Question

Solve the system by the addition method.
2x^2-3y^2=-19
3x^2+2y^2=30
Solution set is ???

Answer 1

2*(2x^2-3y^2)+3(3x^2+2y^2)=-19*2+30*3
13x^2=52;
x=2,-2
2y^2=30-12
y=3,-3
solution set is \[(\pm 2,\pm 3)\]
i.e. (2,3) (2,-3) (-2,3) (-2,-3)