The length of a rectangle is 5 inches more than its vedth. The area of the rectangle is equal to 2 inches more than 4 times the perimeter. Find the length and width of the rectangle.

Step-by-step explanation:

Let  l  = length,  w = width,  A = area =  l×w , and  P  = perimeter =  2(l+w)

Let 'x' be the width

As the length 'l' of a rectangle is 5 inches more than its width.

so the length will be = x+5

As the Area of the rectangle is equal to  2 inches more than 4 times the perimeter.

A = 4P + 2

so the equation becomes

l × w = 4×2(l+w)+2

substituting w=x, l = x+5,

(x+5)x = 8(x+5+x)+2

x²+5x = 8(5+2x)+2

x²+5x = 40+16x+2

x²+5x = 16x+42

x²+5x-16x-42 =0

x²-11x-42=0

solving

similarly solving

so

x = 14, or x = -3

As the width 'x' can not be negative.

so x = 14

Thus,

i think it is x = 1.2