SOLUTION: Find three consecutive odd integers such that four times the sum of the first and second is 11 more than 3 times the third.

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Question 1201890: Find three consecutive odd integers such that four times the sum of the first and second is 11 more than 3 times the third.
Found 2 solutions by Edwin McCravy, josgarithmetic:
Answer by Edwin McCravy(20054) About Me  (Show Source):
You can put this solution on YOUR website!
Find three consecutive odd integers...
1st odd integer = x
2nd odd integer = x+2
3rd odd integer = x+4
such that four times the sum of the first and second
4[x + (x+2)]

is 

   =
11 more than 3 times the third.
3(x+4) + 11

4[x + (x+2)] = 3(x+4) + 11
4[x + x + 2] = 3x + 12 + 11
   4[2x + 2] = 3x + 23
      8x + 8 = 3x + 23
          5x = 15
           x = 3

1st odd integer = x = 3
2nd odd integer = x+2 = 3+2 = 5
3rd odd integer = x+4 = 3+4 = 7

Checking:
four times the sum of the first and second
4(3 + 5) = 4(8) = 32
is 11 more than 3 times the third.
3(7) + 11 = 21 + 11 = 32

So it checks.

Edwin

Answer by josgarithmetic(39616) About Me  (Show Source):
You can put this solution on YOUR website!
Because of this part, "the sum of the first and second is", and the three integers are odd,
variables can be assigned as
n-1, n+1, n+3
.

4%28%28n-1%29%2B%28n%2B1%29%29=11%2B3%28n%2B3%29
-
4%28n-1%2Bn%2B1%29=11%2B3n%2B9
4%282n%29=3n%2B20
8n=3n%2B20
5n=20
n=4

The consecutive odd integers: 3, 5, 7