How do you solve the system $-3x^2+y^2-3x=0$ and $x^2-y^2+27=0$ ?

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Answer 1 · 2016-12-15T23:24:05

$((x = -9/2, y = -3 sqrt[21]/2), (x = -9/2, y = 3 sqrt[21]/2), (x = 3, y = -6), (x = 3, y = 6))$

Considering

${(-3x^2+y^2=3x),(x^2-y^2=-27):}$

Adding term to term the two equations we get

$-2x^2=3x-27$ . Now solving for $x$

$x =-9/2$ and $x = 3$ . Now using the second equation

$(-9/2)^2-y^2=-27->y = pm3 sqrt[21]/2$

$3^2-y^2=-27->y = pm 6$

so the solutions are:

$((x = -9/2, y = -3 sqrt[21]/2), (x = -9/2, y = 3 sqrt[21]/2), (x = 3, y = -6), (x = 3, y = 6))$

How do you solve the system -3x^2+y^2-3x=0-3x^2+y^2-3x=0 and x^2-y^2+27=0x^2-y^2+27=0?