00:01
For this problem, we're going to let x represent the cost of a rose, y represent the cost of a carnation, and z represent the cost of a tulip.
00:07
And we're going to use these variables to figure out how much of each of these items cost, based off of the three bouquets of flowers that are ordered at a florist.
00:14
We had three roses, two carnations, and one tulip, and that cost was $14.
00:21
We also had an order or another bouquet order that had six roses, two carnations again, and six tulips, and that cost.
00:32
And our last bouquet had just one rose, 12 carnations, and then one tulip, and that cost was $18.
00:41
So the way we're going to solve this is by trying to use elimination or substitution to get rid of the same variable twice.
00:49
And i'm going to use elimination.
00:51
So the easiest elimination to start would be eliminating z's, or one of the easiest eliminations, would be eliminating z's from the first and third equations.
00:59
And i can do that by distributing a negative to my top equation, and that would give me negative 3x minus 2y minus z equals negative 14, and i can combine these bottom two equations to get a new equation that just has x and y.
01:15
So x minus 3x would be negative 2x, 12y minus 2y would be 10y, the zs eliminate, and 18 minus 14 would be 4.
01:23
So that's one new equation that has x and y.
01:26
If we can come up with another combination that also eliminates z, then we can set those to the side and solve that as a system to itself, and then worry about the last variable.
01:36
So i'm going to try to get rid of z again.
01:39
This time i'm going to multiply to try to get rid of z between the top two equations, which means i'll have to multiply the top equation, not only just by a negative, but also a 6...