Now try writing and solving your own systems of linear equations one 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 of linear equations with one solution:
Scenario: Bob is a farmer who sells apples and oranges. Each day, he counts the number of apples and oranges he sells and records the total amount of money he earns. On Monday, Bob sold 6 apples and 4 oranges for a total of $10. On Tuesday, he sold 3 apples and 5 oranges for a total of $8. We can set up the following system of equations to represent this scenario:
Equation 1: 6a + 4o = 10 (where 'a' represents the price of each apple and 'o' represents the price of each orange)
Equation 2: 3a + 5o = 8
To solve this system of equations, we can use either substitution or elimination method. Let's use the elimination method:
Multiply Equation 1 by 5 and Equation 2 by 4 to make the coefficients of 'o' equal:
Equation 1: 30a + 20o = 50
Equation 2: 12a + 20o = 32
Subtract Equation 2 from Equation 1:
(30a + 20o) - (12a + 20o) = 50 - 32
18a = 18
a = 1
Now substitute the value of 'a' into either Equation 1 or Equation 2:
6(1) + 4o = 10
6 + 4o = 10
4o = 4
o = 1
Therefore, the solution to this system of equations is a = 1 and o = 1. This means each apple costs $1 and each orange also costs $1. Bob sells each apple and orange for $1 each.
System of linear equations with no solution:
Scenario: John and Sarah are planning a trip to the zoo. They are deciding between two types of tickets: adult tickets which cost $30 each, and child tickets which cost $20 each. John wants to buy 2 adult tickets and 3 child tickets, while Sarah wants to buy 4 adult tickets and 5 child tickets. They have a total budget of $150. We can set up the following system of equations to represent this scenario:
Equation 1: 2a + 3c = 150 (where 'a' represents the price of each adult ticket and 'c' represents the price of each child ticket)
Equation 2: 4a + 5c = 150
To solve this system of equations, let's use the elimination method:
Multiply Equation 1 by 4 and Equation 2 by 2 to make the coefficients of 'a' equal:
Equation 1: 8a + 12c = 600
Equation 2: 8a + 10c = 300
Subtract Equation 2 from Equation 1:
(8a + 12c) - (8a + 10c) = 600 - 300
2c = 300
c = 150
Now substitute the value of 'c' into either Equation 1 or Equation 2:
2a + 3(150) = 150
2a + 450 = 150
2a = -300
a = -150
Since 'a' represents the price of each ticket, it cannot be negative. Therefore, there is no solution to this system of equations. John and Sarah cannot buy tickets that satisfy their requirements and their total budget.
System of linear equations with infinite solutions:
Scenario: Mike is a car rental manager who offers two types of vehicles for rent: sedans and SUVs. He charges $40 per day for sedans and $50 per day for SUVs. On one particular day, he rented out a total of 5 vehicles and earned a total of $220. However, he forgot to keep track of how many sedans and SUVs he rented out individually. We can set up the following system of equations to represent this scenario:
Equation 1: s + u = 5 (where 's' represents the number of sedans and 'u' represents the number of SUVs)
Equation 2: 40s + 50u = 220
To solve this system of equations, let's use the elimination method:
Multiply Equation 1 by 40 and subtract Equation 1 from Equation 2:
40s + 40u = 200
40s + 50u = 220
10u = 20
u = 2
Now substitute the value of 'u' into Equation 1:
s + 2 = 5
s = 3
Therefore, the solution to this system of equations is s = 3 and u = 2. However, since the problem stated that Mike forgot how many sedans and SUVs were rented, there can be any number of sedans and SUVs that satisfy the total number of vehicles rented and the total earnings of $220. Hence, there are infinite possible solutions in this scenario.