00:01
This is a problem that involves a real world scenario and it will involve working with the concept of solving a system of linear equations.
00:13
In this problem, we are going to work with solving a system of three linear equations with three variables.
00:23
We will use the method of elimination to solve this system.
00:28
Now what this problem involves is three purchases at a restaurant for hamburgers, order of fries, and a cola.
00:44
Okay, so we are given information about the cost of each purchase and the items that were purchased.
00:52
So we have one, two, three purchases with the items purchased and the cost of each purchase.
00:58
And our job is going to be to find the price of each item.
01:03
As with any application or word problem, you should identify your verbles and tell what each one represents.
01:11
So for this problem, since we are dealing with hamburgers, french fries, and colas, i'm going to use verbals that are easily identified with each one of those items.
01:23
So i'm going to let each stand for the price of one hamburger.
01:39
I'm going to let f stand for the price of an order of fries, and i will let c stand for the price of a coli.
02:00
So after you identify your variables, then you use the information in the problem to write your equations.
02:08
Let's take the first purchase.
02:11
There are eight hamburgers, six orders of fries, and six colas that were purchased for $26 .10.
02:18
Cents so i'm going to take eight times the price of a hamburger plus six times the price of order fries plus six times the price of a cola and set that equal to the cost of that purchase twenty six dollars and ten cents now i'll do the same thing for the second purchase there's ten hamburgers six orders of fries and eight colas and that cost $31 and 60 cents and then for the third purchase we have three hamburgers two orders of fries and four colors for a total of ten dollars and 95 cents so each of the three equations are in standard form and i want to use elimination to solve to help us reference these equations i'm call this equation 1, equation 2, and equation 3.
03:37
When you solve with elimination, you need to have a variable in which the coefficients of that variable are opposites.
03:48
So when i look at these equations, if i look at equation 1 and equation 2, the coefficients of f are both positive 6.
04:02
I can get these to become opposites very easily if i will take equation 1 and multiply every term in it by a negative 1.
04:15
Now there are other ways to have done this problem but that's what hit me.
04:18
So i'm going to take equation 1 and i will multiply every term in it by negative 1 and when i put that with equation 2 i will be able to eliminate the variable f.
04:32
So i've taken equations 1 and 2 and i'm going to go through the process eliminate the variable f.
04:39
So when i multiply equation 1 by negative 1, i will have a negative 8h minus 6f minus 6c will equal a negative $26 .10.
04:58
I'll leave equation 2 as it is.
05:02
So it's 10h, 6f, 8c, 8c, is 31 60 and i add these two equations together.
05:18
So it's 2h i eliminate my f's plus 2c is equal to 550.
05:29
The next step would be to pick a different pair of equations and eliminate the same variable f so what i'm going to do now is i'm going to use equation one and equation three and i have to eliminate the from those two...