SSC CGL 2024 PYQ — Question 51

The correct answer is option 2 i.e. a = 3, b ≠–1.

x – ay  + a - 2 = 0 ...(1)

(1 – a)x + 6y - (a + b) = 0 ...(2)

Conditions for no solution $\frac {x_1}{ x_2}=\frac {y_1}{ y_2} ≠\frac {c_1}{ c_2}$

Here from equation 1 and 2:

x₁ = 1, y₁ = -a, c₁ = a - 2

x₂ = 1 - a, y₂ = 6, c₂ = -(a + b)

1. $\frac {x_1}{ x_2}=\frac {y_1}{ y_2}$ 

⇒ $\frac { 1}{ 1-a}=\frac {-a}{6}$

⇒ 6 = a² - a

⇒ a² - a - 6 = 0

⇒ (a - 3)(a + 2) = 0

⇒ a = 3, -2

Take a = 3

2. $\frac {y_1}{ y_2} ≠\frac {c_1}{ c_2}$

⇒ $\frac {-3}{ 6} ≠\frac {3-2}{ -(3+b)}$

⇒ 9 + 3b ≠ 6

⇒ 3b ≠ -3

⇒ b ≠ -1