## anonymous 5 years ago Solve the system by the addition method. 2x^2-3y^2=-19 3x^2+2y^2=30 Solution set is ???

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)