To verify the claims made by Tamara, Payton, Krish, and Haruki, we need to solve each of the equations and investigate their solution sets.
1. Solve Equation A: \[ 3(4 - 5h) - 7 = -12 + 2h \]
First, simplify the left side: \[ 3(4 - 5h) - 7 = 12 - 15h - 7 = 5 - 15h \]
Now we have: \[ 5 - 15h = -12 + 2h \]
Rearranging gives: \[ 5 + 12 = 15h + 2h \] \[ 17 = 17h \]
Dividing both sides by 17: \[ h = 1 \]
2. Solve Equation B: \[ 2(p - 3) - 1 = \frac{6p + 14}{2} \]
First, simplify the left side: \[ 2(p - 3) - 1 = 2p - 6 - 1 = 2p - 7 \]
Now simplify the right side: \[ \frac{6p + 14}{2} = 3p + 7 \]
So we have: \[ 2p - 7 = 3p + 7 \]
Rearranging gives: \[ 2p - 3p = 7 + 7 \] \[ -p = 14 \]
Dividing both sides by -1: \[ p = -14 \]
3. Solve Equation C: \[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \]
First, simplify the left side: \[ \frac{2}{3}v - 3 + \frac{8}{3} \]
Convert -3 to a fraction: \[ -3 = -\frac{9}{3} \] So we have: \[ \frac{2}{3}v - \frac{9}{3} + \frac{8}{3} = -1 \] Combining terms gives: \[ \frac{2}{3}v - \frac{1}{3} = -1 \]
Multiply through by 3 to eliminate the fraction: \[ 2v - 1 = -3 \]
Rearranging gives: \[ 2v = -3 + 1 \] \[ 2v = -2 \]
Dividing both sides by 2 gives: \[ v = -1 \]
Summary of Solutions:
- \( h = 1 \) (Equation A)
- \( p = -14 \) (Equation B)
- \( v = -1 \) (Equation C)
Now evaluate the truth of the statements made:
- Tamara: Says Equations A and C have the same solution. (False, because \( h = 1 \) and \( v = -1 \)).
- Payton: Says Equations B and C have the same solution. (False, because \( p = -14 \) and \( v = -1 \)).
- Krish: Says all three have the same solution. (False, they all have different solutions).
- Haruki: Says that none of the equations have the same solution. (True).
Conclusion: Haruki is correct!