Hi wonderful math people. We're going to do solving systems of equation by substitution and elimination. I'm going to use the same equation and show how to solve it in two different methods. The 1st is going to be substitution. What we're going to do with substitution is we're going to take one of the two equations, it doesn't matter which one, and choose one of the two variables to solve for. We usually look for things that have one coefficients. So I'm going to look at this top equation and I'm going to solve for X by just adding the Y to the other side. Once I take one of the two equations and solve for a variable, then I'm going to substitute this into the other equation. So what's going to happen is every time I see that X, I'm going to put in y + 6, so I'm going to have two, and instead of X I'm going to put in y + 6 and then plus three y = 7. When I distribute, we get two y + 12 + 3. Y equals seven or five Y equal -5 Y equal -1. Once I can come up with my Y value, then I just substitute it back in to find my ** was y + 6 or Y is -1 + 6 X is five, so we have a .5 negative one. That means these two lines have to be consistent because there's at least one solution and they're independent of each other. Another way to do this equation or this solving is through elimination. With elimination, I'm going to look to solve for X or for Y and I usually look for things like if one is positive and one is negative. I also want to see if I can get the same coefficients. So in this example, I'm going to multiply that top equation by three. If I multiply by three, then I'm going to get the top equation to have a -3 Y while the bottom equation had a +3 Y. And if I add those two equations together, what happens if I have a -3 Y and a +3 Y right here? Those are going to cancel. So we're going to end up with 5X equaling 25 or X equal 5. Now to get the Yi want to use or to get the Y value, I'm going to use elimination again. This time I'm going to try to get rid of the X's. So I'm going to take that top equation now and multiply by -2. If I multiply by -2, I'm going to get -2 X +2 Y equaling -12. We're going to leave that second equation the way it is, 2X plus three y = 7. Now, when we add these together, the XS are going to cancel because one's a positive 2X and one's a negative 2X and we're going to end up with five Y equaling -5 or Y equal -1. Now, the elimination method is much more powerful when we get into three equations or four equations. So make sure that you can do both methods, but elimination is going to be more powerful in the long run with helping us solve more complex problems. Let's look at a couple others. The next one is going to be Y equal negative two X + 4 and -4 X -2 Y equal -8 substitution. This first equation is already solved for Y, so we're going to put it down here into the other. Every time we see the Y, we're going to put in negative two X + 4. So now if we multiply out negative four X + 4 X -8 equal -8 negative 8 equal -8, this is a true statement, but there are no variables left. If it's a true statement, it means that these two lines are the same line, so we're going to write it in set notation. It's all the points X&Y such that they're on one or the other of the two lines. So I can put either of the two lines right here. It means they're consistent and dependent. If we look at the elimination method, we want to get the X's and Y's to one side. So I'm going to have two X + y equal 4, and then I need to have the variables become the same. Well, if I look at the YS, I'd multiply this bottom equation here by two. If I multiply that by two, I'm going to get 4X plus 2Y equal 8 and then the original we had -4 X -2 Y equal -8. What happens when we add these two equations? Well we get 0 = 0. Zero equals 0 is a true statement, so it means that it's all the points XY such that -4 X -2 Y equal -8, and once again, they're consistent and dependent. The last example is going to be three X -, y equal 4 and -6 X +2 Y equal 5:00. I'm going to take this top equation and solve for Y. So I'm going to get Y equaling three X -, 4. I'm going to substitute it into the second equation. So every time I see AY, I'm going to put in three X -, 4. When I multiply this out and combine like terms negative six X + 6 X -8 equal 5 does -8 equal five? No, not a true statement. So that means there's no solution. And these are in consistent and independent. Graphically, it would mean that these two lines are parallel, so inconsistent independent. If we look over here at this equation, let's say we want to get the X's the same. This time we're going to take that top equation and multiply by two. If we take that top equation and multiply by two, we get six X -, 2 Y equal 8 negative six X + 2 Y equal 5. Does 0 equal 13? No solution, Inconsistent independent. You've been awesome. Have a great day.