# RS Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3C

R S Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables, contains solutions for exercise 3C questions. All questions are solved by experts at BYJU’S based on CBSE latest syllabus using step by step method. R S Aggarwal solutions on exercise 3c helps studentâ€™s solving simultaneous linear equations using cross multiplication method. Download RS Aggarwal Solutions of Class 10Â  maths and clear all your doubts related to linear equations.

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### Access other exercise solutions of Class 10 Maths Chapter 3 Linear Equations in Two Variables

Exercise 3A Solutions: 29 Questions (Long Answers)

Exercise 3B Solutions: 20 Questions (Short Answers)

Exercise 3D Solutions: 14 Questions (Short Answers)

Exercise 3E Solutions: 20 Questions (Long Answers)

Exercise 3F Solutions: 10 Questions (6 Short Answers and 4 Long Answers)

## Exercise 3C Page No: 111

Solve each of the following systems of equations by using the method of cross multiplication:

Question 1:

x + 2y + 1 = 0,

2x â€“ 3y â€“ 12 = 0.

Solution:

x + 2y + 1 = 0 â€¦(1)

2x â€“ 3y â€“ 12 = 0 â€¦(2)

From equation (1): a1 = 1, b1 = 2 and c1 = 1

And from equation (2): a2 = 2, b2 = – 3 and c2 = – 12

Using cross multiplication method:

Answer: x = 3 and y = – 2

Question 2:

3x â€“ 2y + 3 = 0,

4x + 3y â€“ 47 = 0.

Solution:

3x â€“ 2y + 3 = 0 â€¦(1)

4x + 3y â€“ 47 = 0 â€¦(2)

From equation (1): a1 = 3, b1 = – 2 and c1 = 3

From equation (2): a2 = 4, b2 = 3 and c2 = – 47

Using cross multiplication method:

Question 3:

6x â€“ 5y â€“ 16 = 0,

7x â€“ 13y + 10 = 0.

Solution:

6x â€“ 5y â€“ 16 = 0 â€¦(1)

7x â€“ 13y + 10 = 0 â€¦(2)

From equation (1): a1 = 6, b1 = – 5 and c1 = – 16

From equation (2): a2 = 7, b2 = – 13 and c2 = 10

Using cross multiplication method:

Question 4:

3x + 2y + 25 = 0,

2x + y + 10 = 0.

Solution:

3x + 2y + 25 = 0 â€¦(1)

2x + y + 10 = 0 â€¦(2)

From equation (1): a1 = 3, b1 = 2 and c1 = 25

From equation (2): a2 = 2, b2 = 1 and c2 = 10

Using cross multiplication method:

Question 5:

2x + 5y = 1,

2x + 3y = 3.

Solution:

2x + 5y â€“ 1 = 0 â€¦(1)

2x + 3y â€“ 3 = 0 â€¦(2)

From equation (1): a1 = 2, b1 = 5 and c1 = – 1

From equation (2): a2 = 2, b2 = 3 and c2 = – 3

Using cross multiplication method:

Question 6:

2x + y = 35,

3x + 4y = 65.

Solution:

2x + y â€“ 35 = 0 â€¦(1)

3x + 4y â€“ 65 = 0 â€¦(2)

From equation (1): a1 = 2, b1 = 1 and c1 = – 35

From equation (2): a2 = 3, b2 = 4 and c2 = – 65

Using cross multiplication method:

Question 7:

7x â€“ 2y = 3,

22x â€“ 3y = 16.

Solution:

7x â€“ 2y â€“ 3 = 0 â€¦(1)

22x â€“ 3y â€“ 16 = 0 â€¦(2)

From equation (1): a1 = 7, b1 = – 2 and c1 = – 3

From equation (2): a2 = 22, b2 = – 3 and c2 = – 16

Using cross multiplication method:

Question 8.

x/6 + y/15 = 4

x/3 â€“ y/12 = 19/4

Solution:

Simplify given equations:

x/6 + y/15 = 4

(5x + 2y)/30 = 4

5x + 2y â€“ 120 = 0

And x/3 â€“ y/12 = 19/4

(4x â€“ y)/12 = 19/4

4x â€“ y â€“ 57 = 0

New set of equations is:

5x + 2y â€“ 120 = 0 â€¦..(1)

4x â€“ y â€“ 57 = 0 â€¦â€¦â€¦.(2)

From equation (1): a1 = 5, b1 = 2 and c1 = – 120

From equation (2): a2 = 4, b2 = – 1 and c2 = – 57

Using cross multiplication,

Question 9:

1/x + 1/y = 7

2/x + 3/y = 17 (xâ‰ 0 and yâ‰  0)

Solution:

Simplify given equations:

Let 1/x = u and 1/y = v

1/x + 1/y = 7 implies

u + v = 7

2/x + 3/y = 17 implies:

2u + 3v = 17

New set of equations is:

u + v = 7 â€¦â€¦(1)

2u + 3v = 17 â€¦â€¦â€¦(2)

From equation (1): a1 = 1, b1 = 1 and c1 = – 7

From equation (2): a2 = 2, b2 = 3 and c2 = – 17

Using cross multiplication method:

Question 10:

Solution:

Consider 1/(x + y) = a and 1/(x – y) = b

Given equations turn as:

5a – 2b + 1 = 0 â€¦.(1)

15a + 7b – 10 = 0 â€¦(2)

From equation (1): a1 = 5, b1 = -2 and c1 = 1

From equation (2): a2 = 15, b2 = 7 and c2 = – 10

Using cross multiplication method:

So a = 13/65 = 1/5 and b = 65/65 = 1

Putting back values and write equations in the form of x and y, we get

x + y = 5 and x â€“ y = 1

Again we get a set of two linear equations with two variables. Solve again to find the value of x and y.

Rearranging them again,

x + y â€“ 5 = 0 â€¦(3)

x â€“ y â€“ 1 = 0 â€¦(4)

From equation (3): a1 = 1, b1 = 1 and c1 = – 5

From equation (4): a2 = 1, b2 = – 1 and c2 = – 1

Using cross multiplication method,

Answer: x = 3 and y = 2

Question 11:

From equation (2): a2 = a, b2 = – b and c2 = – 2ab

Using cross multiplication method, we get

Answer: x = b and y = -a

Question 12.

2ax + 3by = (a + 2b),

3ax + 2by = (2a + b).

Solution:

Question 13:

## RS Aggarwal Solutions for Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3C

Class 10 Maths Chapter 3 Linear Equations in Two Variables Exercise 3C is based on the topic: Solving simultaneous linear equations using the method of cross multiplication.