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, \({a^3 + 6a^2 \over a^2} = {a^3 \over a^2} + {6a^2 \over a^2} = a + 6\).

To fill a 20 gallon tank exactly halfway you'll need 10 gallons of fuel. Each fuel can holds 2 gallons so:

cans = \( \frac{10 \text{ gallons}}{2 \text{ gallons}} \) = 5

To subtract these fractions, first find the lowest common multiple of their denominators. The first few multiples of 4 are [4, 8, 12, 16, 20, 24, 28, 32, 36, 40] and the first few multiples of 10 are [10, 20, 30, 40, 50, 60, 70, 80, 90]. The first few multiples they share are [20, 40, 60, 80] making 20 the smallest multiple 4 and 10 share.

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

\( \frac{9 x 5}{4 x 5} \) - \( \frac{9 x 2}{10 x 2} \)

\( \frac{45}{20} \) - \( \frac{18}{20} \)

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

\( \frac{45 - 18}{20} \) = \( \frac{27}{20} \) = 1\(\frac{7}{20}\)

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.

A factorial is the product of an integer and all the positive integers below it. To solve a fraction featuring factorials, expand the factorials and cancel out like numbers:

\( \frac{5!}{3!} \)
\( \frac{5 \times 4 \times 3 \times 2 \times 1}{3 \times 2 \times 1} \)
\( \frac{5 \times 4}{1} \)
\( 5 \times 4 \)
20