WEBVTT mathematics/pre-calculus/selhorst-jones
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Hi--welcome back to Educator.com.
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Today, we are going to talk about systems of linear equations.
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A **linear equation** in two variables is something of the form Ax + By = C, where A, B, and C are all constant numbers.
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So, all of the capital letters that we will be seeing in this represent constant values.
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Notice that each of the variables is just a power of 1: we have x¹, effectively, y¹, effectively...
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x without an exponent is the same thing as just saying x to the power of 1.
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So, since each of these variables has a power of 1, that is similar to a linear polynomial, which has a degree of 1.
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And so, that is where we get this idea of calling it a linear equation--because everything has a degree of 1.
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It is not really a degree, technically, because it is not a polynomial; but there is sort of this parallel idea going on.
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All right, here are some examples: 3x + 2y = 4 and x - 5y = 7.
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Both of these would be linear equations, because they are just some constant times a variable,
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plus some other constant times a variable, until we equal, finally, some constant at the end.
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A **system of equations** is a group of multiple equations that are all true at the same time.
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So, we have this equation here and this equation here, and we know that they are both going to be true at the same time.
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For example, the two equations above make up a system of linear equations.
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We put these two things together with each other, and we have a system of linear equations.
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The solution is going to be some pair of x and y, two numbers where one is going to be x, and the other one is going to be y.
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And we plug them both in, and it will satisfy both equations at the same time.
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It is some x and y that is going to be true here and here--in both of our equations--at the same time.
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That is why we are fulfilling a system of equations, as opposed to just one equation.
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Let's talk about graphs to really get into this.
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First, long ago, when we first discussed the concept of a graph, we saw that there were two ways to look at it.
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We can talk about how input x is transformed into output y.
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We can think that, if we plug in x at 1, then y is going to come out at 1, as well.
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If we plug in 2, y is going to come out as 4 (2²).
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If we plug in 3, y is going to come out as 9, because 3² is 9, and so on and so on.
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We could also go backwards: 0, -1...all of that sort of thing.
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But alternatively, we can also think of it as the location of all of the points that make the equation true--where is this going to be true?
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The reason why (0,0) is a point on our graph is because, if we plug that in, 0 = 0²--that checks out.
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If we try plugging in the point (1,1), this point here, if we plug that in, we have 1 = 1²; that checks out.
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If we try plugging in the point (2,4), then we have 4 = 2², and that checks out.
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All of the points on this parabola--the reason why these points are on the parabola,
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the reason why this graph is these points, is because those points make the equation that makes that graph true.
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Any other point--if we were to choose some other point on it, like (2,1)--if we plugged that into our equation, we would have 1 = 2².
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That is not true; this is not a true statement, so this point here does not exist; it is not on our graph.
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So, the location of all of the points that make the equation true is a great way of thinking about this.
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When working with systems of equations, it is useful to think of graphs in this second way.
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It will help us understand the connection between graphs and solutions.
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This way is still useful; it is a great way, in general, whenever we have to graph something.
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But for what we are doing right here with the systems of equations, this second way of thinking of the points that are true
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is going to really help us understand what is going on.
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We first want to put all of the equations from the system into a form that we can easily graph when we are going to graph this.
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So, if we want to graph a system of equations, we first want to change these things that are kind of hard, like 3x + 2y = 4.
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That is not very easy to graph immediately; so we want y = stuff, where that "stuff" is going to involve our x inside of it.
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y =...with our x over there: we can graph that easily--we are used to graphing lines like that all the time.
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So, we have 3x + 2y = 4; we can convert 3x + 2y = 4; this converts into this, -3/2x + 2.
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You can work that out; we can get this as 2y = -3x + 4 by subtracting 3x from both sides,
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and then we divide both sides by 2, and we get -3/2x + 4.
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And that ends up giving us that line right there; that is what we get from this equation, so our red one is this one.
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Similarly, we can solve x - 5y = 7 for y to make it more easy to graph; and that ends up giving us this.
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We have both of these graphed together.
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Now, what does this tell us about a solution to the system?
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Where the graphs intersect, we have a solution to the system; where they intersect, right here, we have a solution.
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Why is that a solution? Well, what we are looking for is...we are looking for a point, some (x,y), that is going to be true here and true here.
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And since this and this are just equivalent to these two equations up here,
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if we can find some (x,y) that is true here and true here,
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then we have found something that is going to be the answer to our system of equations.
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Well, any point that is on the graph of an equation is going to make that equation true.
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So, what we have found here is that this point here is a point that is on both graphs.
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Since it is on both graphs, it must be true for both equations.
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So, since it is on both graphs, it is true for both equations; so we have just found the solution at that point of intersection.
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With this idea in mind, we can see that there are three possibilities for solutions to a system of linear equations.
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The first one is **independent**: we call it independent if we have just one solution--they intersect in just one place.
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Notice that they have to have different slopes for them to intersect in just one place.
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The next one is **inconsistent**: we say that a system of equations is inconsistent if they never intersect each other.
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Notice that they have the same slope, but different intercepts.
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They are parallel lines, but they don't intersect; they are parallel, and they never touch each other in any way.
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They are not on top of each other; they are just parallel.
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The system is inconsistent because, if one equation is true, the other equation can't be true.
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There is no point that will be on both of the graphs; thus, there is no point that is going to fulfill both equations at the same time.
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And finally, a **dependent** system is a system where the two lines are just completely on top of each other.
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They are collinear--they are the same line.
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So, in that case, they have the same slope (they are parallel), but they also have the same intercept; so they are actually just plain the same line.
