How many solutions of the system x - y = 3, x^2 - y = 3 are there such that both x and y are real?
+2666 Convert the second equation into a quadratic: \(y = x^2 - 3\)
Substitute it into the first equation: \(-x^2 + x + 3 = 3\)
Simplify: \(-x^2 + x = 0\)
Note that because the discriminant (\(1 - (4 \times 1 \times 0) = 1 \)) is positive, there will be exactly \(\color{brown}\boxed{2}\) real solutions.
