Now try writing and solving your own systems of linear equations: one with infinite solutions, one with no solutions, and one with one solution. Write a real-world scenario that goes with at least one system of equations. Show your work.
Sure! Here are three scenarios with their corresponding systems of linear equations:
1. Scenario - Infinite Solutions:
Suppose you decide to open a coffee shop and offer two sizes of coffee cups, small cups priced at $2 and large cups priced at $3. In one day, you sell a total of 100 cups and collect $250 in revenue.
Let's denote the number of small and large cups sold as 's' and 'l' respectively. The system of linear equations representing this scenario is:
s + l = 100 (Equation 1)
2s + 3l = 250 (Equation 2)
To solve this system, we'll use the substitution method:
From Equation 1, we have l = 100 - s.
Substituting this value of l into Equation 2, we get:
2s + 3(100 - s) = 250
2s + 300 - 3s = 250
-s + 300 = 250
-s = -50
s = 50
Now, we substitute the value of s back into Equation 1 to find l:
50 + l = 100
l = 100 - 50
l = 50
The solution of this system is s = 50 and l = 50. This means that there are infinitely many ways to sell a total of 100 cups and collect $250 in revenue by selling different combinations of the two sizes of coffee cups.
2. Scenario - No Solutions:
Suppose you are hosting a baking contest, and the contestants can bake either cakes or pies. Each cake takes 2 hours to make, while each pie takes 3 hours. The total time available for baking is 8 hours, and the contestants should bake a total of 7 desserts.
Let's represent the number of cakes and pies baked by 'c' and 'p' respectively. The system of linear equations representing this scenario is:
2c + 3p = 7 (Equation 1)
2c + 3p = 8 (Equation 2)
To solve this system, we'll subtract Equation 2 from Equation 1:
(2c + 3p) - (2c + 3p) = 7 - 8
0 = -1
We obtain -1 = 0, which is not a true statement. Hence, there are no valid solutions to this system. In the given scenario, it is not possible for the contestants to bake a total of 7 desserts within the time frame of 8 hours.
3. Scenario - One Solution:
Suppose you want to buy some apples and oranges from a grocery store. The cost of each apple is $0.50, and the cost of each orange is $0.75. You spend a total of $5.50 and buy a total of 8 fruits.
Let's represent the number of apples and oranges bought as 'a' and 'o' respectively. The system of linear equations representing this scenario is:
0.50a + 0.75o = 5.50 (Equation 1)
a + o = 8 (Equation 2)
To solve this system, we'll use the elimination method. Multiply Equation 2 by 0.50 to make the coefficients of 'a' in both equations equal:
0.50(a + o) = 0.50(8)
0.50a + 0.50o = 4 (Equation 3)
Now, subtract Equation 3 from Equation 1:
(0.50a + 0.75o) - (0.50a + 0.50o) = 5.50 - 4
0.75o - 0.50o = 1.50
0.25o = 1.50
o = 6
Substitute this value of 'o' back into Equation 2:
a + 6 = 8
a = 8 - 6
a = 2
The solution of this system is a = 2 and o = 6. Therefore, you bought 2 apples and 6 oranges from the grocery store for a total cost of $5.50.
Sure! Let's start by writing and solving a system of linear equations with one solution.
System 1: One Solution
Equation 1: 3x + 5y = 11
Equation 2: 2x - y = 4
To solve this system, we will use the method of substitution.
First, solve Equation 2 for x:
2x - y = 4
2x = 4 + y
x = (4 + y)/2
Now substitute x in Equation 1 with (4 + y)/2:
3((4 + y)/2) + 5y = 11
(6 + 3y)/2 + 5y = 11
6 + 3y + 10y = 22
13y = 16
y = 16/13
Now substitute y back into Equation 2 to find x:
2x - (16/13) = 4
2x - 16/13 = 4
2x = 4 + 16/13
2x = (52 + 16)/13
2x = 68/13
x = 34/13
Therefore, the system of equations has one solution: x = 34/13 and y = 16/13.
