Решение систем уравнений второй степени

Решите задачу:

$\left \{ {{x-2y=0} \atop {8x^2-4y^2=252}} \right. \ \ \ \textless \ =\ \textgreater \ \ \ \left \{ {{x=2y} \atop {8x^2-4y^2=252}} \right. \\ \\ 8(2y)^2-4y^2=252 \\ 8*4y^2-4y^2=252 \\ 32y^2-4y^2=252 \\ 28y^2=252 \\ y^2=252/28=9 \\ \\ y_1=3 \\ y_2=-3\\ \\ x_1=2y=2*3=6 \\ x_2=2*(-3)=-6 \\ \\ OTBET: \ (3;6) \ , (-3;-6)$

$\left \{ {{4x^2+8y=100} \atop {y+6x=26}} \right. \ \ \ \textless \ =\ \textgreater \ \ \ \left \{ {{4x^2+8y=100} \atop {y=26-6x}} \right. \\ \\ 4x^2+8(26-6x)=100 \\ 4x^2+ 208-48x=100 \\ 4x^2-48x+108=0\ |:4 \\ x^2-12x+27=0 \\ \\ x_1=3 \\ x_2=9 \\ \\ y_1=26-6x=26-6*3=26-18=8 \\ \\ y_2=26-6*9=26-54=-28 \\ \\ OTBET: (3;8); \ \ (9;-28)$