Answer :
Let's examine the given steps in order to understand the justification for the transformation from Step 1 to Step 2.
### Given:
[tex]\[ \frac{1}{2} r + \frac{1}{2} = -\frac{2}{7} r + \frac{6}{7} - 5 \][/tex]
### Step 1:
[tex]\[ \frac{1}{2} r + \frac{1}{2} = -\frac{2}{7} r - \frac{29}{7} \][/tex]
### Step 2:
[tex]\[ \frac{1}{2} r = -\frac{2}{7} r - \frac{65}{14} \][/tex]
### Justification:
#### Transition from Step 1 to Step 2:
We need to isolate the term with \( r \) on one side of the equation in Step 1. To achieve this, we apply the subtraction property of equality. This means subtracting the same value from both sides of the equation to maintain its equality.
1. Subtract \( \frac{1}{2} \) from both sides:
[tex]\[ \frac{1}{2} r + \frac{1}{2} - \frac{1}{2} = -\frac{2}{7} r - \frac{29}{7} - \frac{1}{2} \][/tex]
Simplify both sides:
[tex]\[ \frac{1}{2} r = -\frac{2}{7} r - \frac{29}{7} - \frac{1}{2} \][/tex]
2. Convert \( \frac{1}{2} \) to a common denominator:
[tex]\[ \frac{1}{2} = \frac{1 \times 7}{2 \times 7} = \frac{7}{14} \][/tex]
3. Combine fractions:
[tex]\[ -\frac{29}{7} - \frac{7}{14} \rightarrow \text{Convert } -\frac{29}{7} \text{ to fourteenths:} -\frac{29 \times 2}{7 \times 2} = -\frac{58}{14} \][/tex]
[tex]\[ -\frac{58}{14} - \frac{7}{14} = -\frac{65}{14} \][/tex]
Therefore:
[tex]\[ \frac{1}{2} r = - \frac{2}{7} r - \frac{65}{14} \][/tex]
#### Conclusion:
The process we used to move from Step 1 to Step 2 involved subtracting \( \frac{1}{2} \) from both sides of the Step 1 equation. This move is justified by the subtraction property of equality.
### Final Answer:
The justification for Step 2 is:
[tex]\[ \boxed{C. \text{the subtraction property of equality}} \][/tex]
### Given:
[tex]\[ \frac{1}{2} r + \frac{1}{2} = -\frac{2}{7} r + \frac{6}{7} - 5 \][/tex]
### Step 1:
[tex]\[ \frac{1}{2} r + \frac{1}{2} = -\frac{2}{7} r - \frac{29}{7} \][/tex]
### Step 2:
[tex]\[ \frac{1}{2} r = -\frac{2}{7} r - \frac{65}{14} \][/tex]
### Justification:
#### Transition from Step 1 to Step 2:
We need to isolate the term with \( r \) on one side of the equation in Step 1. To achieve this, we apply the subtraction property of equality. This means subtracting the same value from both sides of the equation to maintain its equality.
1. Subtract \( \frac{1}{2} \) from both sides:
[tex]\[ \frac{1}{2} r + \frac{1}{2} - \frac{1}{2} = -\frac{2}{7} r - \frac{29}{7} - \frac{1}{2} \][/tex]
Simplify both sides:
[tex]\[ \frac{1}{2} r = -\frac{2}{7} r - \frac{29}{7} - \frac{1}{2} \][/tex]
2. Convert \( \frac{1}{2} \) to a common denominator:
[tex]\[ \frac{1}{2} = \frac{1 \times 7}{2 \times 7} = \frac{7}{14} \][/tex]
3. Combine fractions:
[tex]\[ -\frac{29}{7} - \frac{7}{14} \rightarrow \text{Convert } -\frac{29}{7} \text{ to fourteenths:} -\frac{29 \times 2}{7 \times 2} = -\frac{58}{14} \][/tex]
[tex]\[ -\frac{58}{14} - \frac{7}{14} = -\frac{65}{14} \][/tex]
Therefore:
[tex]\[ \frac{1}{2} r = - \frac{2}{7} r - \frac{65}{14} \][/tex]
#### Conclusion:
The process we used to move from Step 1 to Step 2 involved subtracting \( \frac{1}{2} \) from both sides of the Step 1 equation. This move is justified by the subtraction property of equality.
### Final Answer:
The justification for Step 2 is:
[tex]\[ \boxed{C. \text{the subtraction property of equality}} \][/tex]
