Find three consecutive odd integers such that eight more than the sum of the first two is equal to eleven less than three times the third. Find the THIRD number.
Numbers are x x+2 and x+4
x + x+2 + 8 = 3 (x+4) - 11 solve for 'x' then the third number is x+4
I see now. I put 11 - 3 in the beginning so I had -3 times everything.