What is the justification for step 2 in the solution process?

[tex]\[ \frac{1}{2} r+\frac{1}{2}=-\frac{2}{7} r+\frac{6}{7}-5 \][/tex]

Step 1: [tex]\[ 1 \frac{1}{2} r+\frac{1}{2}=-\frac{2}{7} r-\frac{29}{7} \][/tex]

Step 2: [tex]\[ \frac{1}{2} r=-\frac{2}{7} r-\frac{65}{14} \][/tex]

A. The addition property of equality
B. The subtraction property of equality
C. The division property of equality
D. The multiplication property of equality



Answer :

The justification for Step 2 in solving the given equation:

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]

is the subtraction property of equality.

Here is the detailed, step-by-step explanation:

1. Start with the equation given in Step 1:
[tex]\[ \frac{1}{2} r + \frac{1}{2} = -\frac{2}{7} r - \frac{29}{7} \][/tex]

2. To isolate the terms involving [tex]\( r \)[/tex] on one side, you need to remove [tex]\(\frac{1}{2}\)[/tex] from the left side. To achieve this, you subtract [tex]\(\frac{1}{2}\)[/tex] from both sides of the equation. This step utilizes the subtraction property of equality, which allows you to maintain the equality of the equation while subtracting the same value from both sides.

3. Subtract [tex]\(\frac{1}{2}\)[/tex] from both sides:
[tex]\[ \left( \frac{1}{2} r + \frac{1}{2} \right) - \frac{1}{2} = -\frac{2}{7} r - \frac{29}{7} - \frac{1}{2} \][/tex]

4. This simplifies to:
[tex]\[ \frac{1}{2} r = -\frac{2}{7} r - \frac{29}{7} - \frac{1}{2} \][/tex]

5. Next, to combine the constants on the right side of the equation, convert [tex]\(\frac{1}{2}\)[/tex] to a fraction with a denominator of 14.

[tex]\[ \frac{1}{2} = \frac{7}{14} \][/tex]

6. Combine [tex]\(-\frac{29}{7}\)[/tex] and [tex]\(\frac{7}{14}\)[/tex]. Recall that [tex]\(-\frac{29}{7}\)[/tex] can be written as [tex]\(-\frac{58}{14}\)[/tex] to have common denominators:

[tex]\[ -\frac{29}{7} = -\frac{58}{14} \][/tex]

7. Add [tex]\(-\frac{58}{14}\)[/tex] and [tex]\(-\frac{7}{14}\)[/tex] together:

[tex]\[ -\frac{58}{14} - \frac{7}{14} = -\frac{65}{14} \][/tex]

8. Thus, the equation now is:
[tex]\[ \frac{1}{2} r = -\frac{2}{7} r - \frac{65}{14} \][/tex]

Therefore, the justification for Step 2 is:

B. the subtraction property of equality

This property was used to subtract [tex]\(\frac{1}{2}\)[/tex] from both sides of the equation to isolate the variable terms on one side.