a rectangle is 2 inches longer than it is wide. Numerically,its area exceeds its perimiter by 44 find the peri

Answer 1

w=width w+2=length
area=perimeter+44
w(w+2)=w+w+w+2+w+2+44
w^2+2w=4w+48
w^2-2w-48=0
(w+6)(w-8)=0
w=-6 or w=8
since width must be positive it must be 8

Answer 2

The length is given in terms of the width. So let w = the width. Then the length is w+2. This means the area is (w+2)w, and the perimeter is 2w + 2(w+2). The area in square inches is 44 more than the perimeter in inches, so
(w+2)w = [2w + 2(w+2)] + 44

Solve this for w, and use that to find the perimeter.

Answer 3

x(x+2) = area
4x+4 = perimeter
x(x+2) = 4x+4+44
x^2+2x =4x+48
x^2-2x-48 =0
(x-8)(x+6) =
x=8
perimeter = 4(8) + 4 = 36 inches