SOLUTION: Find three consecutive odd integers such that three times the second minus the third is 13 more than the first.

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Question 216388: Find three consecutive odd integers such that three times the second minus the third is 13 more than the first.
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive odd integers such that three times the second minus the third is 13 more than the first.

Step 1. Let n be one odd integer, let n%2B2 be second consecutive integer and let n%2B4 be the third consecutive odd integers.

Step 2. Let 3%28n%2B2%29 three times the second.

Step 3. Then 3%28n%2B2%29-%28n%2B4%29=n%2B13 since three times the second minus the third is 13 more than the first.

Step 4. The following steps will solve the equation in Step 3.

3%28n%2B2%29-%28n%2B4%29=n%2B13

3n%2B6-n-4=n%2B13

2n%2B2=n%2B13

Subtract n+2 to both sides of the equation

2n%2B2-n-2=n%2B13-2

n=11

Then n%2B2=13 and n%2B4=15

Check...3%28n%2B2%29-%28n%2B4%29=n%2B13 then 3%2A13-15=11%2B13 or 24=24 which is a true statement.

Step 5. The three consecutive odd integers are 11, 13, and 15.

I hope the above steps and explanation were helpful.

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

Also, good luck in your studies and contact me at john@e-liteworks.com for your future math needs.

Respectfully,
Dr J