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Ответ:

Вот ответ это правило Мне поставили 5

[tex]\displaystyle\bf\\\left \{ {{x+y=5} \atop {y^{2} +4xy=33}} \right. \\\\\\\left \{ {{x=5-y} \atop {y^{2} +4\cdot(5-y)\cdot y=33}} \right. \\\\\\\left \{ {{x=5-y} \atop {y^{2} +20y-4y^{2} =33}} \right. \\\\\\\left \{ {{x=5-y} \atop {3y^{2} -20y+33=0}} \right. \\\\\\3y^{2} -20y+33=0\\\\D=(-20)^{2} -4\cdot 3\cdot 33=400-396=4=2^{2} \\\\\\y_{1} =\frac{20-2}{6} =\frac{18}{6} =3\\\\\\y_{2} =\frac{20+2}{6} =\frac{22}{6} =3\frac{2}{3} \\\\\\x_{1} =5-3=2[/tex]

[tex]\displaystyle\bf\\x_{2} =5-3\frac{2}{3} =1\frac{1}{3} \\\\\\Otvet \ : \ \Big(2 \ ; \ 3\Big) \ , \ \Big(1\frac{1}{3} \ ; \ 3\frac{2}{3} \Big)[/tex]