To solve the equations and evaluate the claims made by Tamara, Payton, Krish, and Haruki, let’s simplify each equation one by one.
Equation A:
\[ 3(4 - 5h) - 7 = -12 + 2h \]
Step 1: Distribute on the left side. \[ 12 - 15h - 7 = -12 + 2h \]
Step 2: Combine like terms. \[ 5 - 15h = -12 + 2h \]
Step 3: Rearrange to isolate \(h\). \[ 5 + 12 = 2h + 15h \] \[ 17 = 17h \]
Step 4: Divide by 17. \[ h = 1 \]
Equation B:
\[ 2(p - 3) - 1 = \frac{6p}{2} + \frac{14}{2} \]
Step 1: Simplify the right side. \[ \frac{6p}{2} = 3p \quad \text{and} \quad \frac{14}{2} = 7 \]
So, simplifying the equation we get: \[ 2(p - 3) - 1 = 3p + 7 \]
Step 2: Distribute on the left side. \[ 2p - 6 - 1 = 3p + 7 \] \[ 2p - 7 = 3p + 7 \]
Step 3: Rearrange to isolate \(p\). \[ 2p - 3p = 7 + 7 \] \[ -p = 14 \]
Step 4: Divide by -1. \[ p = -14 \]
Equation C:
\[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \]
Step 1: Combine the constant terms on the left. \[ -3 + \frac{8}{3} = -\frac{9}{3} + \frac{8}{3} = -\frac{1}{3} \]
So we rewrite the equation as: \[ \frac{2}{3}v - \frac{1}{3} = -1 \]
Step 2: Add \(\frac{1}{3}\) to both sides. \[ \frac{2}{3}v = -1 + \frac{1}{3} = -\frac{3}{3} + \frac{1}{3} = -\frac{2}{3} \]
Step 3: Multiply both sides by \(\frac{3}{2}\) to isolate \(v\). \[ v = -1 \]
Summary of Solutions:
- \(h = 1\) (from Equation A)
- \(p = -14\) (from Equation B)
- \(v = -1\) (from Equation C)
Claims Analysis:
- Tamara says Equation A and Equation C have the same solution. (False)
- \(h = 1\), \(v = -1\)
- Payton says Equation B and Equation C have the same solution. (False)
- \(p = -14\), \(v = -1\)
- Krish says all three have the same solution. (False)
- Haruki says that none of the equations have the same solution. (True)
Conclusion:
Haruki is correct, as none of the equations share the same solution.
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