# 10th cbse maths solution for exercise 3.5 part 2

This page 10th cbse maths solution for exercise 3.5 part 2 is going to provide you solution for every problems that you find in the exercise no 3.5

## 10th CBSE maths solution for Exercise 3.5 part 2

(2) For which values of a and b does the following pair of linear equations have an infinite many solution

2 x + 3 y = 7

(a – b) x + (x + b) y = 3 a + b – 2

Solution:

Condition for having infinitely many solutions

a₁/a₂ = b₁/b₂ = c₁/c₂

2 x + 3 y – 7 = 0 --------(1)

(a – b) x + (a + b) y – (3 a + b – 2) = 0 --------(2)

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

a₁ = 2               b₁ = 3                      c₁ = -7

a₂ = (a-b)          b₂ = (a + b)             c₂ = – (3 a + b – 2)

2/(a – b) = 3/(a + b) = -7/-(3a + b – 2)

2/(a – b) = 3/(a + b) = 7/(3a + b – 2)

2/(a – b) = 7/(3a + b – 2)

2(3a + b – 2) = 7(a – b)

6 a + 2 b – 4 = 7 a – 7 b

6 a – 7 a + 2 b + 7 b = 4

-a + 9 b = 4  ----- (3)

3(3a + b – 2) = 7(a + b)

9 a + 3 b – 6 = 7 a + 7 b

9 a – 7 a + 3 b – 7 b = 6

2 a – 4 b = 6 ----- (4)

Substitute b = 1 in the third or fourth equation to get the value of a

-a + 9 (1) = 4

-a + 9 = 4

-a = 4 -9

-a = -5

a = 5

(ii) For which value of k will the following pair of linear equations have no solution

3 x + y = 1

(2k – 1) x + (k – 1) y = 2 k + 1

3 x + y – 1 = 0

(2k – 1) x + (k – 1) y – (2 k + 1) = 0

Condition for having no solution

a₁/a ₂ = b₁/b ₂ ≠ c₁/c ₂

From the above information let us take the values of a₁ , a₂, b₁, b₂, c₁ and c ₂

a₁ = 3                    b₁ = 1                    c₁ = -1

a₂ = (2 k - 1)         b₂ = (k - 1)              c₂ = – (2 k + 1)

3/(2 k – 1) = 1/(k – 1)

3 (k  - 1) = 1 (2 k- 1)

3 k – 3 = 2 k – 1

3 k – 2 k = -1 + 3

K = 2

(3) Solve the following pair of linear equations by substitution and cross multiplication methods:

8 x + 5 y = 9 -------(1)

3 x + 2 y = 4 -------(2)

8x = 9 – 5 y

x = (9 – 5 y)/8

Substitute x =(9 – 5 y)/8 in the second equation

3 [(9 – 5 y)/8] + 2 y = 4

[(27 – 15 y)/8] + 2 y = 4

(27 – 15 y + 16 y)/8 = 4

27 + y = 32

Y = 32 – 27

Y = 5

Substitute y = 5 in the equation x =(9 – 5 y)/8

x =(9 – 5 (5))/8

x = (9 – 25)/8

x = -16/8

x = -2                           10th cbse maths solution for exercise 3.5 part 2 10th cbse maths solution for exercise 3.5 part 2