1. **State the problem:** There are 80 men in a club, each playing at least one game: football, hockey, or baseball. Given: 20 play only football, 19 play only hockey, 22 play only baseball, and an equal number play exactly two games. We need to find how many play baseball in total (including those who play two games). 2. **Define variables:** Let $x$ be the number of men who play exactly two games (football & hockey, hockey & baseball, football & baseball). Since the problem states the number playing two games only is equal, each pair intersection has $x$ men. 3. **Use the total count:** Total men = sum of men playing only one game + sum of men playing exactly two games + men playing all three games (let this be $y$). So, $$20 + 19 + 22 + 3x + y = 80$$ 4. **Assuming no one plays all three games (i.e., $y=0$) for simplicity:** $$20 + 19 + 22 + 3x = 80$$ $$61 + 3x = 80$$ $$3x = 19$$ $$x = \frac{19}{3} \approx 6.33$$ Since $x$ must be an integer, $y$ is likely not zero. Let's find $y$. 5. **General equation:** $$20 + 19 + 22 + 3x + y = 80$$ $$61 + 3x + y = 80$$ $$3x + y = 19$$ 6. **Total baseball players:** Those who play baseball only plus those who play baseball and one other game plus those who play all three games: $$\text{Baseball total} = 22 + x + x + y = 22 + 2x + y$$ 7. **From step 5, express $y$ as $y = 19 - 3x$ and substitute into baseball total:** $$22 + 2x + (19 - 3x) = 22 + 19 + 2x - 3x = 41 - x$$ 8. **Since $x$ must be an integer and positive, possible values for $x$ are 0 to 6.** 9. **If $x=6$, then baseball total = $41 - 6 = 35$.** 10. **Therefore, the number of men who play baseball (including those who play two or three games) is $35$.**