To determine who is correct about the solution sets of the equations, we need to solve each equation step by step.
Equation A:
\[ 3(4 - 5h) - 7 = -12 + 2h \]
Expanding and simplifying:
\[ 12 - 15h - 7 = -12 + 2h \]
\[ 5 - 15h = -12 + 2h \]
Adding \(15h\) to both sides:
\[ 5 = -12 + 17h \]
Adding \(12\) to both sides:
\[ 17 = 17h \]
Dividing by \(17\):
\[ h = 1 \]
Equation B:
\[ 2(p - 3) - 1 = \frac{6p + 14}{2} \]
Expanding the left side:
\[ 2p - 6 - 1 = \frac{6p + 14}{2} \]
Simplifying further:
\[ 2p - 7 = \frac{6p + 14}{2} \]
To eliminate the fraction, multiply both sides by \(2\):
\[ 2(2p - 7) = 6p + 14 \]
This gives:
\[ 4p - 14 = 6p + 14 \]
Rearranging gives:
\[ -14 - 14 = 6p - 4p \]
\[ -28 = 2p \]
Dividing by \(2\):
\[ p = -14 \]
Equation C:
\[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \]
Combining terms on the left:
\[ \frac{2}{3}v + \frac{8}{3} - 3 = -1 \]
Convert \(3\) into thirds:
\[ \frac{2}{3}v + \frac{8}{3} - \frac{9}{3} = -1 \]
This simplifies to:
\[ \frac{2}{3}v - \frac{1}{3} = -1 \]
Adding \(\frac{1}{3}\) to both sides:
\[ \frac{2}{3}v = -1 + \frac{1}{3} \]
Converting \(-1\) into thirds:
\[ \frac{2}{3}v = -\frac{3}{3} + \frac{1}{3} \]
This gives:
\[ \frac{2}{3}v = -\frac{2}{3} \]
Multiplying both sides by \(\frac{3}{2}\):
\[ v = -1 \]
Now we have the following solutions:
- Equation A \(h = 1\)
- Equation B \(p = -14\)
- Equation C \(v = -1\)
Based on the solutions:
- Tamara's claim that Equation A and Equation C have the same solution is incorrect.
- Payton's claim that Equation B and Equation C have the same solution is incorrect.
- Krish's claim that all three have the same solution is incorrect.
- Haruki's claim that none of the equations have the same solution is correct.
Therefore, the correct answer is Haruki.