10. Imagine that you are dividing marbles or dealing cards. Describe a situation for which you could use these kinds of algebraic equations to solve a problem. Write examples for both types of equations you used. Be specific and explain fully.

Let's consider two scenarios: one involving marbles and the other involving dealing cards. In both cases, we can use algebraic equations to solve problems related to distribution. ### Scenario 1: Dividing Marbles **Situation:** Suppose you have a total of 30 marbles, and you want to divide them among three friends: Alice, Bob, and Charlie. You want to give Alice twice as many marbles as Bob, and Charlie will receive 5 more marbles than Bob. **Let:** - \( a \) = number of marbles Alice receives - \( b \) = number of marbles Bob receives - \( c \) = number of marbles Charlie receives **Equations:** 1. The total number of marbles: \[ a + b + c = 30 \] 2. Alice receives twice as many marbles as Bob: \[ a = 2b \] 3. Charlie receives 5 more marbles than Bob: \[ c = b + 5 \] **Solving the Equations:** We can substitute the expressions for \( a \) and \( c \) into the first equation. Substituting \( a \) and \( c \): \[ 2b + b + (b + 5) = 30 \] This simplifies to: \[ 4b + 5 = 30 \] Now, we can solve for \( b \): \[ 4b = 30 - 5 \] \[ 4b = 25 \] \[ b = \frac{25}{4} = 6.25 \] Now substituting \( b \) back to find \( a \) and \( c \): \[ a = 2b = 2 \times 6.25 = 12.5 \] \[ c = b + 5 = 6.25 + 5 = 11.25 \] ### Summary for Marbles: - Alice receives 12.5 marbles - Bob receives 6.25 marbles - Charlie receives 11.25 marbles ### Scenario 2: Dealing Cards **Situation:** Imagine you have a deck of 52 playing cards, and you want to distribute them among four players: Player 1, Player 2, Player 3, and Player 4. Player 1 receives 3 times as many cards as Player 2, and Player 3 receives 2 cards more than Player 2. Player 4 receives the remaining cards. **Let:** - \( p_1 \) = number of cards Player 1 receives - \( p_2 \) = number of cards Player 2 receives - \( p_3 \) = number of cards Player 3 receives - \( p_4 \) = number of cards Player 4 receives **Equations:** 1. The total number of cards: \[ p_1 + p_2 + p_3 + p_4 = 52 \] 2. Player 1 receives three times as many cards as Player 2: \[ p_1 = 3p_2 \] 3. Player 3 receives 2 more cards than Player 2: \[ p_3 = p_2 + 2 \] **Solving the Equations:** Substituting \( p_1 \) and \( p_3 \) into the first equation: \[ 3p_2 + p_2 + (p_2 + 2) + p_4 = 52 \] This simplifies to: \[ 5p_2 + 2 + p_4 = 52 \] Now, we can solve for \( p_4 \): \[ p_4 = 52 - 5p_2 - 2 \] \[ p_4 = 50 - 5p_2 \] ### Summary for Cards: To find specific values, we can choose a value for \( p_2 \) (e.g., \( p_2 = 8 \)): - \( p_1 = 3 \times 8 = 24 \) - \( p_3 = 8 + 2 = 10 \) - \( p_4 = 50 - 5 \times 8 = 50 - 40 = 10 \) ### Final Distribution: - Player 1 receives 24 cards - Player 2 receives 8 cards - Player 3 receives 10 cards - Player 4 receives 10 cards In both scenarios, we used algebraic equations to model the distribution of marbles and cards, allowing us to find the number of items each person receives based on given relationships.