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The sum of three consecutive integers is twice the smallest of the integers. Find the integers?
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The sum of three consecutive integers is twice the smallest of the int...
The problem:
The sum of three consecutive integers is twice the smallest of the integers. We are required to find the integers.

Understanding the problem:
Let's assume the first integer as 'x'. Since the integers are consecutive, the second integer will be 'x + 1' and the third integer will be 'x + 2'.
According to the problem, the sum of these three consecutive integers is twice the smallest integer, which can be expressed as:

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

Solving the equation:
To solve the equation, we will follow these steps:

1. Simplify the equation by combining like terms:
3x + 3 = 2x

2. Subtract 2x from both sides of the equation:
3x - 2x + 3 = 0

3. Simplify the equation further:
x + 3 = 0

4. Subtract 3 from both sides of the equation:
x = -3

Interpreting the result:
The first integer is -3. To find the consecutive integers, we can substitute this value into our assumption.
The second integer is -3 + 1 = -2, and the third integer is -3 + 2 = -1.

Final answer:
The three consecutive integers are -3, -2, and -1.
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