Question

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

Answer 1

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

Explanation:

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))#