Hello wonderful mathematics people. This is Anna Cox from Kellogg Community College Systems of equations. Consistent means there's at least one solution and inconsistent means there's no solution. Parametric equation is used for infinitely many solutions and there will be more than one correct answer. For a parametric equation. If we have five X -, 6 Y equal 1 and six X - 5 Y equal 10, we want to get rid of one or the other of the variables. So I'm going to start by multiplying this top equation through by 5. So we'd get 25 X -30 Y equaling 5. If I thought about multiplying the second equation through by a -6, we'd get -30 six X + 30 Y equaling -60. The logic there is I want to have coefficients that are additive inverses so that when I add one of the variables, they'll cancel. So when we add these two equations together, we can see we end up with -11 X equaling -55 or X equal 5. If we know X, we can substitute it back into either of the two equations in order to find our Y. We get 25 -, 1 equaling 6Y or Y is 4. The fact that this is a single point means that these are consistent equations, and these are actually equations of lines, so they're independent also. Now what we mean by independent is that the two equations don't necessarily depend on each other. If this is one line and we know the solution, the other line could be here or here or here or here. We don't necessarily know just by one equation and the solution exactly where the other line would be. If we look at this next example, if we multiply the top one through by a three, when we get 12 X -6 Y equaling 12, we multiply the bottom one through by a -2 we'd get -12 X plus 6Y equaling -14. When we add this, we get 0 equal -2. This is going to be a no solution. There is no number that I can stick into both equations to get a true statement because zero doesn't ever equal -2 so it's inconsistent. And graphically, what this means is we have parallel lines, so they're also independent. Just because I know one line, and I know that the solution is that there isn't any, it doesn't necessarily tell me I know where the other parallel line would be. Our last type is going to be if they're the same line. If I multiply this top one through by two six X -, 12 Y equaling 24, and I multiply the bottom one through by a -3 negative six X + 12 Y equaling -24. Now when I add these, we get 0 = 0. Zero equals zero is always true. So it tells me it's consistent. And now because I know one line and I know that there's infinitely many solutions, but they're all on that one line, I absolutely do know where the other line has to be. It has to be right on top of the first one. They're called dependent. Now we're going to write these solution answers as parametric. And to do a parametric, we're going to say, let's let Y be some variable T If Y is T, our ordered pair is going to be T for the Y portion. And we'd have three X -, 6 T equaling 12 or 3X would equal six t + 12, or X would equal two t + 4. So this parametric .2 T plus four T is absolutely a possibility. But recall that I said a moment ago that we might have more than one possibility. So instead, what if we let X equal some variable, let's call it M? Then we would have three X - 6 or three M Sorry, three m - 6 Y equaling 12. Negative 6Y would equal 12 + 3. MY would equal -3 halves m - 2. So a different ordered pair that would also be correct would be M, -3 halves m - 2. Assuming I did my math right and I actually didn't, that should have been a - 3 M That's why it was at the end. So a negative divided by a negative is a +1 half m - 2. How about that? So we'd get M, one half m -, 2 as another solution. So both of these are correct answers. They're just depending on where you choose to put your parametric, whether we let the single variable be in the Y location or in the X location. Now two by twos, 2 variables, 2 equations are pretty easy. We're going to look at 3 equations with three unknowns next. So when we look at this one, our first thought process ought to be what variable we do we want to get rid of. And I traditionally look for things that have one coefficients or -1 coefficients. And some of the equations have positives and some have negatives. So in this example, I'm going to choose to get rid of the Z and we need to get rid of the same variable all the way throughout. So if I leave this top equation exactly like it is, and I add together equations one and two, let's label these, we'll call it 1. Oops. We'll call it 1-2 and three. So if I take 1 and add it directly to two, one and two, add them together, I'm going to get a new equation that says 5X plus 12Y. No Z's are going to equal 2341. Now for the next equation I want to get rid of the ZS again and so I'm going to take equation 1 and multiply it by two and then add it to equation 3. So equation 1 * 2. So 3 * 2 is 6/6 plus one is going to give me 7 XS. 2 * 5 is 1010 + 7 is going to give me 17 YS, 2 * -1 is -2 + 2 is going to make that cancel, 2 * 13 is 2626 and 32 is going to give me 58. Now we want to get rid of the YS next. So if we look at this equation, seven X + 17 Y equal 58 and five X + 12 Y equaling 41. I want to choose these two equations because they both have XS and YS in them. And just because I want to keep my numbers small, I actually might choose to get rid of the XS this time. So if I'm going to get rid of the XS, I'm going to have 3X plus five y -, C equal 13. I'm going to multiply this new equation by 7, so I would get 35X plus 84 Y equaling 287 if I multiplied the second equation through by 5. Actually, let's do -5 I'd get -35 X -85 Y equaling -290. So then if I just added the bottom two, I could have 3X plus five y -, C equal 13 if I left the five X + 12 Y equaling 41 because that's just the second equation divided back by 7. But now if I added these two equations together at the bottom, I would get all the X's to cancel. I'd get a -1 Y equaling -3. Or we could see that this bottom one would give me Y equal 3 if I know Yi can then substitute back to get X. So five X + 36 equal 41 subtract 36 so X is going to equal 1 and if I have X&YI can find my Z. So 3 + 15 -, C equal 13 so 1518 subtract 18 to the other side. We can see Z as 5 S With this, we have a single .13 and five for the intersection of three variables with three equations. It's consistent because there's at least one solution. This also tells us that three variables are like a plane, so the planes would intersect in a single point. Let's look and see what happens if we have an inconsistent and or the same or infinitely many solutions. So on this next one, I'm going to look at the variables and I think I'm going to solve for Z again because I've got some that are positive, some that are negative, etcetera. Now I can rewrite the equations in any order I want and I might choose, well, now we'll just leave them in the order they are X -, 3 Y +2 Z equals 6. So now we're going to take this equation 2 and multiply it by two and then add it to equation 1. So here, here's 123. So if I multiply the second equation 3 by 2 and add it, I'm going to end up with 3X, multiply by two, eight, and -3 is going to give me 5 Y -2 Z and +2 Z is going to have the Z's cancel. 