A rational number (or fraction) is represented as a ratio between two integers, a and b, and has the form \({a \over b}\) where a is the numerator and b is the denominator. An improper fraction (\({5 \over 3} \)) has a numerator with a greater absolute value than the denominator and can be converted into a mixed number (\(1 {2 \over 3} \)) which has a whole number part and a fractional part.

First, solve for \( \left|x - 1\right| \):

\( \left|x - 1\right| \) + 9 = 5
\( \left|x - 1\right| \) = 5 - 9
\( \left|x - 1\right| \) = -4

The value inside the absolute value brackets can be either positive or negative so (x - 1) must equal - 4 or --4 for \( \left|x - 1\right| \) to equal -4:

So, x = 5 or x = -3.

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

6\( \sqrt{125} \) + 2\( \sqrt{5} \)
6\( \sqrt{25 \times 5} \) + 2\( \sqrt{5} \)
6\( \sqrt{5^2 \times 5} \) + 2\( \sqrt{5} \)
(6)(5)\( \sqrt{5} \) + 2\( \sqrt{5} \)
30\( \sqrt{5} \) + 2\( \sqrt{5} \)

Now that the radicands are identical, you can add them together:

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:

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

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

-8b² - 8b²
(-8 - 8)b²
-16b²