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| Question 214176This question is from textbook
 :  find two consecutive odd integers such that their product is 15 more than 3 times their sum 
This question is from textbook
 
 Answer by drj(1380)
      (Show Source): 
You can put this solution on YOUR website! Find two consecutive odd integers such that their product is 15 more than 3 times their sum. 
 Step 1.  Let n be an odd integer then n+2 be the next odd consecutive integer
 
 Step 2.  n(n+2) is the product of the two odd integers.
 
 Step 3.  n+n+2 is the sum of the two odd integers.
 
 Step 4.  n(n+2)=15+3(n+n+2)  product is 15 more than sum.
 
 Step 5.  Solve the equation in Step 4 using the following steps.
 
 
 
 | Solved by pluggable solver: EXPLAIN simplification of an expression |  | Your Result: 
 
 
 
  YOUR ANSWER
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This is an equation! Solutions: n=7,n=-3.
Graphical form: Equation  was fully solved.Text form: n*(n+2)=15+3*(n+n+2) simplifies to 0=0Cartoon (animation) form:   For tutors:
 simplify_cartoon( n*(n+2)=15+3*(n+n+2) )If you have a website, here's a link to this solution.  | 
 DETAILED EXPLANATIONLook at
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 Universal Simplifier and SolverDone!
 
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 Step 6. There are two solutions 7 and 9 is one solution set and the other -3 and -1 is another solution set.
 
 I hope the above steps were helpful.
 
 For free Step-By-Step Videos on Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra or for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.
 
 And good luck in your studies!
 
 Respectfully,
 Dr J
 
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