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So, in the case where it is dependent, any point on either of them is going to be a solution,
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because if it is a point on one of them, it is a point on the other one, as well.
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So, that gives us a total of three possibilities for how many solutions a linear system can have:
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one solution in the case where it is independent; zero in the case where it is inconsistent
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(they are parallel--they never touch); or infinitely many in the case where it is a dependent system,
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where it is just the same line on top of itself, where anything that is going to be on it is going to be an answer to both equations.
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Now, notice: I want to point out that, just because it is dependent--just because there are infinitely many solutions--doesn't mean any point is a solution.
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If we consider that point there, it is not on either of our lines, so it is not going to be one of them.
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So, a dependent system doesn't mean that every point is a solution.
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What it means is that all of the points on the line are going to be solutions, because they are both just the same line.
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So, if you find a point on one of the lines, you know it is going to work for the entire system, because it is on both of the lines.
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But that doesn't mean that any point whatsoever is going to be true for both of your equations, for all of your entire system.
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All right, the first method to talk about finding that point--where is that point of intersection located?"--is through substitution.
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Substitution works by plugging in one variable for another; then we solve the resulting equation.
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For example, consider if we have 3x + 2y = 4 and x - 5y = 7.
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We want to begin by solving one equation for one variable in terms of the other.
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We want to be able to plug in y for x, or x for y.
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So, we have to get either y on its own or x on its own.
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Now, either equation and either variable will work; we are happy plugging in x, just as we are happy plugging in y.
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So, just choose whichever one looks easiest.
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Now, to me, it seems really easy to get x on its own, because all we have to do is add 5y to both sides.
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So, we have x - 5y = 7; we add 5y to both sides, and we get x = 5y + 7.
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At this point, we are in a great position, because we have x here; we have x here.
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We can swap out those x's, and we can get an equation that uses just y.
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So, we substitute that into the other equation; we have x = 5y + 7 here; we have 3x + 2y = 4; so we swap out the x;
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and now we have 3 times the quantity--because remember, we have to plug it in as a quantity,
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because it is not just 3 times 5y; it is not just 3 times 7; it is 3 times all of 5y + 7, so we put it in parentheses--
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3 times the quantity (5y + 7), plus 2y, equals 4.
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We simplify that a bit: 3 times 5y is 15y, plus 2y gets us 17y; and 3 times 7 gets us 21.
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So, we have 17y + 21 = 4; subtract 21 on both sides and divide by 17, and finally we get to the answer y = -1.
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Great; how do we get the x?--the same thing--we just substitute our new y-value in for y here or y here; either one will work.
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And look, we actually have this really great thing to use: we have x = 5y + 7.
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So, we have already figured out something where, as soon as you plug in y, you will get x as soon as you just do a little bit of arithmetic.
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So, let's plug it into this one, because we made this equation from other equations we already knew.
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And it is going to be the easiest way to get x, because we won't really have to do any algebra; we will just have to do some basic arithmetic.
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So, we will plug into this equation here.
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We swap out the y that we know here for -1; we have 5(-1) + 7.
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We simplify that: x = -5 + 7; so we get x = 2, and at this point, we have found the point.
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We have found the pair (x,y) that is going to solve both of these equations.
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We know that the point (2,-1) solves both of the equations above; great.
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Of course, the example on the previous slide was an independent system.
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What we just looked at was an independent system of linear equations, because it only had one solution.
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In the end, we only got (2,-1); we didn't get other solutions; we didn't get 0 solutions; so it was an independent system--they had different slopes.
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How do you tell if a system is inconsistent (no solutions) or dependent (infinitely many solutions)?
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Well, in the case where it is inconsistent (that is, it has no solutions), if a system has no solutions,
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you eventually get to a nonsense equation, some sort of equation that has to be clearly false--it can't be true--
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something like 0 = 5 or -8 = 42 or √2 = 2; all of these things are completely ridiculous.
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They are absurd: you can't have 0 equal to 5--we just know that that is not true.
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So, what that tells us is that the only way the system can be solved is if something impossible is true.
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But you can't have something impossible be true--it is impossible.
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So therefore, the only other possibility is that the system has no solutions.
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There are no solutions if we end up getting some statement that is clearly false--something ridiculous like 0 = 1 or 5 = 20.
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At that point, we say, "Oh, it must be impossible for this thing to be true, because we are getting impossible statements out of it."
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So, we know that we have an inconsistent system, a system that has no solutions.
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On the other hand, we could also have something that is a dependent system, something that has infinitely many solutions.
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In that case, if we have a dependent system, one with infinitely many solutions,
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you are eventually going to arrive at an equation that will always be true--
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things that are going to be like 0 = 0, -47 = -47, π = π.
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No matter what we plug in there...there is nothing to plug in: -47 = -47...they are both constants; it is always true.
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It is just a true statement; so since these statements are always true, just as the statement is always true, the system is always true.
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So, we see that the system is always true, meaning that we have infinitely many solutions.
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And remember: I want to point out what I talked about before.
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That doesn't mean that any point whatsoever will solve the system.
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It just means that any point on one of the equations is going to be true for the other equation, as well.
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So, if you find a point on one of the lines, since the other line is just that same line, it is going to be true for the second equation, as well.
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We have infinitely many solutions; we have that entire line of solutions.
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You can also show this by showing that the two equations are equivalent.
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And we will see an example for inconsistent; that will be Example 2 in this lesson.
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And we will see a dependent system in Example 3.
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So, if you are confused about that, you want to watch those specifically; check those out--Example 2 and Example 3.
