How do you solve the system $x^2-y-4=0$ and $x^2+3y^2-4y-10=0$ ?

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Answer 1 · 2016-09-25T13:19:35

$y = -1, x=pmsqrt(3)$
$y = 2,x=pmsqrt(6)$

${(x^2-y-4=0),(x^2+3y^2-4y-10=0):}$

Substracting the first from the second

$3(y^2-y-2)=0$ with solutions

$y = -1, x=pmsqrt(3)$
$y = 2,x=pmsqrt(6)$

How do you solve the system x^2-y-4=0x^2-y-4=0 and x^2+3y^2-4y-10=0x^2+3y^2-4y-10=0?