SOLUTION: The length of a rectangle is 5 less than twice its width. The perimeter is 26 meters. Find the dimensions of the rectangle.
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Question 1049367: The length of a rectangle is 5 less than twice its width. The perimeter is 26 meters. Find the dimensions of the rectangle. Found 3 solutions by ikleyn, addingup, advanced_Learner:Answer by ikleyn(52781) (Show Source):
You can put this solution on YOUR website! .
The length of a rectangle is 5 less than twice its width. The perimeter is 26 meters. Find the dimensions of the rectangle.
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Let W = width.
Then the length is 2W-5.
The perimeter equation is
W + (2W-5) + W + (2W-5) = 26, or
6W - 10 = 26, or
6W = 36.
From this point please complete the solution on your own.
I noticed that this problem is the third today of the same type of problems.
You can put this solution on YOUR website! 2L+2W = 26 (1)
and
L = 2W-5 Substitute for L on (1)
2(2W-5)+2W = 26
4W-10+2W = 26
6W = 36
W = 6
>>>>>>>>>>>>>>
Check:
2(6)-5 = 7 this, according to the problem, is the length. Now:
2(7)+2(6) = 26
14+12 = 26 Correct
:
John
Cartoon (animation) form: For tutors: simplify_cartoon( 6W=36 )
If you have a website, here's a link to this solution.
DETAILED EXPLANATION
Look at . Moved these terms to the left It becomes . Look at . Solved linear equation equivalent to 6*W-36 =0 It becomes . Result: This is an equation! Solutions: W=6.