Question

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

A) 2x + 3/(x + 1)(x - 1)
B) 10/(x + 1)(x - 1)
C) 7/(x + 1)(x - 1)
D)4x + 6/(x + 1)(x - 1)

Answer 1

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

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=(2)/(x+1)` and
`w=(5)/(x^2-1)`.

Replacing values:

`p=2(l+w)`

`p=2((2)/(x+1) +(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((2)/(x+1) +(5)/(x^2-1))`

`p=2((2)/(x+1) +(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((2(x-1)+5)/((x+1)(x-1)) )`

`p=2((2x-2+5)/((x+1)(x-1)) )`

`p=2((2x+3)/((x+1)(x-1)) )`

`p=(4x+6)/((x+1)(x-1))`

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