the length of a rectangle is 1 yd less than twice its width, and the area of the rectangle is 55 yd^2. find the dimensions
length :
width:

Question

the length of a rectangle is 1 yd less than twice its width, and the area of the rectangle is 55 yd^2. find the dimensions
 length : 
width:

anonymous · Answer

A=L*w
L=2w-1 b/c the length is twice the width (2w) and less than it by a yard (-1)

now put this back into your area equation
A=(2w-1)w-->A=2w^2 -w

put 55 in for A and you get 55=2w^2 -w

graph this and find the zeros: -5 and 5.5; -5 is not a possible width possible so
w must be 5.5

then plug 5.5 into the length equation
L=2(5.5)-1--> L=11-1--> L=10

5.5 yds* 10 yds= 55 yds^2