SOLUTION: Find four consecutive integers such that 4 times the sum of the first and the fourth is greater than 6 times the third.

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Question 1123520: Find four consecutive integers such that 4 times the sum of the first and the fourth is greater than 6 times the third.
Found 3 solutions by josgarithmetic, JessLynnLancaster24, ikleyn:
Answer by josgarithmetic(39621) About Me  (Show Source):
You can put this solution on YOUR website!
Integers system%28n%2Cn%2B1%2Cn%2B2%2Cn%2B3%29

4%28n%2B%28n%2B3%29%29%3E6%28n%2B2%29, "... is greater than 6 times the third."
(Literally translated into this inequality according to the description)

4%282n%2B3%29%3E6n%2B12
8n%2B12%3E6n%2B12
2n%3E0
n%3E0


The description of the consecutive integers will work if n is 1 or greater.

Answer by JessLynnLancaster24(2) About Me  (Show Source):
You can put this solution on YOUR website!
4(n +(n +3)) > 6(n +2)
4n + (4n +3) > 6n + 2
8n +3 > 6n +2
2n + 3 > 2
2n > -1
n > -0.5

Answer by ikleyn(52815) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let the numbers are n, n+1, n+2 and n+3.


Then the condition says 


4*(n + (n+3)) > 6*(n+2)


4n + 4n + 12 > 6n + 12


8n + 12 > 6n + 12


8n - 6n > 12 - 12


2n > 0


n > 0.

Answer.   Any four consecutive integers  n,  n+1,  n+2  and  n+3  with  n > 0  satisfy the problem's condition.


Be aware.   The solution by  @JessLynnLancaster24  is totally wrong and contains a lot of errors,
demonstrating extremely low level in Math.

This forum is not the place for such solutions and such  "tutors".