Answer :
- Define variables: Let 'x' be the number of arcade games and 'y' be the number of Ferris wheel rides.
- Identify missing information: We need the cost of each arcade game ('a'), the cost of each Ferris wheel ride ('b'), and Maya's total allowance ('c').
- Formulate the equation: $ax + by = c$.
- Solve for one variable: If we know a, b, c, and one of x or y, we can solve for the other. The additional information needed is Maya's allowance amount, the cost of playing the arcade game, and the cost of riding the Ferris wheel.
### Explanation
1. Defining Variables
To create a two-variable equation for the scenario, we need to define our variables and identify the missing information. Let's define 'x' as the number of times Maya played the arcade game and 'y' as the number of times she rode the Ferris wheel.
2. Identifying Missing Information
The problem states that we need to know the cost of each arcade game, the cost of each Ferris wheel ride, and Maya's total allowance. Let's denote the cost of each arcade game as 'a', the cost of each Ferris wheel ride as 'b', and Maya's total allowance as 'c'.
3. Formulating the Equation
With this information, we can create the equation: $ax + by = c$. This equation represents the total amount Maya spent on arcade games and Ferris wheel rides, which equals her total allowance.
4. Describing the Problem Solved
With this equation, we can solve for one variable (either x or y) if we know the values of the other variable and the constants a, b, and c. For example, if we know how many times Maya played the arcade game (x), we can solve for how many times she rode the Ferris wheel (y).
5. Example Problem
For instance, let's say Maya's allowance is $20 (c = 20)$, the arcade game costs $2 per play (a = 2), and the Ferris wheel costs $3 per ride (b = 3). If she played the arcade game 4 times (x = 4), we can find the number of Ferris wheel rides (y) by solving the equation: $2(4) + 3y = 20$, which simplifies to $8 + 3y = 20$. Subtracting 8 from both sides gives $3y = 12$, and dividing by 3 gives $y = 4$. So, she rode the Ferris wheel 4 times.
6. Conclusion
Therefore, the additional information needed is Maya's allowance amount (c), the cost of playing the arcade game (a), and the cost of riding the Ferris wheel (b). With this information, we can solve for the number of times Maya rode the Ferris wheel (y) given the number of times she played the arcade game (x), or vice versa.
### Examples
Imagine you're planning a school carnival. You need to figure out how many game tickets and ride passes you can offer with a set budget. By using a two-variable equation similar to Maya's situation, you can determine the optimal combination of tickets and passes to maximize participation while staying within budget. This helps in resource allocation and planning events effectively.
- Identify missing information: We need the cost of each arcade game ('a'), the cost of each Ferris wheel ride ('b'), and Maya's total allowance ('c').
- Formulate the equation: $ax + by = c$.
- Solve for one variable: If we know a, b, c, and one of x or y, we can solve for the other. The additional information needed is Maya's allowance amount, the cost of playing the arcade game, and the cost of riding the Ferris wheel.
### Explanation
1. Defining Variables
To create a two-variable equation for the scenario, we need to define our variables and identify the missing information. Let's define 'x' as the number of times Maya played the arcade game and 'y' as the number of times she rode the Ferris wheel.
2. Identifying Missing Information
The problem states that we need to know the cost of each arcade game, the cost of each Ferris wheel ride, and Maya's total allowance. Let's denote the cost of each arcade game as 'a', the cost of each Ferris wheel ride as 'b', and Maya's total allowance as 'c'.
3. Formulating the Equation
With this information, we can create the equation: $ax + by = c$. This equation represents the total amount Maya spent on arcade games and Ferris wheel rides, which equals her total allowance.
4. Describing the Problem Solved
With this equation, we can solve for one variable (either x or y) if we know the values of the other variable and the constants a, b, and c. For example, if we know how many times Maya played the arcade game (x), we can solve for how many times she rode the Ferris wheel (y).
5. Example Problem
For instance, let's say Maya's allowance is $20 (c = 20)$, the arcade game costs $2 per play (a = 2), and the Ferris wheel costs $3 per ride (b = 3). If she played the arcade game 4 times (x = 4), we can find the number of Ferris wheel rides (y) by solving the equation: $2(4) + 3y = 20$, which simplifies to $8 + 3y = 20$. Subtracting 8 from both sides gives $3y = 12$, and dividing by 3 gives $y = 4$. So, she rode the Ferris wheel 4 times.
6. Conclusion
Therefore, the additional information needed is Maya's allowance amount (c), the cost of playing the arcade game (a), and the cost of riding the Ferris wheel (b). With this information, we can solve for the number of times Maya rode the Ferris wheel (y) given the number of times she played the arcade game (x), or vice versa.
### Examples
Imagine you're planning a school carnival. You need to figure out how many game tickets and ride passes you can offer with a set budget. By using a two-variable equation similar to Maya's situation, you can determine the optimal combination of tickets and passes to maximize participation while staying within budget. This helps in resource allocation and planning events effectively.
