Given,
$x-y=2$ ----- (i)
$xy=24$
We know,
$\left(x+y\right)^2=\left(x-y\right)^2+4xy$
বা, $\left(x+y\right)=\sqrt{\left(x-y\right)^2+4xy}$
বা, $\left(x+y\right)=\sqrt{2^2+4\times24}$
বা, $\left(x+y\right)=\sqrt{100}$
বা, $\left(x+y\right)=10$ ----- (ii)
(ii) + (i)
$\left(x+y\right) + \left(x-y\right) = 10+2$
বা, $x+\cancel y+x-\cancel y=12$
বা, $2x=12$
বা, $x=\frac{12}{2}$
$\therefore x=6$
