Solve the system of equations: $x^2 + y^2 = 3$ $x^2 - 18 = y$

1. Problem Description

Solve the system of equations:

$x^2 + y^2 = 3$

$x^2 - 18 = y$

2. Solution Steps

We have the following system of equations:

$x^2 + y^2 = 3$ (1)

$x^2 - 18 = y$ (2)

From equation (2), we can express $x^2$ in terms of $y$:

$x^2 = y + 18$ (3)

Substitute equation (3) into equation (1):

$(y + 18) + y^2 = 3$

$y^2 + y + 18 - 3 = 0$

$y^2 + y + 15 = 0$

Use the quadratic formula to solve for $y$:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

In this case, $a = 1$, $b = 1$, and $c = 15$.

$y = \frac{-1 \pm \sqrt{1^2 - 4(1)(15)}}{2(1)}$

$y = \frac{-1 \pm \sqrt{1 - 60}}{2}$

$y = \frac{-1 \pm \sqrt{-59}}{2}$

Since the discriminant is negative, there are no real solutions for $y$. Therefore, there are no real solutions for the system of equations.

3. Final Answer

No real solutions.