For what value of p will the following pair of linear equations have infinitely many solutions:

(p - 3) x + 3y = p

px + py = 12

given pair of linear equations are

(p - 3) x + 3y = p

⇒ (p - 3) x + 3y - p = 0

px + py = 12

⇒ px + py - 12 = 0

Comparing these with standard equation

a₁x₁ + b₁x₁ + c₁ = 0 and a₂x₂ + b₂x₂ + c₂ = 0

We get

a₁ = (p - 3), b₁ = 3, c₁ = - p

a₂ = p, b₂ = p, c₂ = - 12

Now it is given the linear pair of equations have infinitely many solutions, so the condition for it is

Now substituting the corresponding values, we get

Considering first equality, we get

⇒ p - 3 = 3

⇒ p = 6

Considering the second equality, we get

⇒ 3 × 12 = p²

⇒ p = √36

⇒ p = ±6

Hence p = + 6 satisfies both the conditions.

Hence for p = 6, the given pair of linear equations have infinitely many solutions.