1. **State the problem:** We have 50 players who can only play in 6-player games (3 on 3) or 2-player games (1 on 1). Let $x$ be the number of 6-player games and $y$ be the number of 2-player games. We want to find combinations of $x$ and $y$ that satisfy the total number of players and games. 2. **Write the equations:** Each 6-player game has 6 players, so total players in 6-player games is $6x$. Each 2-player game has 2 players, so total players in 2-player games is $2y$. Since all 50 players participate: $$6x + 2y = 50$$ Also, the total number of games is the sum of 6-player and 2-player games: $$x + y = \text{total games}$$ 3. **Complete the table:** Given some values of $x$, find $y$ using the player equation: For $x=0$: $$6(0) + 2y = 50 \Rightarrow 2y = 50 \Rightarrow y = \frac{50}{2} = 25$$ For $x=1$: $$6(1) + 2y = 50 \Rightarrow 6 + 2y = 50 \Rightarrow 2y = 44 \Rightarrow y = \frac{44}{2} = 22$$ For $x=4$: $$6(4) + 2y = 50 \Rightarrow 24 + 2y = 50 \Rightarrow 2y = 26 \Rightarrow y = \frac{26}{2} = 13$$ So the table is: | $x$ | $y$ | |---|---| | 0 | 25 | | 1 | 22 | | 4 | 13 | 4. **Part b: Find $x$ and $y$ if total games = 13 and all 50 players participate.** We have two equations: $$\begin{cases} 6x + 2y = 50 \\ x + y = 13 \end{cases}$$ From the second equation: $$y = 13 - x$$ Substitute into the first: $$6x + 2(13 - x) = 50$$ Simplify: $$6x + 26 - 2x = 50$$ $$4x + 26 = 50$$ $$4x = 50 - 26 = 24$$ $$x = \frac{24}{4} = 6$$ Then: $$y = 13 - 6 = 7$$ **Answer:** 6 six-player games and 7 two-player games. This satisfies both the total players and total games conditions.