# The area of a rectangle is 16. The length is x^5/ (x+1) and the width is ( x+1)/ x^3 . what is x?

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Given that the length of the rectangle ( L ) = x^5/ (x+1)

Also, given that the width ( w) = (x+1)/ x^3

We need to determine (x) such that the area of the rectangle is 16 square units.

We know that the area of the rectangle (a) is:

a = Length * width = L*w = 16

==> [( x^5 / (x+ 1)] * [(x+1)/ x^3)] = 16

==> (x^5 ( x+ 1) / ( x+ 1) x^3 = 16

We will reduce similar terms.

==> x^5 / x^3 = 16

From exponent properties we know that x^a/ x^b = x^(a-b)

==> x^(5 - 3) = 16

==> x^2 = 16

Now we will take the root for both sides:

**==> x= 4 **

The area of the rectangle is given by length *width.

Length = x^5/(x+1) and width = (x+1)x^3. Area = 16.

Therefore Area = {x^5/(x+1)}{x+1)/x^3} = 16...(1)

The LH S simplifies to x^2.

So area = x^2 = 16.

x= 4 or -4. The negative value is not practical and so we ignore it.

So the length = x^5 /(x+1) = 4^5/(4+1) = 204.8.

Width = x^3/(x+1) = 4^3/(4+1) = 12.8.