Рассмотрим второе уравнение: \(\dfrac{x}{y} + \dfrac{y}{x} = 2.\)
Пусть \(\dfrac{x}{y} = t.\) Тогда:
\(t + \frac{1}{t} = 2\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{t^2}-2t + 1 = 0,}\\{t \ne 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,t = 1.\)
Тогда: \(\dfrac{x}{y} = 1\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,y = x\) и исходная система примет вид:
\(\left\{ {\begin{array}{*{20}{c}}{2x + 3y = 10,}\\{y = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2x + 3x = 10,}\\{y = x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = 2,}\\{y = 2.}\end{array}} \right.\)
Ответ: \(\left( {2;2} \right).\)