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| Question 223562:  Hi, I don't know how to do consecutive odd/even integers.
 Such as this problem:
 
 Find three consecutive integers whose sum is 36.
 
 I know it's x+(x+2)+(x+3).  Am I right? After all, they are 'consecutive', they are not odd or even.
 
 Then the other one is:
 
 Find two consecutive even integers whose sum is 126.
 
 
 I did the first one, x+x+2=126
 Then I subtracted two from both sides, and got x+x=124. Then I added like terms and got 2x=124. And then I simply divided by 2. And got 62.
 I did the last question from another guy who answered another guy. But he didn't explain HOW to get the other one. I know the other answer is 64...but I do not know how to get it!
 Please explain it clearly as I have an Algebra Test tomorrow, and I want to ace it.
 
 Thanks. :)
 
 Found 2 solutions by  brich, drj:
 Answer by brich(7)
      (Show Source): 
You can put this solution on YOUR website! First Question: your logic was correct but your equation was slightly off.
 I think you meant to say:
 x+(x+1)+(x+2)=36
 Second Question:
 you were right about this and the second number is very easy to get:
 If you let the first number = x
 and the next even integer = x+2
 then x+(x+2)=126 like you stated above
 you found x to = 62 so plug it in for x:
 62+(62+2)=126
 remember the entire sum inside the parentheses is equal to the next even number
 do you see the answer you were looking for yet?
 subtract 62 from both sides if you aren't sure
 and the right side of the equation should now = 64
 this means that 64 must be the second number, and that would make sense because its the next even number.
 hope that helped.
 let me know :)
Answer by drj(1380)
      (Show Source): 
You can put this solution on YOUR website! Find three consecutive integers whose sum is 36. 
 Step 1.  Let n be the first integer.
 
 Step 2.  Let n+1 and n+2 be the next two consecutive integers.
 
 Step 3.  Then n+n+1+n+2=36 since the sum is 36.
 
 
 
 | 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+n+1+n+2=36 simplifies to 0=0Cartoon (animation) form:   For tutors:
 simplify_cartoon( n+n+1+n+2=36 )If you have a website, here's a link to this solution.  | 
 DETAILED EXPLANATIONLook at
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  . Result:
  This is an equation! Solutions: n=11.
 
 
 Universal Simplifier and SolverDone!
 
 |  
 
 So
  ,  and  Note integers add up to 36 
 Step 4.  ANSWER:  The integers are  11, 12, and 13.
 
 
 Find two consecutive even integers whose sum is 126.
 
 Step 1. Let n be the first even integer
 
 Step 2. Let n+2 be the next consecutive even integers
 
 Step 3.  Then, n+n+2=126 since the sum is 126.
 
 
 
 | Solved by pluggable solver: EXPLAIN simplification of an expression |  | Your Result: 
 
 
 
  YOUR ANSWER
 | 
This is an equation! Solutions: n=62.
Graphical form: Equation  was fully solved.Text form: n+n+2=126 simplifies to 0=0Cartoon (animation) form:   For tutors:
 simplify_cartoon( n+n+2=126 )If you have a website, here's a link to this solution.  | 
 DETAILED EXPLANATIONLook at
  . Eliminated similar terms
  ,  replacing them with  It becomes
  . 
 Look at
  . Added fractions or integers together
 It becomes
  . 
 Look at
  . Remove unneeded parentheses around factor
  It becomes
  . 
 Look at
  . Moved these terms to the left
  It becomes
  . 
 Look at
  . Added fractions or integers together
 It becomes
  . 
 Look at
  . Removed extra sign in front of
  It becomes
  . 
 Look at
  . Solved linear equation
  equivalent to  2*n-124  =0 It becomes
  . Result:
  This is an equation! Solutions: n=62.
 
 
 Universal Simplifier and SolverDone!
 
 |  
 
 
  and   
 Step 4.  ANSWER: The consecutive even integers are 62 and 64
 
 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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