(A) Given equations are:
$1) 2(3u - v) = 5uv \implies 6u - 2v = 5uv$
$2) 2(u + 3v) = 5uv \implies 2u + 6v = 5uv$
Divide both equations by $uv$ (assuming $u, v \neq 0$):
$1) \frac{6}{v} - \frac{2}{u} = 5$
$2) \frac{2}{v} + \frac{6}{u} = 5$
Let $x = \frac{1}{u}$ and $y = \frac{1}{v}$. The equations become:
$1) -2x + 6y = 5$
$2) 6x + 2y = 5$
Multiply equation $(2)$ by $3$: $18x + 6y = 15$. Subtract equation $(1)$ from this:
$(18x + 6y) - (-2x + 6y) = 15 - 5 \implies 20x = 10 \implies x = \frac{1}{2}$.
Substitute $x = \frac{1}{2}$ into $6x + 2y = 5$:
$6(\frac{1}{2}) + 2y = 5 \implies 3 + 2y = 5 \implies 2y = 2 \implies y = 1$.
Since $x = \frac{1}{u} = \frac{1}{2} \implies u = 2$ and $y = \frac{1}{v} = 1 \implies v = 1$.
Thus,$(u, v) = (2, 1)$.