The area of a rectangle is width (w) x height (h). In this problem we know that the rectangle is twice as long as it is wide so h = 2w. The perimeter of a rectangle is 2w + 2h and we know that the perimeter of this rectangle is 12 meters so the equation becomes: 2w + 2h = 12.

Putting these two equations together and solving for width (w):

2w + 2h = 12
w + h = \( \frac{12}{2} \)
w + h = 6
w = 6 - h

From the question we know that h = 2w so substituting 2w for h gives us:

w = 6 - 2w
3w = 6
w = \( \frac{6}{3} \)
w = 2

Since h = 2w that makes h = (2 x 2) = 4 and the area = h x w = 2 x 4 = 8 m²

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

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 divide fractions, invert the second fraction and then multiply:

\( \frac{4}{5} \) ÷ \( \frac{1}{9} \) = \( \frac{4}{5} \) x \( \frac{9}{1} \)

To multiply fractions, multiply the numerators together and then multiply the denominators together:

\( \frac{4}{5} \) x \( \frac{9}{1} \) = \( \frac{4 x 9}{5 x 1} \) = \( \frac{36}{5} \) = 7\(\frac{1}{5}\)

First, solve for \( \left|b + 2\right| \):

\( \left|b + 2\right| \) - 2 = -3
\( \left|b + 2\right| \) = -3 + 2
\( \left|b + 2\right| \) = -1

The value inside the absolute value brackets can be either positive or negative so (b + 2) must equal - 1 or --1 for \( \left|b + 2\right| \) to equal -1:

So, b = -1 or b = -3.