SOLUTION: find the three consecutive odd integers such that the sum of the first two odd integers exceeds the third by 9.

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Question 213912: find the three consecutive odd integers such that the sum of the first two odd integers exceeds the third by 9.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find the three consecutive odd integers such that the sum of the first two odd integers exceeds the third by 9.

Step 1. Let n be an odd integer, n+2 and n+4 be the next two consecutive odd integers.

Step 2. Let n+n+2 is the sum of the first two integers

Step 3. Then, n+n+2=n+4+9 Since the sum of n+n+2 is 9 more than the third integer given as n+4

Step 4. Solve the above equation using the following steps.

Solved by pluggable solver: EXPLAIN simplification of an expression
Your Result:


YOUR ANSWER


  • This is an equation! Solutions: n=11.
  • Graphical form: Equation n%2Bn%2B2=n%2B4%2B9 was fully solved.
  • Text form: n+n+2=n+4+9 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+n%2Bn%2B2=n%2B4%2B9+%29
    For tutors: simplify_cartoon( n+n+2=n+4+9 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at highlight_red%28+n+%29%2Bhighlight_red%28+n+%29%2B2=n%2B4%2B9.
Eliminated similar terms highlight_red%28+n+%29,highlight_red%28+n+%29 replacing them with highlight_green%28+%281%2B1%29%2An+%29
It becomes highlight_green%28+%281%2B1%29%2An+%29%2B2=n%2B4%2B9.
Look at %28highlight_red%28+1+%29%2Bhighlight_red%28+1+%29%29%2An%2B2=n%2B4%2B9.
Added fractions or integers together
It becomes %28highlight_green%28+2+%29%29%2An%2B2=n%2B4%2B9.
Look at highlight_red%28+%28highlight_red%28+2+%29%29%2An+%29%2B2=n%2B4%2B9.
Remove unneeded parentheses around factor highlight_red%28+2+%29
It becomes highlight_green%28+2+%29%2An%2B2=n%2B4%2B9.
Look at 2%2An%2B2=n%2Bhighlight_red%28+4+%29%2Bhighlight_red%28+9+%29.
Added fractions or integers together
It becomes 2%2An%2B2=n%2Bhighlight_green%28+13+%29.
Look at 2%2An%2B2=highlight_red%28+n%2B13+%29.
Moved these terms to the left highlight_green%28+-n+%29,highlight_green%28+-13+%29
It becomes 2%2An%2B2-highlight_green%28+n+%29-highlight_green%28+13+%29=0.
Look at 2%2An%2Bhighlight_red%28+2+%29-n-highlight_red%28+13+%29=0.
Added fractions or integers together
It becomes 2%2An%2Bhighlight_green%28+-11+%29-n=0.
Look at 2%2An%2Bhighlight_red%28+-11+%29-n=0.
Moved -11 to the right of expression
It becomes 2%2An-n%2Bhighlight_green%28+-11+%29=0.
Look at 2%2An-n%2Bhighlight_red%28+-11+%29=0.
Removed extra sign in front of -11
It becomes 2%2An-n-highlight_green%28+11+%29=0.
Look at highlight_red%28+2%2An+%29-highlight_red%28+n+%29-11=0.
Eliminated similar terms highlight_red%28+2%2An+%29,highlight_red%28+-n+%29 replacing them with highlight_green%28+%282-1%29%2An+%29
It becomes highlight_green%28+%282-1%29%2An+%29-11=0.
Look at highlight_red%28+%282-1%29+%29%2An-11=0.
Remove extraneous '1' from product highlight_red%28+%282-1%29+%29
It becomes n-11=0.
Look at highlight_red%28+n-11+%29=0.
Solved linear equation highlight_red%28+n-11=0+%29 equivalent to n-11 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: n=11.

Universal Simplifier and Solver


Done!



Step 5. ANSWER: So the integers are 11, 13 and 15.

Check 11+13=24 which is 9 more than 15. So it checks!

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