Let's solve the system of equations:

1. $x^2 + 2y - 12 = 0$ (Equation 1)
2. $x^2 + y^2 = 36$ (Equation 2)

Step 1: Express $x^2$ from Equation 1

From Equation 1: $x^2 = 12 - 2y \quad \text{(Equation 3)}$

Step 2: Substitute $x^2$ in Equation 2

Substituting Equation 3 into Equation 2: $(12 - 2y) + y^2 = 36$

Simplify: $y^2 - 2y + 12 = 36$

$y^2 - 2y - 24 = 0$

Step 3: Solve the quadratic equation for $y$

Using the quadratic formula $y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $y = \frac{-(-2) \pm \sqrt{(-2)^2 - 4(1)(-24)}}{2(1)}$

$y = \frac{2 \pm \sqrt{4 + 96}}{2}$

$y = \frac{2 \pm \sqrt{100}}{2}$

$y = \frac{2 \pm 10}{2}$

$y = 6 \quad \text{or} \quad y = -4$

Step 4: Find corresponding $x$ values

For $y = 6$: $x^2 = 12 - 2(6) = 0 \implies x = 0$

For $y = -4$: $x^2 = 12 - 2(-4) = 20 \implies x = \pm \sqrt{20} = \pm 2\sqrt{5}$

Solutions

The solutions are:

• $(x, y) = (0, 6)$
• $(x, y) = (2\sqrt{5}, -4)$
• $(x, y) = (-2\sqrt{5}, -4)$

Would you like more details or have any questions?

Here are some related questions you might explore:

1. How do you solve systems of nonlinear equations in general?
2. What are the implications of having multiple solutions in such systems?
3. How can you verify the solutions to a system of equations?
4. What methods can be used if one equation is not easily solvable for a variable?
5. How do changes in coefficients affect the number and type of solutions?

Tip: Always check your solutions by substituting them back into the original equations to ensure they satisfy all given conditions.