Question

How do you solve x2+y2=4 and y2=3x?

Answer 1

$x {}^{2} + 3x - 4 = 0 \\ x {}^{2} + 4x - x - 4 = 0 \\ x(x + 4) - 1(x + 4) = 0 \\ (x + 1)(x + 4) = 0 \\ x + 1 = 0 \\ x = 1 \\ x + 4 = 0 \\ x = - 4$