Question

X^2 + 4y^2 =100; 4y - x^2 = -20

Answer 1

$x^2+4y^2=100\implies x^2=100-4y^2$
$\implies 4y-(100-4y^2)=-20$
$\implies 4y^2+4y-80=0$
$\implies y^2+y-20=0$
$\implies (y-4)(y+5)=0$
$\implies y=4,y=-5$

When $y=4$, you have

$4(4)-x^2=-20\implies x^2=36\implies x=\pm6$

When $y=-5$,

$4(-5)-x^2=-20\implies x^2=0\implies x=0$

So there are three solutions, $(x,y)\in\{(6,4),(-6,4),(0,-5)\}$.