He's WRONG, one AGAIN!! Integers DO exist!! 

Answer by josgarithmetic(39617)   (Show Source): You can put this solution on YOUR website!
  consecutive integers such that five times the third integer equals twice the first integer plus seven more than the second integer?

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The integers are 4,5,6

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


Oh, my........


Tutor @CubeyThePenguin: wrong equation, leading to wrong answer.


Tutor @MathTherapy: nothing of use to you the student....


Tutor @josgarithmetic: algebra errors, leading to wrong answer... (which she could have seen if she had bothered to check to see if her answers were right).


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While there are many problems involving three consecutive integers where it is advantageous to call the numbers x-1, x, and x+1, I don't see that being helpful here.  So let the numbers be x, x+1, and x+2.  Then


"...five times the third integer equals twice the first integer plus seven more than the second integer"


5 times the third integer: 5(x+2)


twice the first integer plus 7 more than the second integer: 2x+((x+1)+7)











ANSWER: The three consecutive integers are -1, 0, and 1.


CHECK:

5 times the third integer: 5(1) = 5


twice the first integer plus 7 more than the second integer: 2(-1)+(0+7) = -2+7 = 5




 

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