SOLUTION: the length of a rectangle is 7 m less than twice the width, if the length is decreased by 1 m and the width by 4 m, the perimeter will be 66 m. find the dimensions of the original

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Question 13186: the length of a rectangle is 7 m less than twice the width, if the length is decreased by 1 m and the width by 4 m, the perimeter will be 66 m. find the dimensions of the original rectangle.
Answer by akmb1215(68) About Me  (Show Source):
You can put this solution on YOUR website!
You should first set up equations that describe the problem. For the first part of the problem, you will get the equation L+=+2W+-+7. For the second part, you get the equation 2%28L+-+1%29+%2B+2+%28W+-+4%29+=+66.
To solve the problem, first use the distributive property on the second equation. It changes to 2L+-+2+%2B+2W+-+8+=+66. Since you know L = 2W - 7 (from the first equation), you can plug in 2W - 7 for the L in the equaton. That turns into 2%282W+-+7%29+-+2+%2B+2W+-+8+=+66. You will need to use the distributive property again, which gives you 4W+-+14+-+2+%2B+2W+-+8+=+66. Combine like terms, and you get 6W+-+24+=+66. To solve, move the 24 over to the right side of the equation, which give you 6W+=+90, then divide to get W = 15. To find the length, plug 15 in for W in the first original equation, which gives you L+=+2%2815%29+-+7. Solve for L, and you get 23.