How do you solve the system $-x^2+y^2+10=0$ and $-3y^2+x+1=0$ ?

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Answer 1 · 2016-07-18T08:24:17

${x = 1/6 (1 + sqrt[373]), y = -1/3 sqrt[1/2 (7 + sqrt[373])]}$ and
${x = 1/6 (1 + sqrt[373]), y = 1/3 sqrt[1/2 (7 + sqrt[373])]}$

Suming both sides of

$3(-x^2+y^2+10)=0$ and
$-3y^2+x+1=0$

we obtain

$-3x^2+x+31=0$

solving for $x$

$x=1/6 (1 pm sqrt[373])$

but

$y^2=(x+1)/3$

so

$y = pmsqrt((1+1/6 (1 + sqrt[373]))/3)$

so the solutions are

${x = 1/6 (1 + sqrt[373]), y = -1/3 sqrt[1/2 (7 + sqrt[373])]}$ and
${x = 1/6 (1 + sqrt[373]), y = 1/3 sqrt[1/2 (7 + sqrt[373])]}$

How do you solve the system -x^2+y^2+10=0-x^2+y^2+10=0 and -3y^2+x+1=0-3y^2+x+1=0?