Question

Find the perimeter of a field that has length 2/x + 1 and width 5/x^2 -1.

Answer 1

Answer:

D)4x + 6/(x + 1)(x - 1)

Step-by-step explanation:

A field is basically a rectangle, so to find the perimeter of our field we are using the formula for the perimeter of a rectangle

$p=2(l+w)$

where

$p$ is the perimeter

$l$ is the length

$w$ is the width

We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so $l=\frac{2}{x+1}$ and $w=\frac{5}{x^2-1}$.

Replacing values:

$p=2(l+w)$

$p=2(\frac{2}{x+1} +\frac{5}{x^2-1})$

Notice that the denominator of the second fraction is a difference of squares, so we can factor it using the formula $a^2-b^2=(a+b)(a-b)$ where $a$ is the first term and $b$ is the second term. We can infer that $a=x^2$ and $b=1^2$. So, $x^2-1=(x+1)(x-1)$. Replacing that:

$p=2(\frac{2}{x+1} +\frac{5}{x^2-1})$

$p=2(\frac{2}{x+1} +\frac{5}{(x+1)(x-1})$

We can see that the common denominator of our fractions is $(x+1)(x-1)$. Now we can simplify our fraction using the common denominator:

$p=2(\frac{2(x-1)+5}{(x+1)(x-1)} )$

$p=2(\frac{2x-2+5}{(x+1)(x-1)} )$

$p=2(\frac{2x+3}{(x+1)(x-1)} )$

$p=\frac{4x+6}{(x+1)(x-1)}$

We can conclude that the perimeter of the field is D)4x + 6/(x + 1)(x - 1).