\(\left\{ {\begin{array}{*{20}{c}}{{x^2}-y + 2{y^2} = 29,\,\,\,\,\,\,}\\{{y^2}-0,5y + x = 15\left| { \cdot 2} \right.\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{{x^2}-y + 2{y^2} = 29,}\\{2{y^2}-y + 2x = 30}\end{array}} \right.} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{y^2}-y = 29-{x^2},}\\{2{y^2}-y = 30-2x}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{y^2}-y = 29-{x^2}}\\{30-2x = 29-{x^2}}\end{array}\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{y^2}-y = 29-{x^2},}\\{{x^2}-2x + 1 = 0\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{y^2}-y = 29-{x^2},}\\{x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}\,\,\,\,\,\,\, \Leftrightarrow } \right.} \right.\)
\( \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{2{y^2}-y-28 = 0,}\\{x = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\left\{ {\begin{array}{*{20}{c}}{\left[ {\begin{array}{*{20}{c}}{y = 4,\,\,\,\,\,}\\{y = -\dfrac{7}{2},}\end{array}} \right.}\\{x = 1\,\,\,\,\,\,\,\,\,}\end{array}} \right.\,\,\,\,\,\,\,\,\,\, \Leftrightarrow \,\,\,\,\,\,\,\,\,\left[ {\begin{array}{*{20}{c}}{\left\{ {\begin{array}{*{20}{c}}{x = 1,}\\{y = 4,}\end{array}\,\,\,\,\,} \right.}\\{\left\{ {\begin{array}{*{20}{c}}{x = 1,\,\,\,\,}\\{y = -\dfrac{7}{2}.}\end{array}} \right.}\end{array}} \right.\)
Ответ: \(\left( {1;4} \right),\,\,\,\left( {1;-\dfrac{7}{2}} \right).\)