SOLUTION: Need assistance please.
The width of a rectangle is 6 feet shorter than its length. If its perimeter is 82 feet, what is its
width?
I set it up as follows:
82=2(x-6)+2x
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-> SOLUTION: Need assistance please.
The width of a rectangle is 6 feet shorter than its length. If its perimeter is 82 feet, what is its
width?
I set it up as follows:
82=2(x-6)+2x
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Question 1123036: Need assistance please.
The width of a rectangle is 6 feet shorter than its length. If its perimeter is 82 feet, what is its
width?
I set it up as follows:
82=2(x-6)+2x
Which gave me a solution of
47/2=x
23.5=x
I cannot figure out what to do next.
Thank you
In your notations, x is the length (although it would be more logical to make it the width . . . )
82 = 2(x-6) + 2x <<<---=== Correct !
Next steps should be:
41 = x - 6 + x
41 + 6 = 2x
47 = 2x
x = 47/2 = 23.5 <<<---=== it is your LENGTH
Then you should find the WIDTH: W = 23.5 - 6 = 17.5.
The problem is just solved and the solution is completed.
