The sum of the squares of two positive integers is 100 and the difference of their squares is 28. The sum of the numbers is ?

Solution(By Examveda Team)

Let the positive integers be a and b where a > b
According to the question,
$$\eqalign{ & {a^2} + {b^2} = 100.....(\text{i}) \cr & {a^2} - {b^2} = 28.....(\text{ii}) \cr} $$
By adding (i) and (ii), we get :
$$\eqalign{ & \therefore {a^2} + {b^2} + {a^2} - {b^2} = 100 + 28 \cr & \Rightarrow 2{a^2} = 128 \cr & \Rightarrow {a^2} = \frac{{128}}{2} \cr & \Rightarrow a = \sqrt {64} \cr & \Rightarrow a = 8 \cr} $$
From equation (i)
$$\eqalign{ & \Rightarrow {8^2} + {b^2} = 100 \cr & \Rightarrow {b^2} = 100 - 64 \cr & \Rightarrow b = \sqrt {36} \cr & \Rightarrow b = 6 \cr & \therefore a + b = 8 + 6 \cr & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = 14 \cr} $$