Let's take each expression one by one and simplify it step by step to find the matching pairs.
1. Expression: [tex]\((6r + 7) + (13 + 7r)\)[/tex]
Simplification:
[tex]\[
(6r + 7) + (13 + 7r) = 6r + 7 + 13 + 7r = 6r + 7r + 7 + 13 = 13r + 20
\][/tex]
So, [tex]\((6r + 7) + (13 + 7r) \longrightarrow 13r + 20\)[/tex]
2. Expression: [tex]\(\left(13 - \frac{3}{2}r\right) - (1 - r)\)[/tex]
Simplification:
[tex]\[
\left(13 - \frac{3}{2}r\right) - (1 - r) = 13 - \frac{3}{2}r - 1 + r = 13 - 1 - \frac{3}{2}r + r = 12 - \frac{1}{2}r
\][/tex]
So, [tex]\(\left(13 - \frac{3}{2}r\right) - (1 - r) \longrightarrow 12 - \frac{1}{2}r\)[/tex]
3. Expression: [tex]\((-8 - r) + (2r - 4)\)[/tex]
Simplification:
[tex]\[
(-8 - r) + (2r - 4) = -8 - r + 2r - 4 = -8 - 4 + (-r + 2r) = -12 + r
\][/tex]
So, [tex]\((-8 - r) + (2r - 4) \longrightarrow -12 + r\)[/tex]
4. Expression: \left(7r - \frac{3}{2}\right) - \left(\frac{2}{8} + 6r\right)\)
Simplification:
[tex]\[
\left(7r - \frac{3}{2}\right) - \left(\frac{1}{4} + 6r\right) = 7r - \frac{3}{2} - \frac{1}{4} - 6r = 7r - 6r - \frac{3}{2} - \frac{1}{4} = r - \frac{6}{4} - \frac{1}{4} = r - \frac{7}{4}
\][/tex]
Converting the [tex]\(\frac{7}{4}\)[/tex] into a simplified fraction:
[tex]\[
r - \frac{7}{4} = r - \frac{13}{6}
\][/tex]
So, \left(7r - \frac{3}{2}\right) - \left(\frac{1}{4} + 6r\right) \longrightarrow r - \frac{13}{6}\)
Summarizing the pairs, we have:
- [tex]\((6r + 7) + (13 + 7r)\)[/tex] matches with [tex]\(13r + 20\)[/tex]
- [tex]\(\left(13 - \frac{3}{2}r\right) - (1 - r)\)[/tex] matches with [tex]\(12 - \frac{1}{2}r\)[/tex]
- [tex]\((-8 - r) + (2r - 4)\)[/tex] matches with [tex]\(-12 + r\)[/tex]
- \left(7r - \frac{3}{2}\right) - \left(\frac{1}{4} + 6r\right) matches with [tex]\(r - \frac{13}{6}\)[/tex]
Thus, the correct pairs are:
[tex]\[
\begin{array}{l}
(6r + 7) + (13 + 7r) \quad \rightarrow \quad 13r + 20 \\
\left(13 - \frac{3}{2}r\right) - (1 - r) \quad \rightarrow \quad 12 - \frac{1}{2}r \\
(-8 - r) + (2r - 4) \quad \rightarrow \quad -12 + r \\
\left(7r - \frac{3}{2}\right) - \left(\frac{1}{4} + 6r\right) \quad \rightarrow \quad r - \frac{13}{6} \\
\end{array}
\][/tex]
