a) 8

\(\displaystyle{b}{)}{\frac{{{12}}}{{{7}}}}\)

\(\displaystyle{c}{)}{\frac{{{5}}}{{{3}}}}\)

For d and e, I would convert them into improper fractions. You would do this by multiplying the denominator by the whole number (in d's case, the whole number is 1), and then add the product by the numerator (2(denominator)*1(whole number) = 2 + 1(numerator)). However, you would keep the denominator the same.

\(\displaystyle{d}{)}{1}{\frac{{{1}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}\rightarrow{\frac{{{2}}}{{{3}}}}\)

\(\displaystyle{e}{)}{3}{\frac{{{3}}}{{{4}}}}={\frac{{{15}}}{{{4}}}}\rightarrow{\frac{{{4}}}{{{15}}}}\)

\(\displaystyle{f}{)}{\frac{{{1}}}{{{6}}}}\)

As you can see, if you multiply each corresponding problem with their reciprocal, they equal 1. In the end, you want the reciprocal of the problem to equal 1.

\(\displaystyle{b}{)}{\frac{{{12}}}{{{7}}}}\)

\(\displaystyle{c}{)}{\frac{{{5}}}{{{3}}}}\)

For d and e, I would convert them into improper fractions. You would do this by multiplying the denominator by the whole number (in d's case, the whole number is 1), and then add the product by the numerator (2(denominator)*1(whole number) = 2 + 1(numerator)). However, you would keep the denominator the same.

\(\displaystyle{d}{)}{1}{\frac{{{1}}}{{{2}}}}={\frac{{{3}}}{{{2}}}}\rightarrow{\frac{{{2}}}{{{3}}}}\)

\(\displaystyle{e}{)}{3}{\frac{{{3}}}{{{4}}}}={\frac{{{15}}}{{{4}}}}\rightarrow{\frac{{{4}}}{{{15}}}}\)

\(\displaystyle{f}{)}{\frac{{{1}}}{{{6}}}}\)

As you can see, if you multiply each corresponding problem with their reciprocal, they equal 1. In the end, you want the reciprocal of the problem to equal 1.