Hello. I'm Professor Von Schmohawk and welcome to Why U.
When we first introduced systems of linear equations
we initially studied systems of two linear equations in two variables.
And we saw how these systems could be used to model and solve real-world problems.
We then looked at systems of three linear equations in three variables.
But what type of real-world problems can these systems model and solve?
In the lecture entitled "Using Systems of Equations versus One Equation"
we were presented with a problem
involving two groups of people buying movie tickets and drinks.
The first group spent 69 dollars on four movie tickets and three drinks
and the second group spent 88 dollars on five movie tickets and four drinks.
To calculate the price of a ticket and the price of a drink
we created a system of two equations in two variables
letting t represent the ticket price
and d represent the price of a drink.
By solving this system,we discovered that drinks cost 7 dollars each
and tickets cost 12 dollars each.
With a system of three linear equations in three variables
we could model the same problem with three groups of people
who buy three different items at a different theater
say, tickets, drinks, and popcorn.
In this example, we could choose the variables t, d, and p
to represent the price of a ticket
a drink
and popcorn.
Using the elimination method to solve this system
we start by picking a variable to eliminate
using any two equations which contain that variable.
For example, we can choose to eliminate the variable d and use the first two equations.
To eliminate d, we first multiply the second equation by negative one
and then add the equations.
This gives us the equation
t + 3p = 28 .
Now we eliminate d again using a different pair of equations.
For example, this time we could choose the first and third equations.
Using these equations to eliminate d
we multiply the first equation by five and the third equation by negative three.
We then complete the arithmetic
and add the equations
giving us 8t + 16p = 176
or dividing by eight
t + 2p = 22 .
This system of two equations in two variables
can now be solved to find the values of t and p.
Multiplying the bottom equation by negative one
and adding the equations
we get p = 6 .
Now we can substitute 6 for p in either equation to find the value of t.
Substituting 6 into the top equation
gives us t + 18 = 28
or subtracting 18 from both sides
we get t = 10 .
We can now substitute these values for t and p into any of the original equations
and solve for the remaining variable, d.
Choosing the first equation
we complete the arithmetic
and combine like terms
to get 3d + 70 = 85 .
And subtracting 70 from both sides
gives us 3d = 15
or dividing both sides by 3
d = 5 .
So the price of a drink is 5 dollars
the price of a ticket is 10 dollars
and the price of popcorn is 6 dollars.
In this problem, we had three unknown quantities
the price of a ticket, the price of a drink, and the price of popcorn
and we were given three mathematical relationships involving these unknowns.
Since there were at least as many independent equations as unknowns
it was possible to pin down specific values for each quantity.
Together, these three values formed the single unique solution to this system of equations.
In the next lecture, we will see how a system of linear equations in three variables
can be used to solve a real-world problem described by an equation which is not linear.