4 * 2 is 8 + 6 is 14. So then we're going to take the. I think I'll just add two and three actually directly like they are. So if we add two and three, we can see that the ZS will cancel. So we'd get six X + 10 Y equaling 24. Now when we look at this bottom equation, we could see that everything is divisible by a two. So if we rewrote this, we'd have X -, 3 Y +2 Z equals 6 three X + 5 Y equaling 14. Now, if I divided that last equation all through by two three X + 5 Y equaling 12, if I combine these bottom 2 equations, let's say we're just going to subtract equation 3 from 2. So if we subtract those two bottom equations, three X -, 3 X is going to give me 0X's, five y -, 5 Y is going to give me zero Y's. So the left hand side's really going to give me zero. But if I have 14 take away 12, we're going to get 2. This bottom statement is not true. So this is a no solution. And these are inconsistent. So these three planes do not intersect all three of them in anyone location. Our next example is going to be an example of infinitely many. So if we look at this, when we look at our variable to get rid of, I think this time I'm going to simplify the X's because there's lower coefficients up there. So if we combine one and two, I'm going to multiply 2 by a -2 and add it to 1. So 2X plus three y + 7 Z equaling 15. If I multiply equation 2 by -2 and add the XS are going to cancel. We get a negative eight y + 3 Y so -5 Y -2 Z plus 7Z5 Z -40 + 15 so -25. Now I could see that that's all divisible by A5. So as I'm still working with that equation, I think I'll go ahead and do that. I like having my coefficients as small as possible. So we get a negative y + Z equaling -5 if we combine our two and our three equation. Let's take two and subtract equation 3 maybe. So X -, X is 02 Y four, y - 2, Y is 2 yz -3 Z is -2 Z 20 -, 10 is 10. Now that new equations all divisible by two, so I'd get y -, Z equaling 5. Now if we combine these two bottom equations exactly like they are, so we're going to leave that 2X plus three y + 7 Z equal 15, we're going to have negative y + Z equal -5. But if we take equation 2 and add it to equation 3, the new equation 2 and new equation 3, we can see that 0 = 0. And this is a true statement. So these are going to be consistent. Now we're going to answer this as a parametric. So we're going to let our T be our Z. If Z = t, then we can see that negative y + t equal -5 or negative Y equaling negative t - 5 or Y is just t + 5. Now if we know our Y and our Z, we're going to substitute it in and we're going to have two X + 3 * t + 5 + 7 * t equaling 15. So 2X plus 3T plus 15 + 7 T equaling 15 2X is going to equal -10 T or X is going to equal -5 T. So our ordered pair, as far as a parametric is going to be -5 T, t + 5 and T. So for any value of T, if it has this relationship for X and our Y and our Z points, it will be a solution for the original ordered pair or ordered triplet, the original 3 equations. So there's infinitely many solutions, and now we have the relationship between those. Our last example is going to involve having been given an equation that's going to relate our first and 2nd derivative to some given equation with initial conditions. So if we know that Y of X equals AE to the three X + b E to the 7X, we can find our first derivative. Actually, let's go back a minute. If we know that our initial condition of Y 0 equal 15, we know that the Y is 15 every time the X is 0. So E to the 3 * 0 + b, E to the 7 * 0, E to the 0 powers really just one. So we end up with an equation that says 15 equal A+B. Now if we go back and we take the first derivative, we get Y prime equaling 3AE to the three X + 7 BE to the 7X and we have our initial condition here of the first derivative as 13 whenever our X is 0. So we get a new equation that says 13 equal 3A plus 7B. We want to solve for our A and our B. So we can take this top equation up here and multiply by -3 all the way through. If we multiply by -3, we get -45 equaling -3 A -3 B. Combining those we would get 4B equaling -32. So B would equal -8. If we know that B is -8. Now we can go back and we can substitute in to find our A. So 15 equal a + -8 so our A is 23. So now we actually have an equation and our equation says Y equal 23 E to the three X -, 8 E to the 7X. So we can find our first derivative 23 * 3 is 69 E to the three X -, 56 E to the 7X. We can find our second derivative. Our second derivative 69 * 3 is going to be 207 E to the three X 56 * 7 is going to give us 392 E to the 7X. And if we go back to the original, it said Y double prime. So this Y double prime -10 Y primes. Plus 21Y S should equal 0. So when we compute all this out, we can see 207 E to the three X -, 392 E to the seven X -, 690 E to the three X + 560 E to the seven X + 21 * 23. Yeah, I don't know that one off the top of my head. Twenty 36483 if I did it right, really quick E to the three X -, 168 E to the 7X equaling 0. So if we Add all this up, our two O 7 E to the 3X our -690 and our +483 three and seven is 0 carry one eight and one is 9/4 and two is 6. So that definitely positives and negatives balance -392 positive 560, negative 168. When we combine those, we get 34560 and -560 so that checks. So that was what that equation was asking us to find the A's and the B's so that the original initial condition was valid. Thank you and have a wonderful day.