Welcome. In this lesson, you will learn when a system has infinitely many solutions, how to address that and we'll do this by looking at an example. The example is following. Want to solve this system. The system is x plus y, plus z equals to 10 and 2x minus y plus 2z, equals to 8. These two equations are actually the graph of a play in three space. I know that given two planes in three space, there are only two possibilities. Either the planes are parallel and never intersect, so they'll be no solution to this system in that case. Or the planes intersect at a line, so they'll be infinitely many solution points. Let's see which one this is. I'll start off by writing the augmented matrix. If you remember from the elementary row operations, those are the next things I want to do to write this in row echelon or reduced row echelon form. Remember the idea is that entry in the first row, first column. I want it to be the only non-zero entry in that column. I want to put zeros below that 1. To do that, the first step that I have to take, or I don't have to take but I can take, is multiply minus 2 times row one. Add the results to row two, and that'll be my new row two. That gives me the following. Remember I keep the top row to say, this will give me a 0 in that position, a minus 3 here, because I have a minus 2 times y, which is minus 2. Minus 1 gives me a minus 3. In this spot, I'm going to have another 0. This will be a minus 12 over here. That spine is 2 times 10, which is a minus 20 plus 8, gives me a minus 12. The next step I want to do, this now courses is in row echelon form and I can figure out what everything is from here, but I'm going to go here and put it in reduced row echelon form. The next step I would do is where that line is 3 is. I want to have a 1 there. I take a minus one-third times row two, and that'll give me my new row two. That's going to give me, the top row stays the same. The bottom row will be 0, 1, 0 minus 12 divided by the minus 3 is 4. This still is not in reduced row echelon form. Remember for it to be in reduced row echelon form where I have those leading ones, the column that they are in you can only have one non-zero entry and that will be that leading one. Above that 1, I have to have a 0. How do I do that? Well, all I need to do, is take a minus row two, add that to row one and that'll give me my new row one. Well, this stays the same. It's still going to just be a 1 there. I'll have a 0 here because it's a minus 1 plus 1. This all stay a 1. I have a minus 4 plus 10, that gives me 6. The bottom row stays the same. This matrix now is in reduced row echelon form and I can read off the equation from here and it'll tell me really fairly easily what's going on. From the first row, what I get is x plus z equals to 6. From the bottom row, what I get is y equals to 4. This tells me, is that any solution that I look at for this system, the y-coordinate will have to be 4. What I'll do is from the top equation, I can solve for either one or the variables that I want. In this case here, I'll solve for x. This here implies that x equals to 6 minus z. Simply subtract 6 from our subtract z from both sides. Solutions then are points of the form for the first coordinate. We've got to have 6 minus z. For the second coordinate, it has to be 4 because that's what y is. The last coordinate, is going to be z. If you notice z is a free variable, I can let it be anything that I want. My x coordinate the way I have that depends on what z is, and of course, my y has to be 4. This tells me that, for example, if z equals to 1, then we get the point 5, 4, 1 would be a solution to the equation. You can go out and pick several of these. If z equals 0, you see if z is 0, you get 6, 4, 0. That would also be a solution to that system. How you would write the solution set would be the following. Some people are going to write it in set-builder notation. Here's a solution set. It's going to be the set of all ordered triple that are of this form here, 6 minus z, 4, z such that z is any real number. Now, that open face are that I wrote there is a set of all real numbers. Given any point of that form, it will be a solution to that set. Now it is possible that you could work some of these problems, and instead of having one free variable like I have here, it could be that you have 2,3 or 4. Thank you very much and have a wonderful day.