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

P = 2(L + W)
L = W + 5
A = 4P + 2

P = 2(W + 5 + W)
P = 2(2W + 5)
P = 4W + 10

A = 4P + 2
A = 4(4W + 10) + 2
A = 16W + 42

A = L * W
A = W(W + 5)
A = W^2 + 5W

W^2 + 5W = 16W + 42
W^2 + 5W - 16W - 42 = 0
W^2 - 11W - 42 = 0
(W + 3)(W - 14) = 0

W - 14 = 0
W = 14 <==

L = W + 5
L = 14 + 5
L = 19 <==

P = 2(19 + 14)
P = 2(33)
P = 66

A = L * W
A = 19 * 14
A = 266

answer : length = 19, width = 14perimeter = 66area = 266

Yes i could be the length of the triangle