Question

Given the sum of square of two positive integers is $28$ more than the product of them. Also the ratio of numbers is $2:3$. Then find the numbers.

Answer 1

Let the positive integers be $x,y$.

Then $x:y=2:3$

or, $x=\frac{2y}{3}$......(1).

According to the problem,

$x^2+y^2=xy+28$

or, $\frac{4y^2}{9}+y^2=\frac{2y^2}{3}+28$

or, $\frac{7y^2}{9}=28$

or, $y^2=36$

or, $y=6$ [ Since $y>0$]

Using value of $y$ we get, $x=4$.

So the numbers are $4,6$.