Пусть \(x + y = a,\,\,\,xy = b.\) Тогда система уравнений примет вид:
\(\left\{ {\begin{array}{*{20}{c}}{5a + 2b = -19,}\\{a + 3b = -35\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{5\left( {-3b-35} \right) + 2b = -19,}\\{a = -3b-35\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{-13b = 156,}\\{a = -3b-35}\end{array}\,\,} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{b = -12,}\\{a = 1.\,\,\,\,\,\,}\end{array}} \right.\)
Вернёмся к прежним переменным:
\(\left\{ {\begin{array}{*{20}{c}}{x + y = 1,\,\,\,}\\{x \cdot y = -12}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 1-x,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{x \cdot \left( {1-x} \right) = -12}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 1-x,\,\,\,\,\,\,\,\,\,\,\,}\\{{x^2}-x-12 = 0}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = 1-x,}\\{\left[ {\begin{array}{*{20}{c}}{x = -3,}\\{x = 4\,\,\,}\end{array}} \right.}\end{array}} \right.\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = -3,}\\{y = 4,\,\,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = 4,\,\,}\\{y = -3.}\end{array}} \right.}\end{array}} \right.\)
Ответ: \(\left( {-3;4} \right),\,\,\,\left( {4;-3} \right).\)