Let's analyze and solve both systems of equations to determine the nature of their solutions.

System A:

The equations are: - x + 2y = -4 \tag{1} x = 4 + 2y \tag{2}

Step 1: Substituting $x$ from equation (2) into equation (1):

$- (4 + 2y) + 2y = -4$ Expanding this: $-4 - 2y + 2y = -4$ This simplifies to: $-4 = -4$ This is always true, meaning the system has infinitely many solutions.

Step 2: Finding the general solution:

Since we already know that equation (1) is always true, we can solve for $x$ in terms of $y$. From equation (2), we already have: $x = 4 + 2y$ Thus, the system has infinitely many solutions, and they must satisfy: $x = 4 + 2y$

System B:

The equations are: x - 3y = -3 \tag{1} - x + 3y = -3 \tag{2}

Step 1: Adding equations (1) and (2):

$(x - 3y) + (-x + 3y) = -3 + (-3)$ Simplifying this: $0 = -6$ This is a contradiction, meaning the system has no solution.

Conclusion:

• System A has infinitely many solutions, and they must satisfy $x = 4 + 2y$.
• System B has no solution.

Let me know if you would like further details or explanations. Here are five related questions you can explore:

1. What does it mean when a system of equations has infinitely many solutions?
2. How can you tell if a system has no solution without solving it?
3. What are consistent and inconsistent systems in linear algebra?
4. Can a system of equations have exactly two solutions?
5. How would the graphical interpretation of the solution look for both systems?

Tip: For linear systems, look for contradictions or dependencies between the equations to determine if there are no solutions or infinitely many solutions.