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.

To solve this problem, let's break it down step by step.

Let's assume the first integer is x.
According to the problem, the second integer would be x + 1 (as they are consecutive integers).
And the third integer would be x + 2 (as they are consecutive integers).

The problem states that the sum of the first and second integers is three more than three times the third integer. So we can write this as an equation:

x + (x + 1) = 3(x + 2) + 3

Now we can simplify and solve for x:

2x + 1 = 3x + 6 + 3
2x + 1 = 3x + 9

Moving the variables to one side and constants to the other side:

2x - 3x = 9 - 1
-x = 8

Now, dividing both sides by -1 to solve for x:

x = -8

So the first integer is -8, and the second and third consecutive integers would be -7 and -6, respectively.

Therefore, the three consecutive integers that satisfy the given condition are -8, -7, and -6.

Based on your statement, it seems that you may have made an error in your calculations. Double-check your work to ensure you haven't missed any steps or calculations.