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We will explore that--we will see how that is the case.
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All right, there is another way to solve, though: we don't have to use substitution.
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We can also use a method called **elimination**, where we eliminate one of the variables from the equation.
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Doing this, we add multiples of one equation to the other, which allows us to eliminate variables.
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So, we are able to multiply one of the equations by a number, any constant we want.
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And then, we can add that result to the other equation, and that will allow us to eliminate variables.
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For example, if we have 3x + 2y = 4 and x - 5y = 7, well, we see that I have an x here; I have a 3x here;
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well, if we could get a -3x to show up and then add that, it would cancel it out.
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So, we multiply x - 5y = 7 by -3; we multiply both the left side and the right side--we multiply the whole equation.
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That will come out to be -3x + 15y = -21; and we know that that is equal, because x - 5y = 7 is equal;
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so if we multiply both sides by -3, by algebra, we know that we still have an equal thing.
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So, -3x + 15y = -21; now, we bring these down: 3x + 2y = 4...bring it down here;
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and we add -3x + 15y = -21 to that; we add those together.
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3x and -3x cancel each other out; we have 2y and 15y--they come out to be 17y; 4 + -21 comes out to be -17.
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We divide both sides by 17, and we have y = -1; great.
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And remember: that is the same thing that we got when we were working through substitution.
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So that is good, because both of our methods should give us the same answer.
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At this point, since we have y, we can solve for x by substitution with the other equations.
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We can plug this y = -1 in here or here, and we would be able to figure out what the answer is, just by good, old-fashioned substitution.
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Alternatively, we could do it by further elimination.
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If we know y = -1, then we see over here that there is 2y.
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Well, we could multiply this by -2, and we get -2y = +2, because it is -1 times -2.
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So, we have -2y = +2; we can add that over here, and we have 3x + 2y, what we started with, equals 4.
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And we will add the thing that we just created to that, the multiple we just created: -2y = +2.
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Add that on both; these cancel each other out; we have 3x = 6; divide by 3 on both sides; x = 2, so we end up getting (2,-1).
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All right, both of our values are there; that is the exact same answer that we got through substitution, so it looks like elimination works.
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Cool; just like with substitution, you can tell how many solutions this system has by the same things.
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Independent (that means it has one solution): the system will solve in a normal fashion.
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You will end up just getting values for the variables, and there will be only one solution.
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If it is inconsistent (it has no solutions whatsoever), then when you are solving,
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you will get an impossible equation--something like 5 = 8 that just can never be true.
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And then finally, dependent (infinite solutions, where anything on one of the lines is going to be on the other line,
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so the entire line is going to be answers): while solving,
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you get an always-true equation--something that is just automatically true, like 7 = 7.
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At this point, we understand how to use elimination--we can multiply one of the equations by some constant number,
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and then add that to the other one and eliminate variables.
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But we might wonder why it works--what gives us the ability to add whole equations together.
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Why are we allowed to do this--why does this work?
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OK, imagine you had a simple equation, like something as simple as x = y; we just started out with that.
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Now, if we wanted to, we could add 2 to each side: x + 2 = y + 2, if x = y--just basic algebra.
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But we could also say, "2 is equal to 1 + 1, and I feel like adding 2 on one side and 1 + 1 on the other."
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Now, 2 = 1 + 1; they are the same thing; what algebra says is that you have to do the same thing to both sides.
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If we add the same thing on each side, we still have equality.
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So, if we add 2 on one side and 1 + 1 on the other side, we are still adding the same thing on both sides, so there is nothing wrong with doing that.
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We can add this equation: we can add 2 = 1 + 1, because the important thing is that 2 and 1 + 1 are the same thing.
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Since 1 + 1 becomes 2, that is just another way of expressing 2, and that is what an equation really is.
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We can express it this way, or we can express it this way; both the left-hand side and the right-hand side are the same thing--they are equivalent.
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So, we can take x = y here, and we can add 2 = 1 + 1 to that, and we have x + 2 = y + 1 + 1; it makes sense.
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Cool; if we wanted to, we could also start by multiplying 2 = 1 + 1 by some number on both sides.
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We start, and we have some constant multiple; so we multiply by 3 on both sides, and we get 6 = 3 + 3--distribution on the right side.
00:18:57.400 --> 00:19:04.400
By algebra, we know that this equation is equal, too, since we just multiplied the left side and the right side by the same number, by this 3.
00:19:04.400 --> 00:19:10.600
So, we know that 6 does equal 3 + 3, because we just did algebra on it by multiplying both sides by the same number.
00:19:10.600 --> 00:19:17.500
So, we can add it to x = y by the same logic: we have x = y as what we started with, and we add 6 = 3 + 3,
00:19:17.500 --> 00:19:23.100
because we just figured out that that is true, as well; and we will get x + 6 = y + 3 + 3.
00:19:23.100 --> 00:19:27.100
Great; the method of elimination is working by the exact same reasoning.
00:19:27.100 --> 00:19:34.200
The way we add an equation to both sides is because, since it is an equation, it has equality.
00:19:34.200 --> 00:19:42.500
We know that the left side and the right side are the same thing, so that means that we are adding the same thing to both sides.
00:19:42.500 --> 00:19:46.000
What we are adding on the left side is the same thing as what we are adding to the right side.
00:19:46.000 --> 00:19:51.900
They might look different, but we know, because of that equality, that the left side and the right side are the same thing.
00:19:51.900 --> 00:19:54.800
They are equivalent, so we can add them to both sides.
00:19:54.800 --> 00:20:00.200
And this allows us to have that nice process of elimination, where we can just knock out variables.
