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.

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:

-4a⁶ + 2a⁶
(-4 + 2)a⁶
-2a⁶

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 number of students taking German or Spanish is 7 + 7 = 14. Of that group of 14, 4 are taking both languages so they've been counted twice (once in the German group and once in the Spanish group). Subtracting them out leaves 14 - 4 = 10 who are taking at least one language. 22 - 10 = 12 students who are not taking either language.

First, solve for \( \left|b - 3\right| \):

\( \left|b - 3\right| \) + 5 = 7
\( \left|b - 3\right| \) = 7 - 5
\( \left|b - 3\right| \) = 2

The value inside the absolute value brackets can be either positive or negative so (b - 3) must equal + 2 or -2 for \( \left|b - 3\right| \) to equal 2:

So, b = 1 or b = 5.