1.х^2 - у^2 - 5ху=-62. х+у=3

Решите задачу:

$\left \{ {{x^{2} - y^{2} -5xy=-6} \atop {x+y=3}} \right. \left \{ {{(3-y)^{2}-y^{2}-5y \cdot (3-y) +6=0} \atop {x=3-y}} \right. \\ \left \{ {{{9 -6y+y^{2}-y^{2}-15y+5y^{2}+6=0} \atop {x=3-y}} \right. \left \{ {{5y^{2}-21y+15=0} \atop {x=3-y}} \right. \\ 5y^{2}-21y+15=0 \\ y_{1,2}= \frac{21 \pm \sqrt{441 - 300} }{10}= \frac{21 \pm \sqrt{141} }{10} \; \left {{y_{1}=\frac{21 + \sqrt{141} }{10}} \atop {y_{2}=\frac{21 - \sqrt{141} }{10}}} \right.$
$x_{1}=3 - \frac{21 + \sqrt{141} }{10}}, \; x_{2}=3 - \frac{21 - \sqrt{141} }{10}}} \\ x_{1}= \frac{9 - \sqrt{141} }{10}}, \; x_{2}=\frac{9 + \sqrt{141} }{10}}}$