Now let's move on to the system of equations with infinite solutions.
System 2: Infinite Solutions
Equation 1: 2x + 3y = 5
Equation 2: 4x + 6y = 10
To determine if the system has infinite solutions, we can see if one equation is a multiple or a combination of the other equation.
Let's multiply Equation 1 by 2:
4x + 6y = 10
We can see that Equation 2 is indeed a multiple of Equation 1, meaning these equations represent the same line. Therefore, the system of equations has infinite solutions.
Lastly, let's look at a system of linear equations with no solutions.
System 3: No Solution
Equation 1: 3x + 2y = 7
Equation 2: 6x + 4y = 12
To determine if the system has no solutions, let's first divide Equation 2 by 2:
3x + 2y = 6
We can see that Equation 1 and the modified Equation 2 represent parallel lines as they have the same slope but different y-intercepts. Thus, these lines will never intersect, indicating that the system of equations has no solutions.
Real-world Scenario:
Suppose you are managing a manufacturing company that produces two products: Product A and Product B. The company has two machines: Machine 1 and Machine 2. Each machine has a certain production capacity per hour.
Let's write a system of linear equations representing this scenario.
Equation 1: 4A + 2B = 120 (represents the production capacity of Machine 1)
Equation 2: 3A + 5B = 160 (represents the production capacity of Machine 2)
In this scenario, Equation 1 represents the maximum number of units of Product A and B that Machine 1 can produce in an hour, while Equation 2 represents the same for Machine 2. The goal is to find how many units of Product A and B can be produced using both machines within their given capacities.
Sure! Let's work through each scenario with systems of linear equations.
1. System with Infinite Solutions:
Consider the following set of equations:
Equation 1: 2x + 3y = 7
Equation 2: 4x + 6y = 14
To determine if this system has infinite solutions, we will check if the two equations are dependent (i.e., one equation is a multiple of the other equation). We can do this by comparing the ratios of the coefficients. Dividing Equation 2 by 2, we find:
Equation 2 (divided by 2): 2x + 3y = 7
We can observe that Equation 2 is, in fact, a multiple of Equation 1. As a result, the two equations represent the same line and intersect at infinitely many points, having infinite solutions. Graphically, this scenario would illustrate two overlapping lines.
2. System with No Solutions:
Consider the following set of equations:
Equation 1: 3x + 2y = 5
Equation 2: 6x + 4y = 12
To check if this system has no solutions, we will examine if the equations are inconsistent (i.e., parallel lines that do not intersect). We can do this by comparing the ratios of the coefficients. Dividing Equation 2 by 2, we find:
Equation 2 (divided by 2): 3x + 2y = 6
By comparing the coefficients, we can see that the slopes of the lines represented by the two equations are equal. However, the right-hand sides of the equations (constants) are different. Therefore, when we graph the system, the two lines will be parallel and not intersect, indicating no solution.
3. System with One Solution:
Consider the following set of equations:
Equation 1: 2x + 3y = 7
Equation 2: x - y = 1
To check if this system has a single solution, we can use a variety of methods, such as substitution or elimination. We'll use the elimination method here:
Multiply Equation 2 by 2 to make the coefficients of x's in both equations equal:
Equation 2 (multiplied by 2): 2x - 2y = 2
Now we subtract Equation 1 from Equation 2:
(2x - 2y) - (2x + 3y) = 2 - 7
-5y = -5
y = 1
Substituting the value of y back into Equation 2, we find:
x - 1 = 1
x = 2
So, the solution to this system of equations is x = 2, y = 1. This scenario can be related to a real-world situation where you have the cost of purchasing x apples and y oranges, with each equation representing the total cost and the known relationship between the prices.
I hope this explanation helps you understand how to solve systems of linear equations and how the different scenarios arise based on the coefficients and constant terms.