Question 889245

Find 3 consecutive odd integers such that twice the first plus 3 times the third is 19 more than 4 times the second.
<pre>
Let 1st of 3 integers be F
Then 2nd and 3rd are: F + 2, and F + 4, respectively
Thus, 2F + 3(F + 4) = 4(F + 2) + 19 
Solve for F: the 1st of the 2 integers, then find 2nd and 3rd.
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