00:20:00.200 --> 00:20:07.000
All right, solving by a graphing calculator is the final thing we will talk about for ways to solve systems of linear equations.
00:20:07.000 --> 00:20:13.500
Earlier, we talked about how where the graphs of a system intersect--wherever we have intersection--we have solutions.
00:20:13.500 --> 00:20:15.900
Now, normally that isn't very useful in practice.
00:20:15.900 --> 00:20:18.700
It is a great theory for helping us understand how this works.
00:20:18.700 --> 00:20:26.600
We can either cross once, cross never, or just be on top of each other (one solution/zero solutions/infinite solutions).
00:20:26.600 --> 00:20:32.600
And so, that is really useful as a theory thing, but in practice, it is a pain to graph in a really accurate way.
00:20:32.600 --> 00:20:36.700
If we wanted to find solutions from graphing--if we wanted to use graphing--
00:20:36.700 --> 00:20:40.600
we would have to spend so much time making accurate graphs
00:20:40.600 --> 00:20:43.700
that we would be better off just doing the system directly.
00:20:43.700 --> 00:20:48.200
We have substitution; we have elimination; those are just direct methods that will get us the answer.
00:20:48.200 --> 00:20:52.400
So, we can just use those methods if we want to figure out what the answer is, exactly.
00:20:52.400 --> 00:20:56.400
If we need accuracy, it takes a long time to make a really accurate graph.
00:20:56.400 --> 00:20:59.400
So, we would be better off solving it directly if we are doing it by hand.
00:20:59.400 --> 00:21:05.100
However, it is possible to get around this if you have a graphing calculator.
00:21:05.100 --> 00:21:12.200
If you have a graphing calculator, a graphing calculator allows you to graph both equations very quickly and very accurately.
00:21:12.200 --> 00:21:16.100
Then, there is a way on the graphing calculator to find accurate intersections.
00:21:16.100 --> 00:21:22.700
You can analyze the graph, and it will tell you that these two lines, these two curves, intersect at such-and-such a point.
00:21:22.700 --> 00:21:26.800
That allows you to immediately find the solution; you graph one of them; you graph the other one.
00:21:26.800 --> 00:21:36.800
You tell your graphing calculator to look for the intersection point, and it gives out some number, and there is your point that solves the system of equations.
00:21:36.800 --> 00:21:40.200
If you want more information on this--if this sounds interesting--if you have a graphing calculator,
00:21:40.200 --> 00:21:44.100
or if you are interested in buying a graphing calculator, there is an entire appendix on graphing calculators.
00:21:44.100 --> 00:21:48.200
It has lots of useful information; and even if you don't own an actual graphing calculator,
00:21:48.200 --> 00:21:53.000
a physical calculator, there are lots of programs that will work on computers, like the one you are probably using right now,
00:21:53.000 --> 00:21:57.900
or tablets or smartphones, which you might be...whatever you are watching this on right now,
00:21:57.900 --> 00:22:02.400
there is a way to be able to use graphing programs on that, as well.
00:22:02.400 --> 00:22:06.900
And so, you can go and look those up; we have talked about that in the appendix on graphing calculators, as well.
00:22:06.900 --> 00:22:12.600
There are lots of great things out there; you can probably be able to do this, even without having to buy a graphing calculator.
00:22:12.600 --> 00:22:18.400
Now, of course, I want to caution you: you have to show your work in most math classes.
00:22:18.400 --> 00:22:22.300
So, if you are taking a math class, you still have to learn these other methods of solving,
00:22:22.300 --> 00:22:28.300
because it is not enough to say, "The calculator said so"--you have to be able to show--definitely prove--that this is the case.
00:22:28.300 --> 00:22:30.900
We can't just say, "This piece of technology said it."
00:22:30.900 --> 00:22:34.700
And also, you don't want to become reliant on your calculator, even if you are not taking any tests.
00:22:34.700 --> 00:22:41.700
Even if you are not doing this and taking homework or anything else, if it is just for you, you want to understand what is going on.
00:22:41.700 --> 00:22:45.900
And while the graphing calculator is a useful tool, you get a lot out of being able to do this,
00:22:45.900 --> 00:22:50.300
and understand what is going on through substitution and elimination, because that will show up in other things in math--
00:22:50.300 --> 00:22:55.300
whereas this graphing calculator is really useful, but it is just a trick for finding answers to this one thing.
00:22:55.300 --> 00:22:59.900
So, you don't want to become reliant on your calculator, and you are probably going to have to be able to do this yourself for math class.
00:22:59.900 --> 00:23:02.300
So, you have to know the other methods, as well.
00:23:02.300 --> 00:23:07.400
Nonetheless, it is great for checking your answers, and when you don't need to show work.
00:23:07.400 --> 00:23:11.400
So, if you don't need to show work (like if you are taking a multiple choice exam,
00:23:11.400 --> 00:23:15.100
and you are allowed to use your graphing calculator), or if you just want to check your answers
00:23:15.100 --> 00:23:18.100
after you have written the whole thing out by hand and shown your work,
00:23:18.100 --> 00:23:21.700
it is a really great way to be able to do it quickly and accurately and move on.
00:23:21.700 --> 00:23:26.800
All right, we can also expand this idea of a linear equation to more than two variables.
00:23:26.800 --> 00:23:35.200
So far, we have just talked about systems that are x and y, or maybe a and b, or k and l--just two different variables.
00:23:35.200 --> 00:23:39.100
But we could also have more; we could have more variables, like three variables.
