>> Alright! Here's one that has many solutions. Notice I've already arranged the equation so that the top equation has a leading coefficient of 1 that will help with our calculations a little bit. Now, we'd like to create a situation in the next group of equations, where we have a vacancy in these 2 positions and we don't suspect that anything funny is going to happen at this point. Alright, so what we'll do is we'll play these two off of another. We'll multiplying negative 4 in equation 1 and we're adding equation 2. So, we'll write that down. It's negative 4 in equation 1 and we add equation 2. Alright, this requires high concentration now. Negative 4 times x is negative 4x plus 4x, 0x as expected. Negative 4 times 2y is negative 8y minus 1y is negative 9y. Then negative 4 times equation 1, see our notes are here and we refer back to it every time. So, negative 4 times negative 1 is plus 4z and 5z is 9z and then negative 4 times 5 is negative 20 plus 11 is negative 9. Oh, and I'm noticing here we have 9, 9, 9. I can divide through by 9 and create some efficiency here and I'll do that in the next step. Alright, let's go to this one. We have, we're playing these two off of one another to create a vacancy here. We need to multiply in equation 1 by negative 5 and then add equation 3. So, it's negative 5 times equation 1 plus equation 3 is what we're doing, creating a vacancy here as expected. Then negative 5 times 2y is negative 10y with negative 8y is negative 18y. Then negative 5 times negative 1, that's 5z plus 13z that's 18z. And then negative 5 times 5 is negative 25 plus 7 is negative 18. I'm noticing that I can divide through in equations 2 and 3 in order to create some nice things like leading coefficients of 1. In fact, many coefficients in constants of 1, let's see what happens when I do that. I'm going to take equation 2 and divide it by negative 9. And when I do that divide this by negative 9, I get 1, divide here by negative 9 I get negative 1z and divide this by negative 9 I get 1 -- excuse me, my bad. Okay, and then in the next one I'm dividing in equation 3 I'm dividing by negative 18. Same kind of thing happens, divide here by negative 18, divide here by negative 18, divide here by negative 18 and I get this. Now, I'm noticing that these two equations are identical. Now, let's see what happens if I in the next step I'm just using the same logic pattern that I use before. I'm going to bring down these two equations and decide what to do and what I'm deciding to do is just multiply in equation 2 by negative 1 and add this equation. And when I do and I, you know I'm just changing the signs of these rascals and adding those and the sum will simply be 0 all the way across the board. So, in indicating negative of equation 2 plus equation 3 I get 0 equals 0. Now, here's what we have a true statement after a couple of variables have disappeared from the system. And now we have a system that has an infinite number of solutions. Now, your teacher may require that you only say that you have an infinite number of solutions and you can go on to the next problem, but it's probably the case that you'll need to have some way of generating that infinitude of solutions. And the way to do that is like this. You take this equation and solve it for one of the letters. Now, it's easy here to solve for y. So, if I say that y is equal to z plus 1, then I'm solving for y. Now, you let this letter used in the solution of y. You let that letter take on a value like a, just name it a, some foreign letter a. I'll tell you why in just a minute. Alright, so if z is a, and you see now we have a value for z and then this becomes y equals a plus 1. So, now we have a value for y and since we have a value for z and for y we can find the value of x by using the top equation. Here's the top equation, y is 8 plus 1, z is a, we solve this, we find x to be minus a plus 3. So, if x is this then we have this ordered triplet, x, y, z and a way to represent them we have minus 8 plus 3 for the x and y is 8 plus 1 and z is simply a.