To solve the given system of equations:

1. $y = x^2 + 4x + 6$
2. $y = -x + 2$

We need to find the points of intersection between the parabola and the straight line.

Steps to Solve:

1. Set the equations equal to each other: $x^2 + 4x + 6 = -x + 2$

2. Simplify the equation: $x^2 + 4x + x + 6 - 2 = 0$ $x^2 + 5x + 4 = 0$

3. Factorize the quadratic equation: $(x + 4)(x + 1) = 0$

4. Solve for $x$: $x = -4 \quad \text{and} \quad x = -1$

5. Find the corresponding $y$-values by substituting into $y = -x + 2$:

   • For $x = -4$: $y = -(-4) + 2 = 4 + 2 = 6$ So, one solution is $(-4, 6)$.
   • For $x = -1$: $y = -(-1) + 2 = 1 + 2 = 3$ So, another solution is $(-1, 3)$.

Final Answer:

The solutions are: $(-4, 6) \quad \text{and} \quad (-1, 3)$

Would you like further clarification on any of these steps?

Related Questions:

1. How can you verify these solutions graphically?
2. What happens if the parabola and the line do not intersect?
3. How do you solve systems of equations involving two parabolas?
4. Can you solve this problem using substitution or elimination?
5. What is the geometric interpretation of factoring the quadratic equation?

Tip: Always check your solutions by substituting them back into the original equations to confirm they satisfy both!