Now try writing and solving your own systems 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: A movie theater offers two types of tickets: regular and VIP. The theater sold 120 tickets for a total revenue of $1240. The regular ticket costs $8 and the VIP ticket costs $15.
Let x be the number of regular tickets sold
Let y be the number of VIP tickets sold
From the given information, we can set up the following equations:
1) x + y = 120 (Equation for the total number of tickets sold)
2) 8x + 15y = 1240 (Equation for the total revenue)
To solve this system of equations, we can use the elimination method:
Multiplying equation 1 by 8, we get: 8x + 8y = 960
Subtracting equation 2 from this new equation, we have:
(8x + 8y) - (8x + 15y) = 960 - 1240
-7y = -280
Simplifying the equation further, we divide both sides by -7:
y = 40
Substituting the value of y back into equation 1:
x + 40 = 120
x = 80
Therefore, there are 80 regular tickets sold and 40 VIP tickets sold. This system of equations has an infinite number of solutions because any combination of regular and VIP tickets that add up to 120 will satisfy the equations.
System 2: No Solution
Scenario: A bakery sells cupcakes and cookies. The total number of cupcakes and cookies sold is 100. The cupcakes are sold for $2 each, while the cookies are sold for $4 each. The bakery made a total revenue of $300.
Let x be the number of cupcakes sold
Let y be the number of cookies sold
From the given information, we can set up the following equations:
1) x + y = 100 (Equation for the total number of cupcakes and cookies sold)
2) 2x + 4y = 300 (Equation for the total revenue)
To solve this system of equations, we can again use the elimination method:
Multiplying equation 1 by 2, we get: 2x + 2y = 200
Subtracting equation 2 from this new equation, we have:
(2x + 2y) - (2x + 4y) = 200 - 300
-2y = -100
Simplifying the equation further, we divide both sides by -2:
y = 50
Substituting the value of y back into equation 1:
x + 50 = 100
x = 50
However, this implies that the bakery sold 50 cupcakes and 50 cookies, which contradicts the information given. Therefore, there is no solution to this system of equations.
System 3: One Solution
Scenario: A bookstore sells paperbacks and hardcovers. The total number of books sold is 90. The paperbacks are sold for $10 each, while the hardcovers are sold for $20 each. The bookstore made a total revenue of $1350.
Let x be the number of paperbacks sold
Let y be the number of hardcovers sold
From the given information, we can set up the following equations:
1) x + y = 90 (Equation for the total number of books sold)
2) 10x + 20y = 1350 (Equation for the total revenue)
To solve this system of equations, we can again use the elimination method:
Multiplying equation 1 by 10, we get: 10x + 10y = 900
Subtracting equation 2 from this new equation, we have:
(10x + 10y) - (10x + 20y) = 900 - 1350
-10y = -450
Simplifying the equation further, we divide both sides by -10:
y = 45
Substituting the value of y back into equation 1:
x + 45 = 90
x = 45
Therefore, the bookstore sold 45 paperbacks and 45 hardcovers. This system of equations has one unique solution.
Can you make it a little bit shorter
System 1 (Infinite Solutions):
Scenario: A movie theater sells regular tickets for $8 and VIP tickets for $15. The theater sold a total of 120 tickets and made $1240 in revenue.
Equations:
1) x + y = 120 (Total number of tickets sold)
2) 8x + 15y = 1240 (Total revenue)
The solution is x = 80 and y = 40. There are 80 regular tickets sold and 40 VIP tickets sold. This system of equations has an infinite number of solutions.
System 2 (No Solution):
Scenario: A bakery sells cupcakes for $2 each and cookies for $4 each. The total number of cupcakes and cookies sold is 100, and the bakery made $300 in revenue.
Equations:
1) x + y = 100 (Total number of cupcakes and cookies sold)
2) 2x + 4y = 300 (Total revenue)
There is no solution to this system of equations.
System 3 (One Solution):
Scenario: A bookstore sells paperbacks for $10 each and hardcovers for $20 each. The total number of books sold is 90, and the bookstore made $1350 in revenue.
Equations:
1) x + y = 90 (Total number of books sold)
2) 10x + 20y = 1350 (Total revenue)
The solution is x = 45 and y = 45. The bookstore sold 45 paperbacks and 45 hardcovers. This system of equations has one unique solution.
can you make it a bit more shorter
System 1 (Infinite Solutions):
Scenario: A movie theater sold 120 tickets for a total revenue of $1240. Regular tickets cost $8 and VIP tickets cost $15.
Equations:
1) x + y = 120
2) 8x + 15y = 1240
Solution: x = 80, y = 40. This system of equations has an infinite number of solutions.
System 2 (No Solution):
Scenario: A bakery sold a total of 100 cupcakes and cookies and made $300 in revenue. Cupcakes cost $2 each and cookies cost $4 each.
Equations:
1) x + y = 100
2) 2x + 4y = 300
This system of equations has no solution.
System 3 (One Solution):
Scenario: A bookstore sold a total of 90 books and made $1350 in revenue. Paperbacks cost $10 each and hardcovers cost $20 each.
Equations:
1) x + y = 90
2) 10x + 20y = 1350
Solution: x = 45, y = 45. This system of equations has one unique solution.