00:23:39.100 --> 00:23:43.800
For example, we could have A, B, C, and D--all of these capital letters--be constants;
00:23:43.800 --> 00:23:51.300
and then we could have a three-variable linear equation that had A times x, B times y, C times z, and then finally all equal to some constant.
00:23:51.300 --> 00:23:55.100
We could take it even further to four variables, where the capital letters are still constants:
00:23:55.100 --> 00:24:00.400
A times x, B times y, C times z, D times w, equals some constant E;
00:24:00.400 --> 00:24:07.500
or any arbitrary number, n, of variables, where each of our A-sub-some-number is going to be some constant,
00:24:07.500 --> 00:24:11.000
and x-sub-some number is going to be each of our named variables.
00:24:11.000 --> 00:24:15.100
So, it would be A₁ times x₁ + A₂ times x₂ +...
00:24:15.100 --> 00:24:19.400
up until we get to A<font size="-6">n</font> times x<font size="-6">n</font> = some constant number, A₀.
00:24:19.400 --> 00:24:21.800
So, the idea is that we can just have as many variables as we want;
00:24:21.800 --> 00:24:24.900
and as long as they are just being added together and multiplied by some constant,
00:24:24.900 --> 00:24:31.000
and equal some constant at the end, in this nice linear form, we still have a linear equation.
00:24:31.000 --> 00:24:35.800
We can solve a system of such equations if we have at least the same number of equations as variables.
00:24:35.800 --> 00:24:40.700
If you have the same number of equations as you do variables--if you have a 3-variable system,
00:24:40.700 --> 00:24:42.500
and you have 3 equations--you are good to solve it.
00:24:42.500 --> 00:24:47.700
If you have a 4-variable system, and you have 4 equations, you are good to solve it.
00:24:47.700 --> 00:24:53.400
If you have a 57-variable system, and you have 57 equations, you are good to solve it.
00:24:53.400 --> 00:24:56.400
So, as long as you have that, you can work through it.
00:24:56.400 --> 00:25:00.400
We solve these systems in pretty much the exact same manner as when we did two variables.
00:25:00.400 --> 00:25:03.100
We can use substitution; we can use elimination.
00:25:03.100 --> 00:25:09.000
Graphing gets really difficult, because it is difficult to visualize this stuff in higher than two dimensions.
00:25:09.000 --> 00:25:13.600
We can visualize three dimensions to some extent; it is hard to write it on paper, since paper is two-dimensional.
00:25:13.600 --> 00:25:16.500
But we can visualize it, because we are used to living in a 3-D world.
00:25:16.500 --> 00:25:22.800
But once you talk about a four-dimensional or five-dimensional system, we can't really visualize four dimensions or five dimensions.
00:25:22.800 --> 00:25:25.000
How do you add an extra dimension to space?
00:25:25.000 --> 00:25:32.300
So, living in three-dimensional space, we are not used to thinking in more than three dimensions, so we can't really graph in it.
00:25:32.300 --> 00:25:37.200
Graphing can be hard or just simply impossible; but substitution and elimination still work great.
00:25:37.200 --> 00:25:44.900
So, if you encounter a system of equations that has more than two variables, you can still use substitution; you can still use elimination.
00:25:44.900 --> 00:25:51.300
And we will see that when we get to Example 5: we will see a three-variable system that we will work through.
00:25:51.300 --> 00:25:52.900
All right, we are ready for some examples.
00:25:52.900 --> 00:25:59.800
The first example: Solve the system of equations, if it is possible: 3x + 4y = 6, 2x + y = -1.
00:25:59.800 --> 00:26:05.900
Let's try both substitution and elimination; we will work through it with substitution first.
00:26:05.900 --> 00:26:11.800
If we are working through it with substitution, to me it looks like 2x + y = -1 is easy to get just the y alone.
00:26:11.800 --> 00:26:15.900
So, we subtract by 2 on both sides; we have y = -2x - 1.
00:26:15.900 --> 00:26:24.400
At this point, we can plug that in here; we have 3x (switch to a new color); we bring that down; 3x + 4,
00:26:24.400 --> 00:26:40.600
times what we substitute in, -2x - 1 = 6; work that out: 3x - 8x - 4 = 6; 3x - 8x...we simplify that to -5x;
00:26:40.600 --> 00:26:47.600
we add 4 to both sides: -5x = 10; divide by -5 on both sides; x = -2.
00:26:47.600 --> 00:26:51.200
10 divided by -5 becomes -2; great.
00:26:51.200 --> 00:27:00.200
We have what our x is equal to; and then, to figure out what our y is equal to, we already built this nice equation, y = -2x - 1.
00:27:00.200 --> 00:27:01.900
We will switch colors again at this point.
00:27:01.900 --> 00:27:13.900
y = -2x - 1; we can take this and swap it out here; y = -2 times...what was x?...-2, minus 1.
00:27:13.900 --> 00:27:19.900
That is -2(-2) is positive 4, minus 1 is 3; so y = 3.
00:27:19.900 --> 00:27:27.600
So, the point for this comes out to be (-2,3); great.
00:27:27.600 --> 00:27:34.100
Alternatively, we could do this through the method of elimination.
00:27:34.100 --> 00:27:39.500
At this point, we are looking for how we can get these things to mesh nicely so that we can knock out some things.
00:27:39.500 --> 00:27:43.600
So, 2x and 3x...we are going to have to multiply both of the equations to do it.
00:27:43.600 --> 00:27:48.800
But y...we can easily multiply y by -4, and we will have brought it up to the +4 here.
