The length of a rectangle is 4 inches more than its width. the area of the rectangle is equal to 4 inches less than 5 times the perimeter. find the length and width of the rectangle.

Solution:
Let the width of the rectsngle (b) be x
and the length of the rectangle (l) =x+4
Area of the rectangle (A)=?
Perimeter of the rectangle (P)=?
Now,
P= 2 (l+b)
= 2 (x+4+x)
= 2 (2x+4)
= 4x+8
And,
A= l×b
= x(x+4)
= x^2+4x
According to the question,
A=5P-4
or, x^2+4x =5 (4x+8)-4
or, x^2 +4x =20x+40-4
or, x^2 = 20x-4x+36
or, x^2 = 16x+36
or, x^2-16x-36 = 0
or, x^2-(18-2)x-36 = 0
or, x^2-18x+2x-36 = 0
or, x (x-18)+2 (x-18) = 0
or, (x-18) (x+2) = 0

Either, Or,
x-18 = 0 x+2 = 0
so, x = 18 so, x = -2

Taking the value of x = 18 ( Because the length is always positive)
Length = x+4 = 18+4 = 22 cm
Width = x = 18 cm

2 1/8 portions.

 Hope this helps.