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

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

1 Answer

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Answer 1 · 2016-07-18T08:24:17

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

Explanation:

Suming both sides of

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

we obtain

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

solving for $x$ $x$

$x = \frac{1}{6} (1 \pm \sqrt{373})$ $x=1/6 (1 pm sqrt[373])$

but

$y^{2} = \frac{x + 1}{3}$ $y^2=(x+1)/3$

so

$y = \pm \sqrt{\frac{1 + \frac{1}{6} (1 + \sqrt{373})}{3}}$ $y = pmsqrt((1+1/6 (1 + sqrt[373]))/3)$

so the solutions are

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

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

1 Answer

Explanation:

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How do you solve the system −x2+y2+10=0-x^2+y^2+10=0 and −3y2+x+1=0-3y^2+x+1=0?

1 Answer

Explanation:

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