Sure! Let's match the pairs of equivalent expressions based on our calculations:
1. \(\left(-14+\frac{3}{2} b\right)-\left(1+\frac{8}{2} b\right)\)
This simplifies to \(-15 - \frac{5}{2}b\).
2. \(4 b+\frac{13}{2}\)
This simplifies to \(4 b + \frac{13}{2}\).
3. \((5+2 b)+\left(2 b+\frac{3}{2}\right)\)
This simplifies to \(4 b + \frac{13}{2}\).
4. \(8 b-15\)
This simplifies to \(8 b - 15\).
5. \(\left(\frac{7}{2} b-3\right)-(8+6 b)\)
This simplifies to \(-15 - \frac{5}{2}b\).
6. \(\frac{-5}{2} b-11\)
Not immediately seen; observe carefully.
7. \((-10+b)+(7 b-5)\)
This simplifies to \(8 b - 15\).
8. \(-15-\frac{5}{2} b\)
This simplifies to \(-15 - \frac{5}{2}b\).
Now pair the equivalent expressions:
- \(\left(-14+\frac{3}{2} b\right)-\left(1+\frac{8}{2} b\right)\) pairs with \(-15-\frac{5}{2} b\)
- \(4 b+\frac{13}{2}\) pairs with \((5+2 b)+\left(2 b+\frac{3}{2}\right)\)
- \(8 b-15\) pairs with \((-10+b)+(7 b-5)\)
- \(\left(\frac{7}{2} b-3\right)-(8+6 b)\) pairs with \(-15-\frac{5}{2} b\)
- \(\frac{-5}{2} b-11\) is unpaired
So, the matching pairs are:
[tex]\[
\boxed{(-14+\frac{3}{2} b)-(1+\frac{8}{2} b)}, \boxed{-15-\frac{5}{2} b}
\][/tex]
[tex]\[
\boxed{4 b+\frac{13}{2}}, \boxed{(5+2 b)+(2 b+\frac{3}{2})}
\][/tex]
[tex]\[
\boxed{8 b-15}, \boxed{(-10+b)+(7 b-5)}
\][/tex]
[tex]\[
\boxed{(-14+\frac{3}{2} b)-(1+\frac{8}{2} b)}, \boxed{(-15-\frac{5}{2} b)}
\][/tex]
And the [tex]\(\frac{-5}{2} b - 11\)[/tex] does not have a match in this given set.
