To solve this problem, let's analyze each system of equations separately:

System A

The equations given are: $6x + 7y = 5$ $5x + 7y = 3$

To determine the nature of the solution, we can use the elimination method. Subtract the second equation from the first: $(6x + 7y) - (5x + 7y) = 5 - 3$ This simplifies to: $x = 2$

Now substitute $x = 2$ back into one of the original equations to find $y$. Using the first equation: $6(2) + 7y = 5$ $12 + 7y = 5$ $7y = 5 - 12$ $7y = -7$ $y = -1$

So, the solution for System A is $(x, y) = (2, -1)$, indicating that it has a unique solution.

System B

The equations given are: $y = 4x + 8$ $y = 4x + 5$

Since both equations have the same slope ($4x$) but different y-intercepts (8 and 5), they represent parallel lines. Parallel lines do not intersect, meaning there is no solution for System B.

Summary:

• System A: The system has a unique solution, $(2, -1)$.
• System B: The system has no solution.

Would you like more details or have any questions?

Here are 5 related questions to consider:

1. How do you determine if two lines are parallel from their equations?
2. What other methods could be used to solve System A?
3. How would the solution change if System B's equations had the same y-intercept?
4. What are the implications of a system having infinitely many solutions?
5. Can you always use elimination to solve systems of equations?

Tip: When using elimination, it's helpful to line up similar terms and simplify step by step to avoid errors.