Atypical equations

Since $\dfrac{1}{x}+\dfrac{1}{y}= 2 \implies \dfrac{x+y}{xy} = 2 \implies x+y = 2xy.$

Since $x^2 + y^2 = 30 \implies x^2+2xy + y^2 = 30 +2xy \implies (x+y)^2 = 30+2xy.$

Thus, we have $(2xy)^2 = 30 + 2xy.$

Substitute $w = xy,$ then $4w^2 -2w -30 = 0 \implies 2(w-3)(2w+5) = 0.$

For positive numbers, we have $xy=w = \boxed{3}.$

Let the two positive numbers be x and y.

x²+y²= 30 -----------(1a)

1/x + 1/y= 2---------(2a)

For (1a), (x+y)²= x²+2xy+y²

So (x+y)²-2xy= 30-----(1b)

For (2a), establishing a common denominator of xy, we get (x+y)/xy= 2------(2b)

So (x+y)= 2xy and, substituting this into (1b), we get (x+y)²-(x+y)= 30

So (x+y)²-(x+y)-30= 0.

Substituting (x+y) with u, we get u²-u-30= 0.

Factorising this, we get (u-6)(u+5)= 0.

So u= 6 or -5, implying that (x+y)= 6 or -5.

Since (x+y)= 2xy, xy= (x+y)/2.

If (x+y)= 6, then xy= 6/2= 3.

If (x+y) = -5, then xy= -5/2.

Since only integers are accepted as the answer, xy= 3.