Find three consecutive intergers with a sum of -36

Answer 1

Let the three consecutive integers be represented as \( x \), \( x + 1 \), and \( x + 2 \).

The sum of these integers can be written as:

\[ x + (x + 1) + (x + 2) = -36 \]

Simplifying the left side:

\[ 3x + 3 = -36 \]

Next, subtract 3 from both sides:

\[ 3x = -36 - 3 \] \[ 3x = -39 \]

Now, divide by 3:

\[ x = -13 \]

Thus, the three consecutive integers are:

\[ -13, -12, -11 \]

To verify, we can calculate their sum:

\[ -13 + (-12) + (-11) = -36 \]

The three consecutive integers are \(-13\), \(-12\), and \(-11\).