Question 619071
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Hi,
Find three consecutive odd integers{{{highlight(x)}}}, {{{highlight(x+2)}}}, {{{highlight(x+4)}}}
such that 3 times the second minus the third is 11 more than the first?
 3(x+2) - (x+4) = x+11
 3x + 6 -x-4 = x + 11
    2x + 2 = x + 11
     x = 9 
  9, 11 , 13
CHECKING our Answer***
3·11 - 13 = 20