The problem asks us to solve a system of nonlinear equations:

1. $y = x - 2$
2. $y = x^2 - 2$

We need to find all ordered pairs $(x, y)$ that satisfy both equations.

Step 1: Set the equations equal to each other.

Since both equations equal $y$, we can set them equal to each other: $x - 2 = x^2 - 2$

Step 2: Simplify the equation.

Subtract $x - 2$ from both sides: $0 = x^2 - x$

Factorize: $x(x - 1) = 0$

Thus: $x = 0 \quad \text{or} \quad x = 1$

Step 3: Solve for $y$ for each $x$.

1. When $x = 0$: $y = x - 2 = 0 - 2 = -2$ So, one solution is $(0, -2)$.

2. When $x = 1$: $y = x - 2 = 1 - 2 = -1$ So, another solution is $(1, -1)$.

Final Answer:

The ordered pair solutions are: $(0, -2), (1, -1)$

Let me know if you would like me to provide further clarification or additional steps!

Follow-up Questions:

1. How do you determine if a system of equations is nonlinear?
2. What methods can you use to solve a system of equations involving higher-degree terms?
3. How does factoring help in simplifying equations like the one above?
4. What would happen if no real solutions existed for this system?
5. How can you verify that each solution satisfies both equations?

Tip:

Always double-check your solutions by substituting them back into the original equations.