Решите срочно надо ???

1 Задание
а)
$\left \{ {{2x+y=1} \atop {5x+2y=0}} \right. \\ \left \{ {{y=1-2x} \atop {5x+2*(1-2x)=0}} \right. \\ \left \{ {{y=1-2x} \atop {5x+2-4x=2}} \right. \\ \left \{ {{y=1-2x} \atop {x=0}} \right. \\ \left \{ {{y=1} \atop {x=0}} \right.$
Ответ: y=1, x=0
б)
$\left \{ {{x-y=4} \atop { x^{2}- y^{2} =40}} \right. \\ \left \{ {{x=y+4} \atop { (y+4)^{2}- y^{2} }=40} \right. \\ \left \{ {{x=y+4} \atop { y^{2}+16+8y-y^2 }=40 \right. \\ \left \{ {{x=y+4} \atop {8y=24}} \right. \\ \left \{ {{x=y+4} \atop {y=3}} \right. \\ \left \{ {{x=7} \atop {y=3}} \right.$
Ответ: x=7,y=3

2 Задание
$\left \{ {{y=4-x^2} \atop {x-y+2=0}} \right. \\ \left \{ {{y=4-x^2} \atop {x-4+x^2+2=0}} \right. \\ \left \{ {{y=4-x^2} \atop {x^2+x-2=0}} \right. \\ \left \{ {{y=4-x^2} \atop { x_{1} =1; x_{2} =-2}} \right. \\ \left \{ {{ y_{1}=3; y_{2}=0 } \atop {x_{1} =1; x_{2} =-2}} \right. \\$