Answer
Solutions: $(4, 0)$, $(-3, \sqrt{7})$, and $(-3, -\sqrt{7})$
Work Step by Step
$$x + y^2 = 4$$ $$x^2 + y^2 = 16$$ To solve this exercise, we can use the addition method in the following manner: $$-1(x + y^2 = 4)$$ $$x^2 + y^2 = 16$$ which becomes: $$-x - y^2 = -4$$ $$x^2 + y^2 = 16$$ and, when added: $$x^2 - x + 0y^2 = 12$$ $$x^2 - x - 12 = 0$$ $$(x-4)(x+3) = 0$$ $$x = 4$$ $$OR$$ $$x = -3$$ Using these two values, we can "plug" them into the first equation to find the respective values of $y$: $$x + y^2 = 4$$ $$4 + y^2 = 4$$ $$y^2 = 0$$ $$y = 0$$ $$AND$$ $$-3 + y^2 = 4$$ $$y^2 = 7$$ $$y = \frac{+}{}\sqrt{7}$$ We conclude, therefore, that there are 3 solutions: $(4, 0)$, $(-3, \sqrt{7})$, and $(-3, -\sqrt{7})$ which, when substituted, all check out as true solutions.
