# RS Aggarwal Solutions Class 10 Chapter 3 Linear equations in two variables EX 3B

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.

### 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)
108x + 108y = 540
108(x + y) = 540

—-(3)
Subtracting (2) from (1)
34x – 34y = 253 – 287 = -34
34(x – y) = -34

—(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)
80x + 80y = 240
80(x + y) = 240
x + y =

—-(3)
Subtracting (1) from (2),
6x – 6y = -6
6(x – y) = -6

—-(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

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