The problem is asking which of the given equations have infinitely many solutions. Let's analyze each option:

For two equations to have infinitely many solutions, they must be equivalent, meaning that both sides of the equation are identical after simplification.

1. Option A: $6x + 35 = -6x - 35$
   Simplifying both sides:

   • Left side: $6x + 35$
   • Right side: $-6x - 35$

   Adding $6x$ to both sides: $12x + 35 = -35$ Subtracting 35 from both sides: $12x = -70$ Dividing by 12: $x = -\frac{70}{12} = -\frac{35}{6}$ So, this equation has only one solution, not infinitely many.

2. Option B: $-6x + 35 = -6x + 35$
   Both sides of the equation are already identical. This means it is true for all values of $x$, so this equation has infinitely many solutions.

3. Option C: $6x + 35 = -6x + 35$
   Adding $6x$ to both sides: $12x + 35 = 35$ Subtracting 35 from both sides: $12x = 0$ Dividing by 12: $x = 0$ This equation has only one solution, $x = 0$, so it does not have infinitely many solutions.

4. Option D: $-6x + 35 = -6x - 35$
   Adding $6x$ to both sides: $35 = -35$ This is a contradiction, so there are no solutions to this equation.

Conclusion:

• Option B is the correct answer, as it has infinitely many solutions.

Do you want further details or explanations?
Here are 5 related questions to consider:

1. What is the condition for two equations to have infinitely many solutions?
2. How do you recognize when an equation has no solutions?
3. What happens when both sides of an equation are equal?
4. How do you solve an equation step by step to find its solution?
5. What is the meaning of a contradiction in a system of equations?

Tip: When solving equations, always aim to isolate the variable on one side by performing inverse operations systematically.