System of linear equations
System of linear equations can arise naturally from many real life examples. Generally speaking, those problems come up when there are two unknowns or variables to solve.
In the figure above, there are two variables to solve and they are x and y. This kind of system is called system of linear equations with 2 variables.
Real-life situations that lead to a system of linear equations
Example #1
The sum of two numbers is twenty and their difference is ten. What are the two numbers?
Here is how to set up the system:
Let x be the first number
Let y be the second number
Then,
x + y = 20
x − y = 10
You can also write:
x + y = 20
y − x = -10
You will get the same answers!
Example #2
You have 24 coins in your pockets that are worth 4.50 dollars. How many coins are quarters? How many coins are dimes?
Here is how to set it up:
Let q be the number of quarters.
Let d be the number of dimes.
Then,
q + d = 24
25 × q + 10 × d = 450
The second equation is tricky. How did we get it?
Since 1 quarter equal to 25 cents, q quarter equal to 25 × q.
If you had 6 quarters and you wanted to know how many cents are there for the 6 quarters, would you not do 6 × 25?
Just say to yourself that now instead of 6 quarters you have q quarters. Does that make sense?
In a similar way, since 1 dime equal to 10 cents, d dimes equal 10 × d.
What about the 450? 4.50 dollars times 100 = 450 cents.
Finally, since 25 × q represents how many cents you have for quarters and 10 × d represents how many cents you have for dimes, adding them should equal to the total of 450 cents.
Example #3
A cell phone plan offers 300 free minutes for a flat fee of 20 dollars. If your usage exceed 300 minutes, you pay 50 cents for each minute.
A second cell phone plan offers 500 free minutes for a flat fee of 30 dollars. If your usage exceed 400 minutes, you pay 25 cents for each minute.
Model the cost of both plan with a system.
Here is how to set up the system:
Let x be the number of minutes you talk.
Let y be the cost.
y = 20 + 0.50 (x - 300)
y = 30 + 0.25 (x - 500)
y = 20 + 0.50x - 150
y = 30 + 0.25x - 125
Subtract 0.50x from both sides of the first equation and subtract 0.25x from both sides of the second equation.
We get:
y − 0.50 x = 20 - 150
y − 0.30 x = 30 - 125
y − 0.50 x = -130
y − 0.30 x = -95
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