Question 1061817: Find three consecutive odd integers such that 6 times the sum of the first and second is 16 more than 8 times the third.
Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! x, x+2, x+4
6(2x+2)-16=8(x+4)
12x+12-16=8x+32
12x-4=8x+32
4x=36
x=9
9,11,13
6 times the sum of 9 and 11 is 6(20)=120, and that is 16 more than 8(13)=104
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