SOLUTION: Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 82.

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Question 220978: Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 82.
Answer by drj(1380)   (Show Source): You can put this solution on YOUR website!
Find three consecutive odd integers such that the sum of the first, two times the second, and three times the third is 82.

Step 1. Let n be one odd integer.

Step 2. Let n+2 and n+4 be the next two odd consecutive integer.

Step 3. Let 2(n+2) be two times the second.

Step 4. Let 3(n+4) be three times the third.

Step 5. Then n+2(n+2)+3(n+4)=82 since the sum of the first, two times the second, and three times the third is 82

Step 6. Solving n+2(n+2)+3(n+4)=82 yields 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 was fully solved.
  • Text form: n+2*(n+2)+3*(n+4)=82 simplifies to 0=0
  • Cartoon (animation) form:
    For tutors: simplify_cartoon( n+2*(n+2)+3*(n+4)=82 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at .
Moved these terms to the left highlight_green%28+-82+%29
It becomes .
Look at .
Expanded term 2 by using associative property on %28n%2B2%29
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Moved -78 to the right of expression
It becomes .
Look at .
Removed extra sign in front of -78
It becomes .
Look at .
Eliminated similar terms highlight_red%28+n+%29,highlight_red%28+2%2An+%29 replacing them with highlight_green%28+%281%2B2%29%2An+%29
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Remove unneeded parentheses around factor highlight_red%28+3+%29
It becomes .
Look at .
Expanded term 3 by using associative property on %28n%2B4%29
It becomes .
Look at .
Multiplied numerator integers
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Removed extra sign in front of -66
It becomes .
Look at .
Eliminated similar terms highlight_red%28+3%2An+%29,highlight_red%28+3%2An+%29 replacing them with highlight_green%28+%283%2B3%29%2An+%29
It becomes .
Look at .
Added fractions or integers together
It becomes .
Look at .
Remove unneeded parentheses around factor highlight_red%28+6+%29
It becomes .
Look at .
Solved linear equation highlight_red%28+6%2An-66=0+%29 equivalent to 6*n-66 =0
It becomes .
Result:
This is an equation! Solutions: n=11.

Universal Simplifier and Solver


Done!


and

Check sum...11+2*13+3*15=11+26+45=82... which is a true statement

Step 7. ANSWER: The three consecutive odd integers are 11, 13, and 15.

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J


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