00:27:48.800 --> 00:27:59.200
So, let's start with 3x + 4y = 6; and then we are going to take our 2x + y = -1,
00:27:59.200 --> 00:28:05.600
and we want to knock out that +4, so we are going to multiply everything by -4.
00:28:05.600 --> 00:28:18.100
We multiply the whole thing by -4--everything in that equation; that comes out to be -8x - 4y =...-1 times -4 is +4.
00:28:18.100 --> 00:28:26.600
At this point, we bring this down; we will add that together; we have -8x - 4y = positive 4.
00:28:26.600 --> 00:28:36.400
The positive 4y and the -4y cancel each other out; 3x - 8x becomes -5x; that equals 10; x = -2.
00:28:36.400 --> 00:28:43.000
At this point, we can either plug x = -2 into 2x + y = -1, or 3x + 4y = 6.
00:28:43.000 --> 00:28:46.800
We could use substitution, or we could just continue on our way with more elimination.
00:28:46.800 --> 00:29:03.800
So, let's bring down 2x + y = -1; and notice: if we multiply x = -2 by -2 on both sides, we will have -2x =...-2 times -2 is positive 4.
00:29:03.800 --> 00:29:11.100
We add this whole thing together, 2x - 2x; they cancel; we have y = positive 3, and there are our two answers.
00:29:11.100 --> 00:29:14.400
And so, we get the same thing, (-2,3).
00:29:14.400 --> 00:29:17.400
If you wanted to check this, you could plug it into both equations.
00:29:17.400 --> 00:29:23.600
You would have to plug it into both equations, because you can't be sure that you solved both of them, unless you know it works in both of them.
00:29:23.600 --> 00:29:29.800
But you can also, if you are running low on time, check just one of them; and that will probably help you figure out whether or not you got it right.
00:29:29.800 --> 00:29:46.900
So, 2x + y = -1; we can plug in 2(-2) + 3 = -1; -4 + 3 = -1; and sure enough, that checks out: -4 + 3 = -1.
00:29:46.900 --> 00:30:00.200
We can do the same thing with 3x + 4y; 3(-2) + 4(3) = 6; that gets us -6 + 12 = 6; sure enough, that checks out, as well.
00:30:00.200 --> 00:30:08.200
So, we have performed our check; we know for sure that (-2,3); x = -2, y = 3; is the solution to this system.
00:30:08.200 --> 00:30:10.400
So, checking your work is great if you have the time.
00:30:10.400 --> 00:30:12.900
If you are on an exam, it is really nice to be able to check your work.
00:30:12.900 --> 00:30:18.700
So, I recommend checking your work whenever you have the time, if it is an exam and it really matters that you get this right--
00:30:18.700 --> 00:30:23.100
or if it is an important problem, and you are engineering something, and you can't let it fail.
00:30:23.100 --> 00:30:31.800
All right, the second example: Solve the system of equations, if possible: -a + 3b = 6; -3a + 9b = -27.
00:30:31.800 --> 00:30:35.600
Once again, let's see it both through substitution and through elimination.
00:30:35.600 --> 00:30:49.400
Over here, we will do substitution; -a + 3b = 6--let's add a to both sides: 3b = 6 + a, a + 6; and subtract 6 on both sides, so 3b - 6 = a.
00:30:49.400 --> 00:30:54.500
That is the same thing; we can plug this into -3a + 9b.
00:30:54.500 --> 00:31:05.600
We have -3 times a, which is the same thing as 3b - 6, plus 9b, equals -27.
00:31:05.600 --> 00:31:18.600
-9b...-3(-6) gets us + 18...+ 9b = -27; our 9b and our -9b cancel out, and we get 18 = -27.
00:31:18.600 --> 00:31:21.500
That is impossible--we can never have this be true.
00:31:21.500 --> 00:31:30.600
So, since 18 cannot be equal to -27, we know that there are no solutions to this.
00:31:30.600 --> 00:31:35.000
Similarly, we can work through this with elimination.
00:31:35.000 --> 00:31:42.900
Working through this with the process of elimination, we see -a + 3b, so we can multiply -a + 3b by -3 on both sides.
00:31:42.900 --> 00:32:01.400
So, -a + 3b = 6; we can multiply that by -3 on both sides, and that will end up giving us positive 3a - 9b = -18.
00:32:01.400 --> 00:32:11.000
-3a + 9b = -27; and then, we are going to add what we just figured out is a multiple of -a + 3b = 6.
00:32:11.000 --> 00:32:15.200
We have that; we multiply it by -3 on both sides.
00:32:15.200 --> 00:32:25.200
And we get positive 3a - 9b = -18; we add those two together (method of elimination): -3a + 3a knock each other out;
00:32:25.200 --> 00:32:34.100
9b + -9b knock each other out; -27 + -18 gets us -45.
00:32:34.100 --> 00:32:40.500
So, we have 0 on the left-hand side, because we have nothing left over there; 0 = -45--that is impossible.
00:32:40.500 --> 00:32:48.400
We can never have that be the case, so we have no solutions--this can never be true; great.
00:32:48.400 --> 00:32:56.300
Alternatively, one other way to be able to see what is going on is: we can show that these two things are very similar.
00:32:56.300 --> 00:33:04.700
Notice: -a + 3b = 6, and -3a + 9b = -27.
00:33:04.700 --> 00:33:18.000
Well, notice that if we just multiply this one by 3, then the left-hand side is going to be the same thing as the left-hand side to our other equation.
00:33:18.000 --> 00:33:28.000
You have -3a + 9b = 18; so notice, -3a + 9b matches up to -3a + 9b.
