Posted by **Crystal** on Monday, February 22, 2010 at 1:54pm.

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?

- Math -
**tchrwill**, Monday, February 22, 2010 at 3:04pm
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.

- Math -
**crystal**, Monday, February 22, 2010 at 4:04pm
How do you get -8 as a?

- Math -
**tchrwill**, Monday, February 22, 2010 at 6:28pm
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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