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:
Let's consider a real-world scenario where a salesperson is selling two different types of products: Product A and Product B. The salesperson sells the products at different prices and wants to determine how many units of each product they need to sell to achieve a certain revenue goal.
Let:
x = number of units of Product A sold
y = number of units of Product B sold
The salesperson knows that Product A is sold at $10 per unit and Product B is sold at $20 per unit. They want to earn a total revenue of $300. The equations representing this scenario are:
10x + 20y = 300 (Equation 1)
x + y = 25 (Equation 2)
To solve this system of equations:
Multiply Equation 2 by 10: 10x + 10y = 250 (Equation 3)
Now, subtract Equation 3 from Equation 1:
10x + 20y - (10x + 10y) = 300 - 250
10y = 50
y = 5
Substituting y = 5 into Equation 2:
x + 5 = 25
x = 20
Therefore, the solution to this system of equations is:
x = 20, y = 5.
In the real-world scenario, the salesperson needs to sell 20 units of Product A and 5 units of Product B in order to achieve a revenue of $300.
System 2:
Consider a scenario where a manufacturing company produces two types of products: Product X and Product Y. The company has limited resources and wants to optimize their production. They have the following constraints:
Product X requires 4 units of raw material per unit and Product Y requires 6 units of raw material per unit.
The total amount of raw material available is 100 units.
The equations representing this scenario are:
4x + 6y = 100 (Equation 1)
2x + 3y = 50 (Equation 2)
To solve this system of equations, we will multiply Equation 2 by 2 and subtract it from Equation 1:
4x + 6y - (4x + 6y) = 100 - 100
0 = 0
Since the equation simplifies to 0 = 0, the system has infinitely many solutions. This indicates that the constraints are not restrictive enough to determine a unique solution. In this scenario, the manufacturing company can produce any combination of Product X and Product Y as long as the total raw material used does not exceed 100 units.
System 3:
Consider a scenario where a bakery sells two types of pastries: muffins and croissants. The bakery wants to determine the price for each pastry in order to maximize their revenue. They have collected the following information:
On Monday, the bakery sold 30 muffins and 20 croissants, earning $120.
On Tuesday, the bakery sold 20 muffins and 15 croissants, earning $100.
Let:
x = price of a muffin
y = price of a croissant
The equations representing this scenario are:
30x + 20y = 120 (Equation 1)
20x + 15y = 100 (Equation 2)
To solve this system of equations, we can use the substitution method. Solving Equation 2 for x:
x = (100 - 15y) / 20
Substituting x into Equation 1:
30((100 - 15y) / 20) + 20y = 120
Simplifying the equation:
600 - 90y + 20y = 120
-70y = -480
y = 6
Substituting y = 6 into Equation 2:
20x + 15(6) = 100
20x + 90 = 100
20x = 10
x = 0.5
Therefore, the solution to this system of equations is:
x = 0.5, y = 6.
In the real-world scenario, the bakery should price muffins at $0.5 and croissants at $6 in order to maximize their revenue.
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:
Consider a real-world scenario where a coffee shop offers two different sizes of coffee cups: Small and Large. The total cost of a coffee order depends on the number of cups of each size and their respective prices. The equations representing this scenario are:
Let:
x = number of small cups
y = number of large cups
The coffee shop charges $2 for each small cup and $3 for each large cup. They want to earn a total revenue of $16. The equations are:
2x + 3y = 16 (Equation 1)
x + y = 8 (Equation 2)
To solve this system of equations, first multiply Equation 2 by 2:
2x + 2y = 16 (Equation 3)
Next, subtract Equation 3 from Equation 1:
2x + 3y - (2x + 2y) = 16 - 16
y = 0
Substituting y = 0 into Equation 2:
x + 0 = 8
x = 8
Therefore, the solution to this system of equations is:
x = 8, y = 0.
In the real-world scenario, the coffee shop needs to sell 8 small cups and 0 large cups in order to achieve a revenue of $16. This represents a situation with only one solution.
System 2:
Consider a scenario where a manufacturer produces two types of toys: Action Figures and Dolls. The company wants to optimize their production based on the available materials and labor. They have the following constraints:
Producing an Action Figure requires 3 units of material and 2 units of labor, while producing a Doll requires 4 units of material and 3 units of labor.
The company has 10 units of material and 8 units of labor available.
The equations representing this scenario are:
3x + 4y = 10 (Equation 1)
2x + 3y = 8 (Equation 2)
To solve this system of equations, we can use the elimination method. By multiplying Equation 1 by 2 and Equation 2 by 3, we can eliminate the x variable:
6x + 8y = 20 (Equation 3)
6x + 9y = 24 (Equation 4)
Now, subtract Equation 3 from Equation 4:
6x + 9y - (6x + 8y) = 24 - 20
y = 4
Substituting y = 4 into Equation 2:
2x + 3(4) = 8
2x + 12 = 8
2x = -4
x = -2
Therefore, the solution to this system of equations is:
x = -2, y = 4.
