Three consecutive odd integers have a sum of –21. Which integers are they?

1 answer

Let's denote the three consecutive odd integers as \( x \), \( x + 2 \), and \( x + 4 \).

We know that their sum is given by:

\[ x + (x + 2) + (x + 4) = -21 \]

Now, simplify the left side:

\[ 3x + 6 = -21 \]

Next, we will isolate \( x \) by first subtracting 6 from both sides:

\[ 3x = -21 - 6 \]

\[ 3x = -27 \]

Now, divide both sides by 3:

\[ x = -9 \]

Now we can find the three consecutive odd integers:

The first integer is \( x = -9 \).

The second integer is \( x + 2 = -9 + 2 = -7 \).

The third integer is \( x + 4 = -9 + 4 = -5 \).

Thus, the three consecutive odd integers are:

\[ \boxed{-9}, \quad \boxed{-7}, \quad \text{and} \quad \boxed{-5} \]

To verify, we check their sum:

\[ -9 + (-7) + (-5) = -9 - 7 - 5 = -21 \]

The sum is indeed \(-21\), confirming the integers are correct.