To determine who is correct about the solution sets for the equations, we will solve each equation step by step.
Equation A:
\[ 3(4 - 5h) - 7 = -12 + 2h \]
- Distribute on the left side: \[ 12 - 15h - 7 = -12 + 2h \] \[ 5 - 15h = -12 + 2h \]
- Add \(15h\) to both sides: \[ 5 = -12 + 17h \]
- Add \(12\) to both sides: \[ 17 = 17h \]
- Divide by \(17\): \[ h = 1 \]
Equation B:
\[ 2(p - 3) - 1 = \frac{6p + 14}{2} \]
- Distribute on the left side: \[ 2p - 6 - 1 = \frac{6p + 14}{2} \] \[ 2p - 7 = \frac{6p + 14}{2} \]
- Multiply everything by \(2\) to eliminate the fraction: \[ 2(2p - 7) = 6p + 14 \] \[ 4p - 14 = 6p + 14 \]
- Subtract \(4p\) from both sides: \[ -14 = 2p + 14 \]
- Subtract \(14\) from both sides: \[ -28 = 2p \]
- Divide by \(2\): \[ p = -14 \]
Equation C:
\[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \]
- Combine constants on the left side: \[ \frac{2}{3}v + \frac{8}{3} - 3 = -1 \] First, convert \(3\) to thirds: \[ 3 = \frac{9}{3} \] So: \[ \frac{2}{3}v + \frac{8}{3} - \frac{9}{3} = -1 \] \[ \frac{2}{3}v - \frac{1}{3} = -1 \]
- Add \(\frac{1}{3}\) to both sides: \[ \frac{2}{3}v = -1 + \frac{1}{3} \] Convert \(-1\) to thirds: \[ -1 = -\frac{3}{3} \] So: \[ \frac{2}{3}v = -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3} \]
- Multiply both sides by \(\frac{3}{2}\): \[ v = -1 \]
Summary of Solutions:
- Equation A: \(h = 1\)
- Equation B: \(p = -14\)
- Equation C: \(v = -1\)
Conclusion:
- Tamara claims Equation A and C have the same solution (Incorrect).
- Payton claims Equation B and C have the same solution (Incorrect).
- Krish claims all three have the same solution (Incorrect).
- Haruki claims none of the equations have the same solution (Correct).
Final statement: Haruki is correct; none of the equations have the same solution.