# For what value of

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.

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