To determine which of the following equations have infinitely many solutions, we will solve each equation step-by-step:
### Equation A: [tex]\(6x + 35 = -6x - 35\)[/tex]
1. Move all terms involving [tex]\(x\)[/tex] to one side and constants to the other side.
[tex]\[6x + 6x = -35 - 35\][/tex]
2. Combine like terms:
[tex]\[12x = -70\][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[x = \frac{-70}{12} = \frac{-35}{6}\][/tex]
4. This equation has a unique solution, not infinitely many solutions.
### Equation B: [tex]\(6x + 35 = -6x + 35\)[/tex]
1. Move all terms involving [tex]\(x\)[/tex] to one side:
[tex]\[6x + 6x = 35 - 35\][/tex]
2. Combine like terms:
[tex]\[12x = 0\][/tex]
3. Solve for [tex]\(x\)[/tex]:
[tex]\[x = 0\][/tex]
4. This equation has a single unique solution, not infinitely many solutions.
### Equation C: [tex]\(-6x + 35 = -6x + 35\)[/tex]
1. Move all terms involving [tex]\(x\)[/tex] to one side:
[tex]\[-6x + 6x = 35 - 35\][/tex]
2. Combine like terms:
[tex]\[0 = 0\][/tex]
3. Since [tex]\(0 = 0\)[/tex] holds true for all [tex]\(x\)[/tex], this equation is an identity.
4. This equation has infinitely many solutions.
### Equation D: [tex]\(-6x + 35 = -6x - 35\)[/tex]
1. Move all terms involving [tex]\(x\)[/tex] to one side and constants to the other side:
[tex]\[-6x + 6x = -35 - 35\][/tex]
2. Combine like terms:
[tex]\[0 = -70\][/tex]
3. This results in a contradiction since [tex]\(0 \neq -70\)[/tex].
4. This equation has no solutions.
Therefore, the equation that has infinitely many solutions is option C: [tex]\(-6x + 35 = -6x + 35\)[/tex]. The answer is:
[tex]\[ \boxed{\text{[3]}} \][/tex]