00:33:28.000 --> 00:33:35.300
So, the left-hand side is the same, but the right-hand sides are totally different.
00:33:35.300 --> 00:33:44.600
In one world, in one equation, -3a + 9b = 18; in the other world, our other equation, -3a + 9b, the same thing, is equal to -27.
00:33:44.600 --> 00:33:48.300
So, in one world, we have a totally different meaning than the other world.
00:33:48.300 --> 00:33:54.700
Thus, the worlds can never match up; it means that our lines never touch--we have no solutions, based on this.
00:33:54.700 --> 00:34:00.600
So, one other way to be able to see what is going on is the fact that we can make the left-hand sides equal,
00:34:00.600 --> 00:34:04.100
but the right-hand sides won't be equal, even when the left-hand sides are equal.
00:34:04.100 --> 00:34:08.800
Therefore, the left-hand sides of the equations are the same thing; the right-hand sides will be different.
00:34:08.800 --> 00:34:13.000
Therefore, it must be never possible for the two things to meet.
00:34:13.000 --> 00:34:21.200
Solve the system of equations, if possible: -3/7p + 2/7q = -6/7 and 9p - 6q = 18.
00:34:21.200 --> 00:34:27.700
The first thing I would do is say, "Wow, 7...divided by 7...divided by 7...divided by 7; I don't like fractions."
00:34:27.700 --> 00:34:32.700
So, what I am going to do is just multiply the whole thing by 7.
00:34:32.700 --> 00:34:42.400
At that point, we have -3p + 2q = -6; it is easier to work from this point on, I think.
00:34:42.400 --> 00:34:53.000
So now, let's do substitution first: -3p + 2q = -6, so we can work out...let's go with q.
00:34:53.000 --> 00:35:05.500
2q = 3p - 6; divide both sides by 2; we have 3/2p; 6 divided by 2 becomes 3, so 3/2p - 3.
00:35:05.500 --> 00:35:26.900
At this point, we have 9p - 6q = 18; we can substitute in our q there: 9p - 6 times what we swap out for q, (3/2p - 3), equals 18.
00:35:26.900 --> 00:35:38.700
Simplify this: 9p - 6(3/2)...half of our 6 goes away, so we can break this into 3 times 2: 6 breaks into 3 times 2;
00:35:38.700 --> 00:35:46.800
so that knocks out the 2 down here, and we have 3 times 3, or 9p, here; and -6 times -3 got a little confusing,
00:35:46.800 --> 00:35:50.300
because I was distributing, but it didn't cancel out both of them...sorry about that;
00:35:50.300 --> 00:35:57.200
I hope that it wasn't too confusing...-6 times -3 gets us positive 18 equals 18.
00:35:57.200 --> 00:36:05.100
9p - 9p means that we have 18 = 18; and that is always true, so we have infinite solutions.
00:36:05.100 --> 00:36:18.400
We have a dependent system: infinite solutions, as long as one of these equations is true, so let's go with 9p - 6q = 18.
00:36:18.400 --> 00:36:22.600
So, as long as one of our equations is true, we know that the entire system is going to work,
00:36:22.600 --> 00:36:26.000
because we know, from what we just figured out, that they are actually just the same line.
00:36:26.000 --> 00:36:28.000
They make the same linear thing.
00:36:28.000 --> 00:36:31.700
All right, we can also do this through elimination.
00:36:31.700 --> 00:36:41.400
I think this way is actually easier; once we get to this point, the -3p + 2q = -6, we see that -3p + 2q...we can eliminate the p's pretty easily.
00:36:41.400 --> 00:36:52.500
So, we multiply this one, -3p + 2q = 6, by 3 on both sides: 9p - 6q = 18.
00:36:52.500 --> 00:37:04.500
And then, we multiply -3p + 2q = -6 by 3 on both, so we get -9p; 2 times -3 becomes -6q; -6 times -3 becomes -18.
00:37:04.500 --> 00:37:11.600
We add those together, and we get that these cancel; these cancel; these cancel; 0 = 0.
00:37:11.600 --> 00:37:17.900
And that always ends up working out; so we know that we have infinite solutions over here, as well.
00:37:17.900 --> 00:37:21.600
It always ends up being the case; great.
00:37:21.600 --> 00:37:28.900
One last way to see what is going on here: we can actually show that these two things are just the same equation.
00:37:28.900 --> 00:37:42.800
If you have -3p + 2q = -6, well, if we want, we can just multiply both sides by 3, at which point we have -9p + 6q = -18.
00:37:42.800 --> 00:37:55.300
At this point, we multiply both sides by -1; we have 9p - 6q (because we multiplied by -1) = +18.
00:37:55.300 --> 00:38:04.000
Oh, that is what we started with here; the equations are equivalent.
00:38:04.000 --> 00:38:09.000
They are just the same equation, put in different ways of phrasing it.
00:38:09.000 --> 00:38:13.500
This equation here is the exact same thing as this equation here.
00:38:13.500 --> 00:38:17.700
The left equation and the right equation--the two equations in our system--are just the same equation.
00:38:17.700 --> 00:38:21.800
So, if the equations are equivalent, that means we have infinitely many solutions,
00:38:21.800 --> 00:38:25.400
because anything that fulfills one of the equations will fulfill the other.
00:38:25.400 --> 00:38:29.000
Now once again, remember: that doesn't mean that all points are true.
00:38:29.000 --> 00:38:37.800
But anything on the line created by 9p - 6q = 18, which is the same thing as the line -3/7p + 2/7q = -6/7,
00:38:37.800 --> 00:38:44.700
which is the same thing as the line -3p + 2q = -6--since they are all equivalent equations, they make the same line.
