To determine whose assertion about the equations A, B, and C is correct, we need to solve each equation and find their respective solution sets.
Equation A:
\( 3(4 - 5h) - 7 = -12 + 2h \)
- Distribute: \( 12 - 15h - 7 = -12 + 2h \)
- Combine like terms: \( 5 - 15h = -12 + 2h \)
- Add \( 15h \) and \( 12 \) to both sides: \( 5 + 12 = 15h + 2h \)
- Simplify: \( 17 = 17h \)
- Divide by 17: \( h = 1 \)
Equation B:
\( 2(p - 3) - 1 = \frac{6p + 14}{2} \)
- Distribute: \( 2p - 6 - 1 = \frac{6p + 14}{2} \)
- Combine like terms: \( 2p - 7 = \frac{6p + 14}{2} \)
- Multiply both sides by 2 to eliminate the fraction: \( 2(2p - 7) = 6p + 14 \)
- Distribute: \( 4p - 14 = 6p + 14 \)
- Rearranging gives: \( 4p - 6p = 14 + 14 \)
- Combine like terms: \( -2p = 28 \)
- Divide by -2: \( p = -14 \)
Equation C:
\( \frac{2}{3}v - 3 + \frac{8}{3} = -1 \)
- Combine like terms: \( \frac{2}{3}v - 3 + \frac{8}{3} = -1 \)
- Convert -3 to thirds: \( -3 = -\frac{9}{3} \), so the equation becomes: \[ \frac{2}{3}v - \frac{9}{3} + \frac{8}{3} = -1 \] \[ \frac{2}{3}v - \frac{1}{3} = -1 \]
- Add \(\frac{1}{3}\) to both sides: \[ \frac{2}{3}v = -1 + \frac{1}{3} \] \[ \frac{2}{3}v = -\frac{3}{3} + \frac{1}{3} \] \[ \frac{2}{3}v = -\frac{2}{3} \]
- 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:
- Equation A, B, and C all have different solutions (h=1, p=-14, and v=-1 respectively).
- Haruki is correct: none of the equations have the same solution.
So the answer is Haruki.
1 answer