If d is HCF of 56 and 72, then find x,y satisfying the equation d=56x+72y. Also show that x and y are not unique.

Let d = gcd(56,72) = 12. We know d = 12 = 56x + 72y. Subtracting 72y from both sides, d - 72y = 56x ⇒ 12 - 72y = 56x Dividing both sides by 56, (12/56) - (72/56)y = x ⇒ (1/4) - (3/4)y = x Since (1/4) - (3/4)y is a linear equation in y, its solution is y = a, where a is some real number. Substituting y = a in the linear equation, (1/4) - (3/4)a = x ⇒ x = (1/4) - (3/4)a Therefore, if d = 12, then x = (1/4) - (3/4)a and y = a, where a is some real number. Since a is some real number, x and y are not uniquely determined. Therefore, x and y are not unique.