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
[tex]p=2(l+w)[/tex]
where
[tex]p[/tex] is the perimeter
[tex]l[/tex] is the length
[tex]w[/tex] is the width
We know from our problem that the field has length 2/x + 1 and width 5/x^2 -1, so [tex]l=\frac{2}{x+1}[/tex] and [tex]w=\frac{5}{x^2-1}[/tex].
Replacing values:
[tex]p=2(l+w)[/tex]
[tex]p=2(\frac{2}{x+1} +\frac{5}{x^2-1})[/tex]
Notice that the denominator of the second fraction is a difference of squares, so we can factor it using the formula [tex]a^2-b^2=(a+b)(a-b)[/tex] where [tex]a[/tex] is the first term and [tex]b[/tex] is the second term. We can infer that [tex]a=x^2[/tex] and [tex]b=1^2[/tex]. So, [tex]x^2-1=(x+1)(x-1)[/tex]. Replacing that:
[tex]p=2(\frac{2}{x+1} +\frac{5}{x^2-1})[/tex]
[tex]p=2(\frac{2}{x+1} +\frac{5}{(x+1)(x-1})[/tex]
We can see that the common denominator of our fractions is [tex](x+1)(x-1)[/tex]. Now we can simplify our fraction using the common denominator:
[tex]p=2(\frac{2(x-1)+5}{(x+1)(x-1)} )[/tex]
[tex]p=2(\frac{2x-2+5}{(x+1)(x-1)} )[/tex]
[tex]p=2(\frac{2x+3}{(x+1)(x-1)} )[/tex]
[tex]p=\frac{4x+6}{(x+1)(x-1)} [/tex]
We can conclude that the perimeter of the field is D)4x + 6/(x + 1)(x - 1).
