$\text{Ta có: $\dfrac{x}{y}$ = $\dfrac{2}{3}$ ⇔ $\dfrac{x}{2}$ = $\dfrac{y}{3}$}$
$\text{Đặt: $\dfrac{x}{2}$ = $\dfrac{y}{3}$ = k ⇒ $\left \{ {{x = 2k} \atop {y = 3k}} \right.$ }$
$\text{Có: x.y = 24 ⇔ 2k.3k = 24}$
$\text{6$k^{2}$ = 24}$
$\text{$k^{2}$ = 24 ÷ 6}$
$\text{$k^{2}$ = 4}$
$\text{$k^{2}$ = $2^{2}$}$
$\text{⇒ k = ± 2}$
$\text{Với k = 2 ⇒ $\left \{ {{x = 2.2 = 4} \atop {y = 2.3 = 6}} \right.$ }$
$\text{Với k = - 2 ⇒ $\left \{ {{x = (-2).2 = - 4} \atop {y = (-2).3 = - 6}} \right.$ }$
$\text{Vậy $\left \{ {{x = 4} \atop {y = 6}} \right.$ hoặc $\left \{ {{x = - 4} \atop {y = - 6}} \right.$}$
