SOLUTION: 148 ft of fencing was used to make a rectangular fence in a yard. the length of the yard is 6 less than 3 times it's width. What is the length of the yard.
how would i begin to
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how would i begin to
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Question 642736: 148 ft of fencing was used to make a rectangular fence in a yard. the length of the yard is 6 less than 3 times it's width. What is the length of the yard.
how would i begin to solve this?
tried Y = L + 6<3xW
also tried:
148= 2L+2W Y = 2L - 6
.. would it be 31... did it too many different ways AND CAN'T SEE how to work it out. also came up w. 42... thank you so much for your help. Answer by MathTherapy(10551) (Show Source):
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148 ft of fencing was used to make a rectangular fence in a yard. the length of the yard is 6 less than 3 times it's width. What is the length of the yard.
how would i begin to solve this?
tried Y = L + 6<3xW
also tried:
148= 2L+2W Y = 2L - 6
.. would it be 31... did it too many different ways AND CAN'T SEE how to work it out. also came up w. 42... thank you so much for your help.
Let width be W
Then length = 3W - 6
2W + 2L = 148 ----- 2(W + L) = 2(74) ----- Factoring out GCF, 2. This means that:
W + L = 74
Now, since W = W, and L = 3W - 6, then W + L = 74 becomes:
W + 3W - 6 = 74
4W = 74 + 6
4W = 80
W, or width = , or units
20 + L = 74 ------ Substituting 20 for W
L = 74 - 20
L, or length = units
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