Помогите решить №1423

Решите задачу:

$\left \{ {{y+5=x^2} \atop {x^2+y^2=25}} \right. \\ \left \{ {{y+5=x^2} \atop {x^2=25-y^2}} \right. \\ \left \{ {{y+5=x^2} \atop {x^2=(5-y)(5+y)}} \right. \\ \left \{ {{y+5=x^2} \atop {5+y=(5-y)(5+y)}} \right. \\ \left \{ {{y+5=x^2} \atop 1=5-y}} \right. \\ \left \{ {{y+5=x^2} \atop y=4}} \right. \\ \left \{ {{4+5=x^2} \atop y=4}} \right. \\ \left \{ {{x^2=9} \atop y=4}} \right. \\ \left \{ {{x=\pm3} \atop y=4}} \right. \\$

$\left \{ {{xy=16} \atop { \frac{x}{y} =4}} \right. \\ \left \{ {{xy=16} \atop {x =4y} \right. \\\left \{ {{4y*y=16} \atop {x =4y} \right. \\ \left \{ {{y^2=4} \atop {x =4y} \right\\ \left \{ {{y=\pm2} \atop {x =\pm8} \right. \\$

$\left \{ {{ x^{2} +2y^2=96} \atop {x=2y}} \right. \\ \left \{ {{ (2y)^{2} +2y^2=96} \atop {x=2y}} \right. \\ \left \{ {{ 4y^{2} +2y^2=96} \atop {x=2y}} \right. \\ \left \{ {{ 6y^2=96} \atop {x=2y}} \right. \\ \left \{ {{ y^2=16} \atop {x=2y}} \right. \\ \left \{ {{ y=\pm 4} \atop {x=2y}} \right. \\ \left \{ {{ y=\pm 4} \atop {x=\pm8}} \right. \\$