Рассмотрим первое уравнение системы:
\(xy-3y + x = 3\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,y\left( {x-3} \right) + \left( {x-3} \right) = 0\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left( {x-3} \right)\left( {y + 1} \right) = 0.\)
Тогда исходная система равносильна совокупности двух систем:
\(\left[ \begin{array}{l}\left\{ {\begin{array}{*{20}{c}}{x-3 = 0,\,\,\,\,\,\,\,\,}\\{{x^2} + {y^2} = 10,}\end{array}} \right.\\\left\{ {\begin{array}{*{20}{c}}{y + 1 = 0,\,\,\,\,\,\,\,\,\,}\\{{x^2} + {y^2} = 10.}\end{array}} \right.\end{array} \right.\)
Рассмотрим первую систему совокупности:
\(\left\{ {\begin{array}{*{20}{c}}{x-3 = 0,\,\,\,\,\,\,\,}\\{{x^2} + {y^2} = 10}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,\,\,\,\,\,\,\,\,\,}\\{9 + {y^2} = 10}\end{array}} \right.\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 3,}\\{y = 1,}\end{array}\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = 3,\,\,\,\,}\\{y = -1.}\end{array}} \right.}\end{array}} \right.\)
Рассмотрим вторую систему совокупности:
\(\left\{ {\begin{array}{*{20}{c}}{y + 1 = 0,\,\,\,\,\,\,\,}\\{{x^2} + {y^2} = 10}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{y = -1,\,\,\,\,\,\,\,\,}\\{{x^2} + 1 = 10}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 3,\,\,\,}\\{y = -1,}\end{array}} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = -3,}\\{y = -1.}\end{array}} \right.}\end{array}} \right.\)
Ответ: \(\left( {3;1} \right),\,\,\,\left( {3;-1} \right),\,\,\,\left( {-3;-1} \right).\)