Đặt: $a=\frac{1}{x};$ $b=\frac{1}{y}$
Ta có hpt:
$\left \{ {{a+b=\frac{5}{24}} \atop {a=\frac{3b}{2}}} \right.$
$⇔$$\frac{5b}{2}=$ $\frac{5}{24}$
$⇔120b=10$
$⇔b=$$\frac{1}{12}$
$⇔y=12$
$⇔a=$$\frac{3}{2}.$ $\frac{1}{12}$
$⇔a=$$\frac{1}{8}$
$⇔x=8$
Vậy $(x,y)∈${$8;12$}