SOLUTION: If the sum of three consecutive odd integers is 411, what is the first integer? A 131 B 133 C 135 D 137 E 139

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Question 223722: If the sum of three consecutive odd integers is 411, what is the first integer?


A 131
B 133
C 135
D 137
E 139
Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
If the sum of three consecutive odd integers is 411, what is the first integer?

A 131
B 133
C 135
D 137
E 139

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. Then n+(n+2)+(n+4)=411 be the sum of three odd integers

Step 4. Solving n+(n+2)+(n+4)=411 yields the following steps.

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


YOUR ANSWER


  • This is an equation! Solutions: n=135.
  • Graphical form: Equation n%2B%28n%2B2%29%2B%28n%2B4%29=411 was fully solved.
  • Text form: n+(n+2)+(n+4)=411 simplifies to 0=0
  • Cartoon (animation) form: simplify_cartoon%28+n%2B%28n%2B2%29%2B%28n%2B4%29=411+%29
    For tutors: simplify_cartoon( n+(n+2)+(n+4)=411 )
  • If you have a website, here's a link to this solution.

DETAILED EXPLANATION

Look at n%2Bhighlight_red%28+%28n%2B2%29+%29%2B%28n%2B4%29=411.
Remove unneeded parentheses around terms highlight_red%28+n+%29,highlight_red%28+2+%29
It becomes n%2Bhighlight_green%28+n+%29%2Bhighlight_green%28+2+%29%2B%28n%2B4%29=411.
Look at n%2Bn%2B2%2Bhighlight_red%28+%28n%2B4%29+%29=411.
Remove unneeded parentheses around terms highlight_red%28+n+%29,highlight_red%28+4+%29
It becomes n%2Bn%2B2%2Bhighlight_green%28+n+%29%2Bhighlight_green%28+4+%29=411.
Look at n%2Bn%2Bhighlight_red%28+2+%29%2Bn%2Bhighlight_red%28+4+%29=411.
Added fractions or integers together
It becomes n%2Bn%2Bhighlight_green%28+6+%29%2Bn=411.
Look at n%2Bn%2Bhighlight_red%28+6+%29%2Bn=411.
Moved 6 to the right of expression
It becomes n%2Bn%2Bn%2Bhighlight_green%28+6+%29=411.
Look at highlight_red%28+n+%29%2Bhighlight_red%28+n+%29%2Bhighlight_red%28+n+%29%2B6=411.
Eliminated similar terms highlight_red%28+n+%29,highlight_red%28+n+%29,highlight_red%28+n+%29 replacing them with highlight_green%28+%281%2B1%2B1%29%2An+%29
It becomes highlight_green%28+%281%2B1%2B1%29%2An+%29%2B6=411.
Look at %28highlight_red%28+1+%29%2Bhighlight_red%28+1+%29%2Bhighlight_red%28+1+%29%29%2An%2B6=411.
Added fractions or integers together
It becomes %28highlight_green%28+3+%29%29%2An%2B6=411.
Look at highlight_red%28+%28highlight_red%28+3+%29%29%2An+%29%2B6=411.
Remove unneeded parentheses around factor highlight_red%28+3+%29
It becomes highlight_green%28+3+%29%2An%2B6=411.
Look at 3%2An%2B6=highlight_red%28+411+%29.
Moved these terms to the left highlight_green%28+-411+%29
It becomes 3%2An%2B6-highlight_green%28+411+%29=0.
Look at 3%2An%2Bhighlight_red%28+6+%29-highlight_red%28+411+%29=0.
Added fractions or integers together
It becomes 3%2An%2Bhighlight_green%28+-405+%29=0.
Look at 3%2An%2Bhighlight_red%28+-405+%29=0.
Removed extra sign in front of -405
It becomes 3%2An-highlight_green%28+405+%29=0.
Look at highlight_red%28+3%2An-405+%29=0.
Solved linear equation highlight_red%28+3%2An-405=0+%29 equivalent to 3*n-405 =0
It becomes highlight_green%28+0+%29=0.
Result: 0=0
This is an equation! Solutions: n=135.

Universal Simplifier and Solver


Done!



n=135 n%2B2=137 and n%2B4=139

Check sum...135+137+139... which is a true statement

Step 5. ANSWER: The three consecutive odd integers are 135, 137, and 139. The first integer is 135 so the solution is C.

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