1. **State the problem:** We need to find three consecutive odd integers such that the sum of the first and second is 31 less than 3 times the third. 2. **Define variables:** Let the first odd integer be $x$. Since the integers are consecutive odd numbers, the second integer is $x + 2$ and the third integer is $x + 4$. 3. **Write the equation:** The sum of the first and second integers is $x + (x + 2) = 2x + 2$. The problem states this sum is 31 less than 3 times the third integer, so: $$2x + 2 = 3(x + 4) - 31$$ 4. **Simplify the equation:** $$2x + 2 = 3x + 12 - 31$$ $$2x + 2 = 3x - 19$$ 5. **Isolate $x$:** Subtract $2x$ from both sides: $$\cancel{2x} + 2 = 3x - 19 - \cancel{2x}$$ $$2 = x - 19$$ Add 19 to both sides: $$2 + 19 = x - 19 + 19$$ $$21 = x$$ 6. **Find the integers:** First integer: $x = 21$ Second integer: $x + 2 = 23$ Third integer: $x + 4 = 25$ 7. **Check the solution:** Sum of first and second: $21 + 23 = 44$ Three times the third: $3 \times 25 = 75$ $75 - 31 = 44$, which matches the sum. **Final answer:** The three consecutive odd integers are $21$, $23$, and $25$.