SOLUTION: The length of a rectangle is 1 yd less than twice the width, and the area of the rectangle is 21 yd^2. Find the dimensions of the rectangle

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Question 1151179: The length of a rectangle is 1 yd less than twice the width, and the area of the rectangle is 21 yd^2. Find the dimensions of the rectangle

Found 3 solutions by MathLover1, Alan3354, ikleyn:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

if the length of a rectangle is 1 yd less than twice the width, we have
L=2W-1....eq.1
and, if the area of the rectangle is 21 yd^2, we have
L%2AW=21.......solve for L
L=21%2FW....eq.2
from eq.1 and eq.2 we have
2W-1=21%2FW......solve for W
2W%2AW-W=21
2W%5E2-W-21=0.........factor
2W%5E2%2B6W-7W-21=0
%282W%5E2%2B6W%29-%287W%2B21%29=0
2W%28W%2B3%29-7%28W%2B3%29=0
%28W+%2B+3%29+%282W+-+7%29+=+0

solutions:
if %28W+%2B+3%29++=+0=> W=-3=> since width, disregard negative solution
if +%282W+-+7%29+=+0=> W=3.5 yd
L=6 yd

so, the dimensions of the rectangle are:
the length=6 yd
the width=3.5 yd





Answer by Alan3354(69443) About Me  (Show Source):
Answer by ikleyn(52780) About Me  (Show Source):
You can put this solution on YOUR website!
.

            You can solve the problem in much shorter way, practically in your head, without solving quadratic equation. 


If W is the width, then the length is  L = (2W-1).


From the given area, you get the equation

    W*(2W-1) = 21.


Multiply both sides by 2. You will get

    2W(*2W-1) = 42.


So, the product of two numbers  (2w)  and  (2W-1) is 42 and their difference is 1.


It needs 2 seconds (and knowing the multiplication table) to guess/(to recognize)  the numbers:  2W = 7,  2W-1 = 6.


Hence, the width is 3.5 yards,  the length is 2*3.5-1 = 6 yards.    ANSWER

Solved.