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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.

Two systems of equations are given. System A: -x + 2y = -4 and x = 4 + 2y. System B: x - 3y = -3 and -x + 3y = -3. Choose the best description of their solution.

Analyze and solve two systems of linear equations. Learn how to identify whether the system has infinitely many solutions, no solutions, or a unique solution. Suitable for students in Grades 9-12, this problem covers key algebraic concepts such as consistency and substitution methods.

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