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.
1. System of linear equations with infinite solutions:
Let's consider a scenario where two friends, Alice and Bob, are collecting marbles. Alice has twice as many marbles as Bob. Let x be the number of marbles Alice has, and y be the number of marbles Bob has. We can write the following equations:
x = 2y (1)
Now let's say that the total number of marbles they have combined is T. So,
x + y = T (2)
We can substitute (1) into (2) to eliminate x:
2y + y = T
3y = T
Since T can be any value, there are infinite solutions to this system.
2. System of linear equations with no solution:
Let's consider a scenario where two people are trying to divide a pizza between them. Anna wants 3 slices and Chris wants 4 slices. Each slice is of equal size. However, the pizza was accidentally sliced into unequal pieces: some pieces are larger than others. Let x represent the larger pieces and y represent the smaller pieces. We can write the following equations:
x + y = 3 (Anna's slices)
x + y = 4 (Chris's slices)
This system has no solution, as the two equations contradict each other – there cannot be two different total numbers of slices.
3. System of linear equations with one solution:
Let's consider a scenario where a store is selling two kinds of fruit: apples and oranges. Apples cost $2 each and oranges cost $3 each. A customer buys a combination of apples and oranges, and the total cost is $17. Let x be the number of apples and y be the number of oranges. We can write the following equations:
2x + 3y = 17 (1)
Now let's say that the customer buys a total of 5 fruits. So,
x + y = 5 (2)
We can solve this system to find the number of apples (x) and oranges (y) that the customer purchased. We can rearrange (2) to isolate x:
x = 5 - y
Substitute this into (1) to eliminate x:
2(5-y) + 3y = 17
10 - 2y + 3y = 17
y = 7
Now that we have the value of y, we can find the value of x:
x = 5 - 7 = -2
There is an error in this system because there cannot be a negative number of apples. The scenario and system of equations are relevant, but the numerical values chosen led to an incorrect solution. To fix this, we can change the total cost to $19:
2x + 3y = 19 (1')
x + y = 5 (2)
Substitute x = 5-y into (1'):
2(5-y) + 3y = 19
10 - 2y + 3y = 19
y = 9
Now we can find the value of x:
x = 5 - 9 = -4
Again, this is an error. Let's now change the quantity equation to:
x + y = 7 (2')
x = 7 - y
Substitute into (1'):
2(7-y) + 3y = 19
14 - 2y + 3y = 19
y = 5
Now we can find the value of x:
x = 7 - 5 = 2
Finally, the correct system with one solution is:
2x + 3y = 19
x + y = 7
The customer purchased 2 apples and 5 oranges.