Let's analyze the two systems of equations and determine the type of solutions they have:

System A

$6x - y = 5 \quad \text{and} \quad -6x + y = 5$

1. Step 1: Add the two equations
   $(6x - y) + (-6x + y) = 5 + 5$ Simplify:
   $0 = 10$

   This is a contradiction, meaning the system has no solution. The lines are parallel.

System B

$x + 4y = 7 \quad \text{and} \quad -x + 4y = 1$

1. Step 1: Add the two equations
   $(x + 4y) + (-x + 4y) = 7 + 1$ Simplify:
   $8y = 8$ Solve for $y$:
   $y = 1$

2. Step 2: Substitute $y = 1$ into the first equation $x + 4y = 7$:
   $x + 4(1) = 7$ Simplify:
   $x + 4 = 7$ Solve for $x$:
   $x = 3$

   Therefore, the system has a unique solution:
   $(x, y) = (3, 1)$

Final Answers:

1. System A: The system has no solution.
2. System B: The system has a unique solution: $(3, 1)$.

Do you need further clarification or a detailed breakdown for any step? Let me know!

Here are 5 related questions for practice:

1. How do you determine if two lines are parallel in a system of equations?
2. What does it mean for a system to have infinitely many solutions?
3. Solve $2x + 3y = 6$ and $4x + 6y = 12$. What type of solution does it have?
4. Solve a system where the two lines are identical.
5. What are the steps to solve a system using substitution?

Tip: To quickly check for contradictions, try adding or subtracting equations in a system to see if the result is valid.