1. **State the problem:** We need to find two consecutive integers such that twice the smaller integer plus half the larger integer equals 33. 2. **Define variables:** Let the smaller integer be $x$. Then the larger consecutive integer is $x+1$. 3. **Write the equation:** Twice the smaller integer plus half the larger integer equals 33: $$2x + \frac{x+1}{2} = 33$$ 4. **Solve the equation:** Multiply both sides by 2 to clear the fraction: $$2 \times 2x + 2 \times \frac{x+1}{2} = 2 \times 33$$ $$4x + (x+1) = 66$$ 5. **Simplify:** $$4x + x + 1 = 66$$ $$5x + 1 = 66$$ 6. **Isolate $x$:** $$5x = 66 - 1$$ $$5x = 65$$ $$x = \frac{65}{5} = 13$$ 7. **Find the two integers:** The smaller integer is 13, and the larger integer is $13 + 1 = 14$. **Final answer:** The two consecutive integers are 13 and 14.