In this chapter, we provide RS Aggarwal Solutions for Class 10 Chapter 3 Linear equations in two variables EX 3B Maths for English medium students, Which will very helpful for every student in their exams. Students can download the latest RS Aggarwal Solutions for Class 10 Chapter 3 Linear equations in two variables EX 3B Maths pdf, free RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables EX 3B Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 3 |

Chapter Name | Linear equations in two variables |

Exercise | 3 B |

Category | RS Aggarwal Solutions |

**RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables Ex 3B**

**Question 1:**

**Question 2:**

**Question 3:**

**Question 4:**

**Question 5:**

**Question 6:**

**Question 7:**

**Question 8:**

**Question 9:**

**Question 10:**

**Question 11:**

**Question 12:**

**Question 13:**

**Question 14:**

**Question 15:**

**Question 16:**

**Question 17:**

**Question 18:**

**Question 19:**

**Question 20:**

**Question 21:**

**Question 22:**

Taking L.C.M, we get

Multiplying (1) by 1 and (2) by

Subtracting (4) from (3), we get

Substituting x = ab in (3), we get

Therefore solution is x = ab, y = ab**Question 23:**

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

6ax + 6by = 3a + 2b —(1)

6(bx – ay) = 3b – 2a

6bx – 6ay = 3b- 2a —(2)

6ax + 6by = 3a + 2b —(1)

6bx – 6ay = 3b – 2a —(2)

Multiplying (1) by a and (2) by b

Adding (3) and (4), we get

Substituting in (1), we get

Hence, the solution is

**Question 24:**

**Question 25:**

The given equations are

71x + 37y = 253 —(1)

37x + 71y = 287 —(2)

Adding (1) and (2)

108x + 108y = 540

108(x + y) = 540

—-(3)

Subtracting (2) from (1)

34x – 34y = 253 – 287 = -34

34(x – y) = -34

—(4)

Adding (3) and (4)

2x = 5 – 1= 4

Subtracting (4) from (3)

2y = 5 + 1 = 6

Hence solution is x = 2, y = 3**Question 26:**

37x + 43y = 123 —-(1)

43x + 37y = 117 —-(2)

Adding (1) and (2)

80x + 80y = 240

80(x + y) = 240

x + y =

—-(3)

Subtracting (1) from (2),

6x – 6y = -6

6(x – y) = -6

—-(4)

Adding (3) and (4)

2x = 3 – 1 = 2

Subtracting (4) from (3),

2y = 3 + 1 = 4

Hence solution is x = 1, y = 2**Question 27:**

217x + 131y = 913 —(1)

131x + 217y = 827 —(2)

Adding (1) and (2), we get

348x + 348y = 1740

348(x + y) = 1740

x + y = 5 —-(3)

Subtracting (2) from (1), we get

86x – 86y = 86

86(x – y) = 86

x – y = 1 —(4)

Adding (3) and (4), we get

2x = 6

x = 3

putting x = 3 in (3), we get

3 + y = 5

y = 5 – 3 = 2

Hence solution is x = 3, y = 2**Question 28:**

41x – 17y = 99 —(1)

17x – 41y = 75 —(2)

Adding (1) and (2), we get

58x – 58y = 174

58(x – y) = 174

x – y = 3 —(3)

subtracting (2) from (1), we get

24x + 24y = 24

24(x + y) = 24

x + y = 1 —(4)

Adding (3) and (4), we get

2x = 4 x = 2

Putting x = 2 in (3), we get

2 – y = 3

-y = 3 – 2 y = -1

Hence solution is x =2, y = -1

**All Chapter RS Aggarwal Solutions For Class 10 Maths**

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**All Subject NCERT Solutions For Class 10**

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