 4y = 10

27 - 10y = 10

-10y = 10 - 27 = -17

y = -17 / -10 = 17 / 10

Now, substitute y = 17 / 10 in x = 9 - 2y:

x = 9 - 2(17 / 10) = 9 - 34 / 10 = 90 / 10 - 34 / 10 = 56 / 10 = 28 / 5

Thus, the solution is x = 28 / 5 and y = 17 / 10.

4. Solve the following system of linear equations by the elimination method:

6x + 3y = 15

4x - 2y = 6

Solution:

Multiply the second equation by 3 and the first equation by 2 to make the coefficients of y equal:

3(6x + 3y) = 3(15) ⟶ 18x + 9y = 45

2(4x - 2y) = 2(6) ⟶ 8x - 4y = 12

Now, add the two equations:

(18x + 9y) + (8x - 4y) = 45 + 12

26x + 5y = 57

Now, solve for x and y.

5. Solve the system of linear equations:

x - y = 3

2x + y = 9

Solution:

We can use the substitution method. From the first equation:

x = y + 3

Substitute x = y + 3 in the second equation:

2(y + 3) + y = 9

2y + 6 + y = 9

3y = 3

y = 1

Now, substitute y = 1 in x = y + 3:

x = 1 + 3 = 4

Thus, the solution is x = 4 and y = 1.

6. Solve the following system of equations using the substitution method:

3x + 5y = 14

2x - 3y = 4

Solution:

From the second equation, solve for x:

2x = 4 + 3y

x = (4 + 3y) / 2

Now, substitute x = (4 + 3y) / 2 in the first equation:

3((4 + 3y) / 2) + 5y = 14

(12 + 9y) / 2 + 5y = 14

Multiply the entire equation by 2 to eliminate the fraction:

12 + 9y + 10y = 28

12 + 19y = 28

19y = 28 - 12 = 16

y = 16 / 19

Substitute y = 16 / 19 in x = (4 + 3y) / 2:

x = (4 + 3(16 / 19)) / 2

x = (4 + 48 / 19) / 2 = (76 / 19 + 48 / 19) / 2 = (124 / 19) / 2 = 124 / 38 = 62 / 19

Thus, the solution is x = 62 / 19 and y = 16 / 19.

7. Points A and B are 70km. apart on a highway. A car starts from A, and another car starts from B simultaneously. If they travel in the same direction, they meet in 7hrs, but if they travel towards each other, they meet in one hour. Find the speed of two cars.

Solution:

Let’s consider the car starting from point A as X and its speed as x km/hr.

And, the car starts from point B as Y and its speed as y km/hr.

It’s seen that there are two cases in the question:

Case 1: Car X and Y are moving in the same direction

Case 2: Car X and Y are moving in the opposite direction

Let’s assume that the meeting point in case 1 is P and in case 2 is Q.

Now, solving for case 1:

The distance travelled by car X = AP

And, the distance travelled by car Y = BP

As the time taken for both the cars to meet is 7 hours,

The distance travelled by car X in 7 hours = 7x km [∵ Distance = Speed x Time]

⇒ AP = 7x

Similarly,

The distance travelled by car Y in 7 hours = 7y km

⇒ BP = 7Y

As the cars are moving in the same direction (i.e., away from each other), we can write

AP – BP = AB

So, 7x – 7y = 70

x – y = 10 ………………………. (i) [after taking 7 common out]

Now, solving for case 2:

In this case, as it’s clearly seen that

The distance travelled by car X = AQ

And,

The distance travelled by car Y = BQ

As the time taken for both the cars to meet is 1 hour,

The distance travelled by car x in 1 hour = 1x km

⇒ AQ = 1x

Similarly,

The distance travelled by car y in 1 hour = 1y km

⇒ BQ = 1y

Now, since the cars are moving in the opposite direction (i.e., towards each other), we can write

AQ + BQ = AB

⇒ x + y = 70 …………… (ii)

Hence, by solving (i) and (ii), we get the required solution

From (i), we have x = 10 + y……. (iii)

Substituting this value of x in (ii),

⇒ (10 + y) + y = 70

⇒ y = 30

Now, using y = 30 in (iii), we get

⇒ x = 40

Therefore,

– Speed of car X = 40km/hr.

– Speed of car Y = 30 km/hr.