In the real-world scenario, the manufacturer needs to produce -2 (or 2 less than zero) Action Figures and 4 Dolls, which is not feasible. Thus, this system of equations has no solution.
System 3:
Consider a scenario where a bookstore sells two types of books: Fiction and Non-Fiction. The bookstore wants to determine the price for each book category to maximize their profit. They have collected the following information:
On Monday, the bookstore sold 20 fiction books and 10 non-fiction books, earning $180.
On Tuesday, the bookstore sold 10 fiction books and 5 non-fiction books, earning $80.
Let:
x = price of a fiction book
y = price of a non-fiction book
The equations representing this scenario are:
20x + 10y = 180 (Equation 1)
10x + 5y = 80 (Equation 2)
To solve this system of equations, we can use the substitution method. Solving Equation 2 for x:
x = (80 - 5y) / 10
Substituting x into Equation 1:
20((80 - 5y) / 10) + 10y = 180
Simplifying the equation:
160 - 10y + 10y = 180
160 = 180
Since the equation simplifies to 160 = 180, which is not true, this system has no solution. In this scenario, the bookstore cannot determine a unique combination of prices for the fiction and non-fiction books to maximize their profit.
Sure! Let's start with a real-world scenario that represents a system of linear equations with one solution:
Real-World Scenario: The cost of admission to a zoo is $5 for children and $10 for adults. A group of people consisting of both children and adults paid a total of $50 for admission. The total number of people in the group is 7. How many children and how many adults are in the group?
Let's assign variables to represent the unknowns:
Let's say 'c' represents the number of children and 'a' represents the number of adults.
From the given scenario, we can create two equations:
Equation 1: The total cost of admission is $50:
5c + 10a = 50
Equation 2: The total number of people in the group is 7:
c + a = 7
To solve this system of equations, we can substitute the value of ācā from Equation 2 into Equation 1:
5c + 10a = 50
5(7 - a) + 10a = 50
35 - 5a + 10a = 50
5a = 50 - 35
5a = 15
a = 15/5
a = 3
Substituting the value of 'a' back into Equation 2:
c + 3 = 7
c = 7 - 3
c = 4
Therefore, there are 4 children and 3 adults in the group.
Now let's move on to a system of linear equations with no solution:
Real-World Scenario: Two friends are selling cookies for a fundraiser. The first friend sells chocolate chip cookies for $2 each, and the second friend sells oatmeal raisin cookies for $3 each. Together, they sold a total of 20 cookies and earned $55. How many of each type of cookie did they sell?
Let's assign variables to represent the unknowns:
Let's say 'c' represents the number of chocolate chip cookies, and 'o' represents the number of oatmeal raisin cookies.
The two equations based on the given scenario are:
Equation 1: The total number of cookies sold is 20:
c + o = 20
Equation 2: The total amount earned from selling the cookies is $55:
2c + 3o = 55
To solve this system of equations, we can use the method of substitution.
From Equation 1, we can rewrite it as c = 20 - o.
Now we substitute this value into Equation 2:
2(20 - o) + 3o = 55
40 - 2o + 3o = 55
40 + o = 55
o = 55 - 40
o = 15
Substituting the value of 'o' back into Equation 1:
c + 15 = 20
c = 20 - 15
c = 5
Therefore, they sold 5 chocolate chip cookies and 15 oatmeal raisin cookies.
Finally, let's consider a system of linear equations with infinite solutions:
Real-World Scenario: You have a bag of coins consisting of only quarters and dimes. The total value of the coins in the bag is $3.75. If there are a total of 15 coins in the bag, how many quarters and dimes are there?
Let's assign variables to represent the unknowns:
Let's say 'q' represents the number of quarters, and 'd' represents the number of dimes.
The two equations based on the given scenario are:
Equation 1: The total value of the coins is $3.75:
0.25q + 0.10d = 3.75
Equation 2: The total number of coins is 15:
q + d = 15
To solve this system of equations, we can use the substitution method.
From Equation 2, we can rewrite it as q = 15 - d.
Now we substitute this value into Equation 1:
0.25(15 - d) + 0.10d = 3.75
3.75 - 0.25d + 0.10d = 3.75
-0.15d = 0
No matter what values we substitute in for 'd', the equation will always hold true since 0 multiplied by any number is still equal to 0. Therefore, this system of equations has infinite solutions.
In this scenario, the number of quarters and dimes can vary as long as their total value is $3.75 and the total number of coins is 15.