Tamara, Payton, Krish, and Haruki are analyzing the solution sets to the following linear equations. Tamara says that Equation A and Equation C have the same solution. Payton says Equation B and Equation C have the same solution. Krish says all three have the same solution. Haruki says that none of the equations have the same solution. Identify who is correct about the linear equations.

Question

Equation A: 3(4−5h)−7=−12+2h

Equation B: 2(p−3)−1=6p+142

Equation C: 23v−3+83=−1

(1 point)
Responses

Payton
Payton

Krish
Krish

Tamara
Tamara

Haruki

Answer 1

To determine who is correct about the given linear equations, we need to solve each equation and identify their solution sets.

Equation A: \[ 3(4 - 5h) - 7 = -12 + 2h \] Distributing 3: \[ 12 - 15h - 7 = -12 + 2h \] Combining like terms: \[ 5 - 15h = -12 + 2h \] Adding 15h to both sides: \[ 5 = -12 + 17h \] Adding 12 to both sides: \[ 17 = 17h \] Dividing by 17: \[ h = 1 \]

Equation B: \[ 2(p - 3) - 1 = 6p + 142 \] Distributing 2: \[ 2p - 6 - 1 = 6p + 142 \] Combining like terms: \[ 2p - 7 = 6p + 142 \] Subtracting 2p from both sides: \[ -7 = 4p + 142 \] Subtracting 142 from both sides: \[ -149 = 4p \] Dividing by 4: \[ p = -\frac{149}{4} \]

Equation C: \[ 23v - 3 + 83 = -1 \] Combining terms: \[ 23v + 80 = -1 \] Subtracting 80 from both sides: \[ 23v = -81 \] Dividing by 23: \[ v = -\frac{81}{23} \]

Now we have the solutions:

• Equation A has the solution \( h = 1 \).
• Equation B has the solution \( p = -\frac{149}{4} \).
• Equation C has the solution \( v = -\frac{81}{23} \).

Comparing the solutions:

• Equation A has a solution of \( h = 1 \).
• Equation B has a solution of \( p = -\frac{149}{4} \).
• Equation C has a solution of \( v = -\frac{81}{23} \).

Since all three equations yield different solutions, Haruki is correct in stating that none of the equations have the same solution.

Final Answer: Haruki