There are 2 2-passenger taxis available so that's 2 x 2 = 4 total seats. There are 5 people needing transportation leaving 5 - 4 = 1 who will have to find other transportation.

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 5 are [5, 10, 15, 20, 25, 30, 35, 40, 45, 50] and the first few multiples of 9 are [9, 18, 27, 36, 45, 54, 63, 72, 81, 90]. The first few multiples they share are [45, 90] making 45 the smallest multiple 5 and 9 share.

Next, convert the fractions so each denominator equals the lowest common multiple:

\( \frac{7 x 9}{5 x 9} \) - \( \frac{2 x 5}{9 x 5} \)

\( \frac{63}{45} \) - \( \frac{10}{45} \)

Now, because the fractions share a common denominator, you can subtract them:

\( \frac{63 - 10}{45} \) = \( \frac{53}{45} \) = 1\(\frac{8}{45}\)

A factorial has the form n! and is the product of the integer (n) and all the positive integers below it. For example, 5! = 5 x 4 x 3 x 2 x 1 = 120.

A ratio of 3:1 means that there are 3 home fans for every one visiting fan. So, of every 4 fans, 3 are home fans and \( \frac{3}{4} \) of every fan in the stadium is a home fan:

35,000 fans x \( \frac{3}{4} \) = \( \frac{105000}{4} \) = 26,250 fans.

The distributive property for division helps in solving expressions like \({b + c \over a}\). It specifies that the result of dividing a fraction with multiple terms in the numerator and one term in the denominator can be obtained by dividing each term individually and then totaling the results: \({b + c \over a} = {b \over a} + {c \over a}\). For example, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).