Решаем способом сложения:
[m]\left\{\begin {matrix}ax^2+4ax-y+7a+2 ≥0 \\ax^2+4ax-x+ay^2-2ay-y+7a+2+4a-1 ≥0 \end {matrix}\right.[/m]
[m]\left\{\begin {matrix}ax^2+4ax-y+7a+2 ≥0 \\ax^2+(4a-1)x+ay^2-(2a+1)y+7a+2+4a-1 ≥0 \end {matrix}\right.[/m]
Выделим полные квадраты во втором неравенстве
[m]\left\{\begin {matrix}ax^2+4ax-y+7a+2 ≥0 \\a(x^2+\frac{4a-1}{a}x)+a(y^2-\frac{2a+1}{a}y)+11a+1 ≥0 \end {matrix}\right.[/m]
[m]\left\{\begin {matrix}ax^2+4ax-y+7a+2 ≥0 \\a(x^2+2\cdot \frac{4a-1}{2a}x +(\frac{4a-1}{2a})^2)+a(y^2-2\cdot \frac{2a+1}{2a}y+(\frac{2a+1}{2a})^2)-(\frac{4a-1}{2a})^2-(\frac{2a+1}{2a})^2)+11a+1 ≥0 \end {matrix}\right.[/m]
[m]\left\{\begin {matrix}ax^2+4ax-y+7a+2 ≥0 \\a(x+ \frac{4a-1}{2a})^2+a(y- \frac{2a+1}{2a})^2-(\frac{4a-1}{2a})^2-(\frac{2a+1}{2a})^2+11a+1 ≥0 \end {matrix}\right.[/m]
Второе неравенство имеет единственное решение:
[m]\left\{\begin {matrix}x+ \frac{4a-1}{2a}=0 \\y- \frac{2a+1}{2a}=0 \end {matrix}\right.[/m]
[m]-(\frac{4a-1}{2a})^2-(\frac{2a+1}{2a})^2+11a+1 =0[/m] ⇒
22a^3-8a^2+2a-1=0