Question 212902: Jack is building a rectangular deck and wants the length to be 3 feet greater than the width. What will the dimensions of the deck be if the perimeter is to be 54ft?
Answer by rapaljer(4671)   (Show Source):
You can put this solution on YOUR website! Let x = width of the deck
x+3 = length of the deck
2W + 2L = Perimeter
2x + 2(x+3)=54
2x +2x+6=54
4x+6=54
4x=48
x=12 Width
x+4= 16 Length
Check:
2W+2L=P
2*12 + 2*16
24+32
56
If you need an informal explanation of Word Problems, together with examples, exercises and answers, please see my own website. Do a "Bing" search for my last name "Rapalje" and look for "Rapalje Homepage" near the top of the search list. Click on the link "Basic, Intermediate and College Algebra: One Step at a Time" near the top of my Homepage. Select "Basic Algebra", and look in "Chapter 1" for the topics "Word Problems". Also, check out my "MATH IN LIVING COLOR" pages on this topic to see solutions to a lot of these problems, IN COLOR of course!
R^2
Dr. Robert J. Rapalje, Retired
Seminole Community College
Altamonte Springs Campus
Florida
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