SOLUTION: Find three consecutive integers such that twice the larger exceeds the smaller by 40

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Question 1099506: Find three consecutive integers such that twice the larger exceeds the smaller by 40
Found 4 solutions by josgarithmetic, addingup, greenestamps, MathTherapy:
Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!

-






36, 37, 38
Answer by addingup(3677)   (Show Source): You can put this solution on YOUR website!
Integers: n, n+1, n+2
and
2(n+2) = n-40
2n+4 = n-40
n = -44
------------------
n = -44
n+1 = -43
n+2 = -42
-----------------------
Check:
twice the larger:
-42 x 2 = -84
exceeds the smaller by 40
n = -44 and the distance from -44 to -84 is 40, this is the correct answer.

Answer by greenestamps(13200)   (Show Source): You can put this solution on YOUR website!

If you are confused with the two answers, you can ignore the response from tutor addingup. His original equation says that twice the largest integer is 40 LESS THAN the smallest -- not 40 MORE THAN.
Answer by MathTherapy(10552)   (Show Source): You can put this solution on YOUR website!

Find three consecutive integers such that twice the largerest exceeds the smaller by 40 
With twice the LARGEST exceeding the smallest by 40, the correct numbers are: 

IGNORE all other RIDICULOUS/INCORRECT/WRONG answers. 

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