Question

How to solve for 3 times the first odd integer plus 2 times the second odd integer equals 4 times the third odd integer minus a constant?

Answer 1

To solve the problem of finding three odd integers based on the equation involving them, let's denote the first three odd integers. We can express them in terms of the first odd integer, which we can call \( n \). Thus, the integers can be written as:

- First odd integer: \( n \)
- Second odd integer: \( n + 2 \) (the next odd integer)
- Third odd integer: \( n + 4 \) (the subsequent odd integer)

Now, we can set up the equation based on the given information:

\[
3 \times (\text{first odd integer}) + 2 \times (\text{second odd integer}) = 4 \times (\text{third odd integer}) - k
\]

Substituting our expressions for the odd integers into the equation:

\[
3n + 2(n + 2) = 4(n + 4) - k
\]

Now, let's simplify both sides of the equation:
1. For the left side:
\[
3n + 2(n + 2) = 3n + 2n + 4 = 5n + 4
\]

2. For the right side:
\[
4(n + 4) - k = 4n + 16 - k
\]

Now we equate both sides of the simplified equation:

\[
5n + 4 = 4n + 16 - k
\]

Next, isolate \( n \) and \( k \):

1. Subtract \( 4n \) from both sides:
\[
5n - 4n + 4 = 16 - k
\]
This simplifies to:
\[
n + 4 = 16 - k
\]

2. Rearrange to isolate \( k \):
\[
k = 16 - n - 4
\]
Simplifying gives us:
\[
k = 12 - n
\]

The result indicates that \( k \) can depend on the value of the first odd integer \( n \). For a specific case, choosing an integer value for \( n \) (keeping in mind \( n \) should be odd), we can find specific values for the integers and \( k \).

### Example:

1. Choose \( n = 1 \):
- Then:
- First odd integer: \( 1 \)
- Second odd integer: \( 3 \)
- Third odd integer: \( 5 \)
- Constant \( k = 12 - 1 = 11 \)

2. Checking:
\[
3 \times 1 + 2 \times 3 = 4 \times 5 - 11
\]
Calculating both sides:
- Left side: \( 3 \times 1 + 2 \times 3 = 3 + 6 = 9 \)
- Right side: \( 4 \times 5 - 11 = 20 - 11 = 9 \)

Both sides are equal, confirming that our integers (\( n = 1, n + 2 = 3, n + 4 = 5 \)) and constant \( k = 11 \) satisfy the initial condition.

In conclusion, you can derive the constant \( k \) as \( k = 12 - n \) based on any odd integer \( n \), and the integers formed around \( n \) (as in the examples) will satisfy the equation.