Помогите с решением пожалуйста. Система неравенств (2x+3)^2=5y (3x+2)^2=5y; След:...

Решите задачу:

$\left \{ {{2x+3)^2=5y} \atop {(3x+2)^2=5y}} \right. \left \{ {{(2x+3)^2=(3x+2)^2} \atop {(3x+2)^2=5y}} \right. \left \{ {{2x+3=3x+2} \atop {(3x+2)^2=5y}} \left \{ {{2x-3x=2-3} \atop {(3x+2)^2=5y}} \left \{ {{-x=-1} \atop {(3x+2)^2=5y}}$

$\left \{ {{x=1} \atop {(3*1+2)^2=5y}} \left \{ {{x=1} \atop {5y=5^2}} \left \{ {{x=1} \atop {y=5}}$

$\left \{ {{3x^2+y=5} \atop {6x^2-y=2}} \right. \left \{ {{9x^2=7} \atop {3x^2+y=5}} \right. \left \{ {{x^2= \frac{7}{9} } \atop {3x^2+y=5}} \right. \left \{ {{x^2= \frac{7}{9} } \atop {3*\frac{7}{9} +y=5}} \right.$

$\left \{ {{x^2=\frac{7}{9} } \atop {3*\frac{7}{9} +y=5}} \right. \left \{ {{x^2=\frac{7}{9} } \atop {2\frac{1}{3} +y=5}} \right. \left \{ {{x= \frac{+}{} \frac{ \sqrt{7} }{3} } \atop {y=2 \frac{2}{3} }} \right.$