Question 578420
find three consecutive odd integers such that 12 more than twice the sum of the first and third is equal to 15 less than five times the second.
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let x = 1st consecutive odd integer
x+2 = 2nd consecutive odd integer
x+4 = 3rd consecutive odd integer
..
2(x+x+4)+12=5(x+2)-15
4x+8+12=5x+10-15
4x+20=5x-5
x=25
x+2=27
x+4=29
ans:
The three consecutive odd integers are: 25, 27, and 29