The Number Puzzle

Let the two positive numbers be x and y and x>y.

x - y = 3 ....(1)

x^2 - y^2 = 51

We know , x^2 - y^2 = (x + y)(x - y)

Or , x + y = 51/3 = 17 ....(2)

Adding eqn (1) & (2),

2x = 20 or x = 10

y = x-3 = 7

Therefore the smallest of the two is 7.

$\text{suppose, the two positive numbers are x \& y and}$ $x>y$

$\text{now,}$ $x-y=3$ .......................[1]

$\text{and}$ $x^2-y^2=51$ .................[2]

$\text{now, dividing [2] by [1] we get,}$

$\frac{x^2-y^2}{x-y}$ $=\frac{51}{3}$

$\text{or,}$ $x+y$ $=\frac{51}{3}$ .........................[3]

$\text{now, doing [1]+[2] we get ,}$

$x-y+x+y=3+$ $\frac{51}{3}$

$\text{or,}$ $2x=20$

$\text{or,}$ $x=10$

$\text{or,}$ $y=10-3=7$ .........[answer]