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)\}`.