(A) Given equations are:
$(1)$ $6x + 3y = 6xy$
$(2)$ $2x + 4y = 5xy$
Divide both equations by $xy$:
From $(1)$: $\frac{6x}{xy} + \frac{3y}{xy} = \frac{6xy}{xy} \implies \frac{6}{y} + \frac{3}{x} = 6$
From $(2)$: $\frac{2x}{xy} + \frac{4y}{xy} = \frac{5xy}{xy} \implies \frac{2}{y} + \frac{4}{x} = 5$
Let $u = \frac{1}{x}$ and $v = \frac{1}{y}$. The equations become:
$(3)$ $3u + 6v = 6$
$(4)$ $4u + 2v = 5$
Multiply $(4)$ by $3$ to eliminate $v$:
$12u + 6v = 15$ $(5)$
Subtract $(3)$ from $(5)$:
$(12u + 6v) - (3u + 6v) = 15 - 6$
$9u = 9 \implies u = 1$
Substitute $u = 1$ into $(3)$:
$3(1) + 6v = 6 \implies 6v = 3 \implies v = \frac{1}{2}$
Since $u = \frac{1}{x} = 1 \implies x = 1$
Since $v = \frac{1}{y} = \frac{1}{2} \implies y = 2$
Thus,the solution is $(x, y) = (1, 2)$.