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.
3 answers
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.
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.
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.