Now try writing and solving your own system of linear equations: one with infinite solutions, one with no solution, and one with one solution. Write a real-world scenario that goes with at least one system of equations. Show your work.
System 1: Infinite Solutions
Scenario: Cost of Movie Tickets
Let's consider a scenario where a group of friends decides to go to the movies. They are charged a fixed fee for a group ticket and an additional fee for each individual ticket. The total cost of the group ticket is the same as the combined cost of each individual ticket, resulting in infinite solutions.
Let's assign variables:
x = cost of the group ticket
y = cost of an individual ticket
Equations:
1. Group ticket cost: x = y + 5
2. Combined individual tickets cost: 5y = y
Solving the system:
From equation 2, we can simplify it as:
5y = y
4y = 0
y = 0
Substituting y = 0 into equation 1:
x = 0 + 5
x = 5
The solution to this system is x = 5 and y = 0. However, in this real-world scenario, it means that the group ticket cost is $5, while the individual tickets have no cost. Thus, as long as the cost of the group ticket is $5, there are infinite possible combinations where the individual tickets cost $0.
System 2: No Solution
Scenario: Coffee Shop Profits
Consider a small coffee shop that offers two sizes of coffee cups: small and large. The shop sells the small cup of coffee for $2 and the large cup for $3. The shopkeeper counts the total revenue for a day, but mistakenly loses the information about the number of small and large cups sold. Consequently, we cannot find a solution as there is no way to determine the number of each size of cup sold from the given information.
Let's assign variables:
x = number of small cups sold
y = number of large cups sold
Equations:
1. Revenue equation: 2x + 3y = 100
2. Unknown information equation: x + y = 50
Solving the system:
Rearranging equation 2 to express x in terms of y:
x = 50 - y
Substituting this expression into equation 1:
2(50 - y) + 3y = 100
100 - 2y + 3y = 100
y = 0
Substituting y = 0 into equation 2:
x + 0 = 50
x = 50
However, these values mean that 50 small cups and 0 large cups were sold, resulting in no revenue at all. This discrepancy shows that there is no solution for this system of equations, as the information provided does not align with the given revenue.
System 3: One Solution
Scenario: Weekly Allowance
Consider a scenario where a parent gives their child a weekly allowance for completing tasks. The child receives $5 for doing their chores and $10 for completing extra tasks. The total allowance received in a week is $50. We can determine the number of chores and extra tasks completed by finding one unique solution.
Let's assign variables:
x = number of chores completed
y = number of extra tasks completed
Equations:
1. Weekly allowance equation: 5x + 10y = 50
2. Task completion equation: x + y = 10
Solving the system:
Rearranging equation 2 to express x in terms of y:
x = 10 - y
Substituting this expression into equation 1:
5(10 - y) + 10y = 50
50 - 5y + 10y = 50
5y = 0
y = 0
Substituting y = 0 into equation 2:
x + 0 = 10
x = 10
So, in this scenario, the child completes 10 chores and 0 extra tasks to receive the total weekly allowance of $50.