[tex]x^2+4y^2=100\implies x^2=100-4y^2[/tex]
[tex]\implies 4y-(100-4y^2)=-20[/tex]
[tex]\implies 4y^2+4y-80=0[/tex]
[tex]\implies y^2+y-20=0[/tex]
[tex]\implies (y-4)(y+5)=0[/tex]
[tex]\implies y=4,y=-5[/tex]
When [tex]y=4[/tex], you have
[tex]4(4)-x^2=-20\implies x^2=36\implies x=\pm6[/tex]
When [tex]y=-5[/tex],
[tex]4(-5)-x^2=-20\implies x^2=0\implies x=0[/tex]
So there are three solutions, [tex](x,y)\in\{(6,4),(-6,4),(0,-5)\}[/tex].
