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.

Question

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.

Answer 1

System with infinite solutions:

Let's consider the situation at a carnival where the ticket prices for different rides are the same.

Let x represent the cost of the Ferris wheel, and y represent the cost of the roller coaster.

Equation 1: x + y = 10
Equation 2: 2x + 2y = 20

To see that there are infinite solutions in this system, we can divide the second equation by 2:

Equation 2: x + y = 10

We can see that both equations are the same, meaning any value of x and y that satisfies one equation will satisfy the other.

Thus, there are infinite solutions for this system.

System with no solution:
Now let's consider a scenario with two different production facilities each trying to meet certain production quotas. Facility A needs to produce x items, and Facility B needs to produce y items.

Equation 1: 3x + 4y = 50
Equation 2: 3x + 4y = 60

There cannot be a solution to this system because both facilities cannot meet the different quotas at the same time.

System with one solution:

Let's consider a real-world scenario where a farmer wants to plant two different types of crops (x and y) and needs to meet specific land requirements.

Equation 1: 2x + 3y = 100 (represents total acreage)
Equation 2: x - y = 10 (represents difference in acreage between crops x and y)

We can solve for x and y using various methods, such as substitution or elimination, but we'll use elimination here:

First, we'll multiply the second equation by 3, so the y coefficients match:

Equation 2 (modified): 3x - 3y = 30

Now we can add both equations:

Equation 1: 2x + 3y = 100
Equation 2 (modified): 3x - 3y = 30
---------------------------
5x = 130

Now we can solve for x:

x = 130 / 5
x = 26

Now we can substitute x back into either equation to find y. We'll use Equation 2:

26 - y = 10
y = 26 - 10
y = 16

So the unique solution to this system is x = 26 and y = 16, meaning the farmer should plant 26 acres of crop x and 16 acres of crop y.