To add these fractions, first find the lowest common multiple of their denominators. The first few multiples of 6 are [6, 12, 18, 24, 30, 36, 42, 48, 54, 60] and the first few multiples of 8 are [8, 16, 24, 32, 40, 48, 56, 64, 72, 80]. The first few multiples they share are [24, 48, 72, 96] making 24 the smallest multiple 6 and 8 share.

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

\( \frac{2 x 4}{6 x 4} \) + \( \frac{3 x 3}{8 x 3} \)

\( \frac{8}{24} \) + \( \frac{9}{24} \)

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

\( \frac{8 + 9}{24} \) = \( \frac{17}{24} \) = \(\frac{17}{24}\)

To subtract these radicals together their radicands must be the same:

6\( \sqrt{20} \) - 9\( \sqrt{5} \)
6\( \sqrt{4 \times 5} \) - 9\( \sqrt{5} \)
6\( \sqrt{2^2 \times 5} \) - 9\( \sqrt{5} \)
(6)(2)\( \sqrt{5} \) - 9\( \sqrt{5} \)
12\( \sqrt{5} \) - 9\( \sqrt{5} \)

Now that the radicands are identical, you can subtract them:

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

The equation for this sequence is:

a_n = a_n-1 + 6

where n is the term's order in the sequence, a_n is the value of the term, and a_n-1 is the value of the term before a_n. This makes the next number:

a₆ = a₅ + 6
a₆ = 25 + 6
a₆ = 31

To add or subtract terms with exponents, both the base and the exponent must be the same. In this case they are so add the coefficients and retain the base and exponent:

6a⁶ + 5a⁶
(6 + 5)a⁶
11a⁶