Question

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

Answer 1

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.