To determine who is correct about the linear equations, we will solve each equation and compare their solution sets.
Equation A:
\[ 3(4 - 5h) - 7 = -12 + 2h \]
First, distribute the left side: \[ 12 - 15h - 7 = -12 + 2h \] \[ 5 - 15h = -12 + 2h \]
Next, add \(15h\) and \(12\) to both sides: \[ 5 + 12 = 2h + 15h \] \[ 17 = 17h \]
Now, divide by 17: \[ h = 1 \]
Equation B:
\[ 2(p - 3) - 1 = 6p + \frac{14}{2} \] First, simplify the right side: \[ \frac{14}{2} = 7 \]
Now we can rewrite the equation: \[ 2(p - 3) - 1 = 6p + 7 \]
Distribute: \[ 2p - 6 - 1 = 6p + 7 \] \[ 2p - 7 = 6p + 7 \]
Next, subtract \(2p\) from both sides: \[ -7 = 4p + 7 \]
Now, subtract 7 from both sides: \[ -14 = 4p \]
Finally, divide by 4: \[ p = -\frac{7}{2} \]
Equation C:
\[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \]
First, simplify the left side: \[ \frac{2}{3}v - 3 + \frac{8}{3} = -1 \]
Convert \(-3\) into a fraction with a denominator of 3: \[ -3 = -\frac{9}{3} \]
Now combine: \[ \frac{2}{3}v - \frac{9}{3} + \frac{8}{3} = -1 \] \[ \frac{2}{3}v - \frac{1}{3} = -1 \]
Add \( \frac{1}{3} \) to both sides: \[ \frac{2}{3}v = -1 + \frac{1}{3} \]
Convert \(-1\) into a fraction: \[ -1 = -\frac{3}{3} \] \[ \frac{2}{3}v = -\frac{3}{3} + \frac{1}{3} \] \[ \frac{2}{3}v = -\frac{2}{3} \]
Now multiply by \(\frac{3}{2}\) to solve for \(v\): \[ v = -1 \]
Summary of Solutions:
- Equation A: \( h = 1 \)
- Equation B: \( p = -\frac{7}{2} \)
- Equation C: \( v = -1 \)
Now we can evaluate the claims:
- Tamara states that Equation A and Equation C have the same solution. This is false.
- Payton states Equation B and Equation C have the same solution. This is false.
- Krish states all three have the same solution. This is false.
- Haruki states that none of the equations have the same solution. This is true.
Thus, the correct answer is Haruki.
1 answer