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Question 652531: The length of a rectangular room is 6 feet longer than twice the width. If the room's perimeter is 152 feet, what are the rooms dimensions?
Answer by Shana-D77(132)   (Show Source):
You can put this solution on YOUR website! Ok, so let's start by translating this problem literally:
"The length of a rectangular room is 6 feet longer than twice the width"
L = 2w + 6 (remember "=" reads as "is")
"If the Perimeter is 152 feet"
P = 152
We know that the perimeter of a rectangle is L + L + W + W or 2L + 2W
So, we have 152 = 2L + 2W
Our two equations are then:
L = 2w + 6
2L + 2w = 152
Since the first equation is solved for L, let's get the second equation also solved for L:
2L + 2w = 152
L + w = 76 (dided everything by 2)
L = 76 - w (subtracted w from both sides)
Now we have:
L = 2w + 6
L = 76 - w
Since they're both =L, they're equal to each other (transitive peoperty):
2w + 6 = 76 - w
3w + 6 = 76 (added w to both sides)
3w = 70 (subtracted 6 from both sides)
w = 70/3 or 23.333 feet (23 feet, 4 inches)
To find L, we can plug our w back into any of the equations we used. The easiest may be the 152 = 2L + 2w equation:
152 = 2L + 2(70/3)
152 = 2L + 140/3
316/3 = 2L (subtracted 140/3 from both sides)
L = 158/3 (52.66 feet or 52 feet and 8 inches)
w = 23 feet, 4 inches
L = 52 feet, 8 inches
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