Now try writing and solving your own systems of linear equations: one with infinite solutions, one with no solution, and one with one solution.

Question

Write a real-world scenario that goes with at least one system of equations.
Show your work.

Answer 1

1) System with Infinite Solutions:
Equations:
2x + 3y = 10
4x + 6y = 20

To solve this system, we can divide the second equation by 2 to get the same equation as the first one:
2x + 3y = 10
2x + 3y = 10

There are infinite solutions to this system because the two equations represent the same line. Any point on this line will satisfy both equations.

2) System with No Solution:
Equations:
3x - y = 5
6x - 2y = 10

To solve this system, we can divide the second equation by 2 to get:
3x - y = 5
3x - y = 5

Subtracting the two equations, we get 0 = 0. This means that the two equations are representing parallel lines and will never intersect, hence no solution.

3) System with One Solution:
Equations:
2x + y = 7
x - y = 1

To solve this system, we can add the two equations:
3x = 8
x = 8/3

Substitute x back into one of the original equations:
2(8/3) + y = 7
16/3 + y = 7
y = 5/3

Therefore, the solution to this system is x = 8/3 and y = 5/3.

Real-world Scenario:
A company sells two types of products, Product A and Product B. The company sold a total of 100 units of the two products combined, generating a revenue of $500. Each unit of Product A sells for $2 and each unit of Product B sells for $3.
Let x be the number of units of Product A sold and y be the number of units of Product B sold.

Equations:
x + y = 100 (total units sold)
2x + 3y = 500 (total revenue generated)

This system of equations can be solved to find how many units of each product were sold.