Problema Solution
The area of a rectangle is 33 yd^2, and the length of the rectangle is 5 yd
less than twice the width. Find the dimensions of the rectangle.
Answer provided by our tutors
let
l = the length of the rectangle, l>0
w = the width of the rectangle, w>0
A = 32 yd^2 the area of the rectangle
the length of the rectangle is 5 yd less than twice the width:
l = 2w - 5
the area A = l*w thus
l*w = 32
plug l = 2w - 5 into the last equation:
(2w - 5)*w = 32
by solving we find:
w = 5.44 yd
click here to see the step by step solution of the system of equations:
l = 2*5.44 - 5
l = 5.88 yd
the dimensions of the rectangle are: the length is 5.88 yards and the width is 5.44 yd.