SOLUTION: The length of a rectangle is 7 yards less than twice its width. If the length was increased by 11 yards and the width decreased by 6 yards, the area would be decreased by 40 square
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Question 890799: The length of a rectangle is 7 yards less than twice its width. If the length was increased by 11 yards and the width decreased by 6 yards, the area would be decreased by 40 square yards. Find the original dimensions of the lot.
Please help me, thank you!
Answer by ankor@dixie-net.com(22740) (Show Source): You can put this solution on YOUR website!
The length of a rectangle is 7 yards less than twice its width.
If the length was increased by 11 yards and the width decreased by 6 yards, the area would be decreased by 40 square yards.
Find the original dimensions of the lot.
:
let w = the original width
then "length of a rectangle is 7 yards less than twice its width.". therefore:
(2w-7) = the original length
and
w(2w-7) = original area
:
"If the length was increased by 11 yards and the width decreased by 6 yards, the area would be decreased by 40 square yards."
:
Original area - new area = 40 sq/yds
w(2w-7) - (2w-7+11)(w-6) = 40
2w^2 - 7w - (2w+4)(w-6) = 40
2w^2 - 7w - (2w^2 - 12w + 4w - 24) = 0
2w^2 - 7w - (2w^2 - 8w - 24) = 0
Removing the brackets changes the signs
2w^2 - 7w - 2w^2 + 8w + 24 = 40
Combine like terms
2w^2 - 2w^2 - 7w + 8w = 40 - 24
w = 16 yds is the original width
then
2(16) - 7 = 25 yds is the original length
:
:
See if this checks out
25*16 = 400 sq yds origiinal area
36*10 = 360 sq yds new area (added 11 to the length, subtracted 6 from the width
-------------
diff: 40 sq yds
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