Let's analyze the two systems of equations to determine their solutions.

System A

The equations are:

1. $y = 2x + 8$
2. $2x - y = -8$

Substitute $y = 2x + 8$ into the second equation:

$2x - (2x + 8) = -8$

Simplifying:

$2x - 2x - 8 = -8 \quad \Rightarrow \quad -8 = -8$

This is a true statement, indicating that the two equations are dependent, and hence, the system has infinitely many solutions. The solution set satisfies $y = 2x + 8$.

System B

The equations are:

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

Add the two equations:

$(2x + y) + (2x - y) = -4 - 4$

Simplifying:

$4x = -8 \quad \Rightarrow \quad x = -2$

Substitute $x = -2$ back into either equation, for example, $2x + y = -4$:

$2(-2) + y = -4 \quad \Rightarrow \quad -4 + y = -4 \quad \Rightarrow \quad y = 0$

So, the unique solution for System B is $(x, y) = (-2, 0)$.

Conclusion

• System A: The system has infinitely many solutions that satisfy the equation $y = 2x + 8$.
• System B: The system has a unique solution $(x, y) = (-2, 0)$.

Would you like more details on these solutions or have any questions? Here are some related questions:

1. How can you determine if a system of equations has no solution?
2. What does it mean if a system of equations has infinitely many solutions?
3. What is the significance of consistent and inconsistent systems in linear equations?
4. How do you use substitution to solve a system of linear equations?
5. What are the different methods to solve systems of linear equations (e.g., elimination, graphing)?

Tip: When solving systems of linear equations, always check if substituting the solution into both original equations satisfies them. This ensures the solution is valid.