To determine which of the given equations have infinitely many solutions, we need to analyze each equation step-by-step.
Equation A: [tex]\(6x + 35 = -6x - 35\)[/tex]
1. Combine like terms to isolate the [tex]\(x\)[/tex] terms on one side of the equation:
[tex]\[
6x + 6x + 35 = -35
\][/tex]
[tex]\[
12x + 35 = -35
\][/tex]
2. Subtract 35 from both sides:
[tex]\[
12x = -70
\][/tex]
3. Divide both sides by 12:
[tex]\[
x = -\frac{70}{12} \quad (\text{Simplifies to } x = -\frac{35}{6})
\][/tex]
Equation A has a unique solution [tex]\(-\frac{35}{6}\)[/tex], so it does not have infinitely many solutions.
Equation B: [tex]\(6x + 35 = -6x + 35\)[/tex]
1. Combine like terms to isolate the [tex]\(x\)[/tex] terms on one side of the equation:
[tex]\[
6x + 6x + 35 = 35
\][/tex]
[tex]\[
12x + 35 = 35
\][/tex]
2. Subtract 35 from both sides:
[tex]\[
12x = 0
\][/tex]
3. Divide both sides by 12:
[tex]\[
x = 0
\][/tex]
Equation B has a unique solution [tex]\(x = 0\)[/tex], so it does not have infinitely many solutions.
Equation C: [tex]\(-6x + 35 = -6x + 35\)[/tex]
1. Notice that both sides of the equation are identical. This means no matter what value of [tex]\(x\)[/tex] we substitute, the equation will always hold true.
Since any value of [tex]\(x\)[/tex] satisfies this equation, Equation C has infinitely many solutions.
Equation D: [tex]\(-6x + 35 = -6x - 35\)[/tex]
1. Combine like terms to isolate the [tex]\(x\)[/tex] terms on one side of the equation:
[tex]\[
-6x + 6x + 35 = -35
\][/tex]
[tex]\[
35 = -35
\][/tex]
Notice that 35 does not equal [tex]\(-35\)[/tex], which means this is a contradiction.
Equation D has no solutions because it leads to a false statement.
Based on the analysis, the only equation that has infinitely many solutions is:
- Equation C [tex]\(-6x + 35 = -6x + 35\)[/tex]
The answer is:
[tex]\[ \boxed{3} \][/tex]
