To determine who is correct about the linear equations, we need to solve each equation one by one.
Equation A: \[ 3(4 - 5h) - 7 = -12 + 2h \] Distributing the 3: \[ 12 - 15h - 7 = -12 + 2h \] Simplifying: \[ 5 - 15h = -12 + 2h \] Adding \(15h\) and \(12\) to both sides: \[ 5 + 12 = 15h + 2h \] \[ 17 = 17h \] Dividing both sides by 17: \[ h = 1 \]
Equation B: \[ 2(p - 3) - 1 = \frac{6p + 14}{2} \] Distributing the 2: \[ 2p - 6 - 1 = 3p + 7 \] Simplifying: \[ 2p - 7 = 3p + 7 \] Subtracting \(2p\) from both sides: \[ -7 = p + 7 \] Subtracting \(7\) from both sides: \[ -14 = p \]
Equation C: \[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \] Combining constants on the left: \[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \] Changing -3 to have a denominator of 3: \[ \frac{2}{3}v - \frac{9}{3} + \frac{8}{3} = -1 \] Combining like terms: \[ \frac{2}{3}v - \frac{1}{3} = -1 \] Adding \(\frac{1}{3}\) to both sides: \[ \frac{2}{3}v = -1 + \frac{1}{3} \] Changing -1 to have a denominator of 3: \[ \frac{2}{3}v = -\frac{3}{3} + \frac{1}{3} \] Thus: \[ \frac{2}{3}v = -\frac{2}{3} \] Multiplying both sides by \(\frac{3}{2}\): \[ v = -1 \]
Summarizing the results:
- Equation A has the solution \(h = 1\).
- Equation B has the solution \(p = -14\).
- Equation C has the solution \(v = -1\).
From this information:
- The solutions are different for all three equations: \(h = 1\), \(p = -14\), \(v = -1\).
Therefore, Haruki is correct in saying that none of the equations have the same solution.
Final answer: Haruki