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))#