Решите систему неравенств

Решите задачу:

$\left \{ {{x^2+y=7} \atop {x-y=3}} \right. \\ \\ \left \{ {{y=7- x^{2} } \atop {x-y=3}} \right. \\ \\ x-7+x^2=3 \\ x^{2} +x-10=0 \\ D=1+40=41 \\ \\ \sqrt{D} = \sqrt{41} \\ \\ x1= \frac{-1+ \sqrt{41} }{2} \\ \\ x2= \frac{-1- \sqrt{41} }{2} \\ \\ y1=x-3= \frac{-1+ \sqrt{41} }{2} -3= \frac{-7+ \sqrt{41} }{2} \\ \\ y2= \frac{-7- \sqrt{41} }{2}$