Math
posted by Crystal .
Find three consecutive integers such that the sum of the first and second is three mose than three times the third.
When I did the problem,I got a negative answer. Does that seem correct? Did I make any mistakes?

Let a = the first number, b = the second number and c = the third number.
Then, b = (a + 1) and c = (a + 2)
Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3 making a = 8 and the three numbers 8, 7 and 6.
8 + 7 = 3(6) + 3
15 = 18 + 3 = 15. 
How do you get 8 as a?

Let a = the first number, b = the second number and c = the third number.
Then, b = (a + 1) and c = (a + 2)
Therefore, a + (a + 1) = 3(a + 2) + 3 yielding 2a + 1 = 3a + 6 + 3.
.........2a + 1 = 3a + 9
.........2a + 1  9 = 3a
.........2a  8 = 3a
............8 = 3a  2a
............8 = a
8 + 7 = 3(6) + 3
15 = 18 + 3 = 15.
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