SOLUTION: the area of a rectangle is 56 and its perimeter is 30. find the difference of its length and width

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Question 845367: the area of a rectangle is 56 and its perimeter is 30. find the difference of its length and width
Found 2 solutions by josgarithmetic, hamsanash1981@gmail.com:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
x and y are the dimensions, one of them length and the other width.

Area description means xy=56 and perimeter description means 2x+2y=30. For just solving the problem asked, solving for x and y is more important and would be done first.

The perimeter equation is easily simplified to x+y=15. This also directly gives y=15-x and can be substituted....
xy=56
x%2815-x%29=56
15x-x%5E2=56
15x-56-x%5E2=0
x%5E2-15x%2B56=0
highlight%28highlight%28%28x-8%29%28x-7%29=0%29%29
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Easily mentally skipping steps, the difference between the length and width is ONE UNIT.
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In case one could not skip the steps, then continuing form the factored quadratic equation, either x=7 or x=8.
Using area equation xy=56, find that y=56/x.
That means either y=56/7=8 or y=56/8=7.
Either way, the difference between x and y is 1.

Answer by hamsanash1981@gmail.com(151) About Me  (Show Source):
You can put this solution on YOUR website!
area = 56
l*b = 56
perimeter = 30
2(l+b) = 30
l+b = 15
squaring on both sides
(l+b)^2 = 225
l%5E2%2Bb%5E2+%2B+2lb+=+225
a%5E2%2Bb%5E2+=+%28%28a%2Bb%29%5E2+%2B+%28a-b%29%5E2%29%2F2
%28%28l%2Bb%29%5E2+%2B%28l-b%29%5E2%29%2F2+%2B2lb=+225
%28%2815%29%5E2%2B+%28l-b%29%5E2+%2B+4%2A56%29%2F2+=225
225%2B%28l-b%29%5E2+%2B+224+=450
%28l-b%29%5E2%2B+449+=+450
%28l-b%29%5E2+=1
l-b+=+1