00:38:44.700 --> 00:38:50.300
So, anything that fulfills one of the equations fulfills both of the equations.
00:38:50.300 --> 00:38:56.800
We have infinitely many solutions; we have an entire line of solutions--not the whole plane, but a line of solutions.
00:38:56.800 --> 00:38:59.900
All right, let's try a word problem here.
00:38:59.900 --> 00:39:04.300
You need to make 100 milliliters of 22% acid solution for chemistry.
00:39:04.300 --> 00:39:09.000
However, the lab only has 10% and 50% solutions available.
00:39:09.000 --> 00:39:13.900
How much of each should you use to arrive at the desired acid solution, without wasting any?
00:39:13.900 --> 00:39:23.700
We are going to have to mix 10% solution acid and 50% solution acid together to finally make a 22% acid solution with a quantity of 100 milliliters.
00:39:23.700 --> 00:39:30.700
So, the first thing: we have to mix some quantity of the low acid and some quantity of the high acid, so let's name those things.
00:39:30.700 --> 00:39:43.600
We have l, which will be the name for the amount, which will be the number of milliliters that we use of low acid, which is the 10% acid.
00:39:43.600 --> 00:39:52.600
And we will use h to describe the amount of the high acid, the 50%.
00:39:52.600 --> 00:40:00.800
So, l is the milliliters of 10% acid; h is the milliliters of 50% acid (for low acid and high acid).
00:40:00.800 --> 00:40:07.200
All right, so if we are not going to waste any low acid or high acid, that means that the amount that we mix of low acid
00:40:07.200 --> 00:40:13.400
and the amount that we mix of high acid--the amount that we are using of both of those must come up to be together exactly 100 milliliters.
00:40:13.400 --> 00:40:17.300
Otherwise, we have gone over the amount that we are going to use eventually, so we are being wasteful.
00:40:17.300 --> 00:40:19.700
We are using too much in creating our thing.
00:40:19.700 --> 00:40:25.800
So, our first piece of knowledge is that h + l = 100.
00:40:25.800 --> 00:40:32.900
The number of milliliters of our high acid, plus the number of milliliters of our low acid, must come out to a total of 100 milliliters.
00:40:32.900 --> 00:40:37.900
Our other piece of information is that we want to make a 22% acid solution.
00:40:37.900 --> 00:40:42.700
So, at this point, we have to start thinking out: what does it mean to be a percent acid solution?
00:40:42.700 --> 00:40:47.500
Well, that means that there is some quantity of acid there; and when divided by the total volume
00:40:47.500 --> 00:40:52.200
of whatever we are dealing with, that is going to give us a percentage number.
00:40:52.200 --> 00:41:02.700
So, amount, divided by volume that we are dealing--the amount of the acid stuff in there,
00:41:02.700 --> 00:41:10.100
divided by the volume--the number of milliliters--is going to be equal to some ratio, 0.something,
00:41:10.100 --> 00:41:13.400
which we can then convert up into a percent, remember.
00:41:13.400 --> 00:41:20.500
Percent is the same thing as the decimal shifted over to the side; so, 10% is the same thing as .10;
00:41:20.500 --> 00:41:25.400
50% is the same thing as .50; and 22% is the same thing as .22.
00:41:25.400 --> 00:41:36.800
So, for every milliliter of high acid that we put in, we will put in .5, times the milliliters of high acid, of acid stuff going in.
00:41:36.800 --> 00:41:41.100
With that idea in mind, we can start creating a formula here.
00:41:41.100 --> 00:41:52.900
.50, 50% acid here, times the number of milliliters, plus .10, the low acid,
00:41:52.900 --> 00:41:57.300
is the total amount of acid stuff that we have put into our mixture.
00:41:57.300 --> 00:42:03.500
And then, we are going to divide it by...we know, at the end, that we are going to end up with 100 milliliters, so that will be our final volume.
00:42:03.500 --> 00:42:07.600
And we want that to come out to be .22.
00:42:07.600 --> 00:42:10.100
So, we know that, at the end, it is going to come out being that.
00:42:10.100 --> 00:42:11.100
These are our two equations.
00:42:11.100 --> 00:42:17.800
Now, alternatively, I am going to tell you a trick that I think is a lot easier way to think about this.
00:42:17.800 --> 00:42:21.600
Alternatively, we can think of this in terms of points.
00:42:21.600 --> 00:42:31.000
Every milliliter of 10% brings 10 points to the table; every milliliter of 50% brings 50 points to the table.
00:42:31.000 --> 00:42:39.200
We know that, in the end, if we have 100 milliliters of 22%, then it is going to be 100 times 22 points.
00:42:39.200 --> 00:42:45.100
We can think of each of the percentages, times the amount of it, as "it is going to be that many points on the table."
00:42:45.100 --> 00:42:50.200
It is a sort of abstract way of thinking about it; but I think it makes it a lot easier to make these equations.
00:42:50.200 --> 00:42:55.500
We know that, if it is 10% for the low and 50% for the high, then we have 50 times h
00:42:55.500 --> 00:43:02.600
(the number of high points that come in), plus 10 times the low (the number of low points that come in,
00:43:02.600 --> 00:43:07.800
the points mixed together)...we know that, in the end, we want to have 22 times 100 points.
00:43:07.800 --> 00:43:14.200
That is going to be our final solution, 22 times 100, which is 2200.
00:43:14.200 --> 00:43:21.500
50h + 10l = 2200; so we can either think in terms of the percentage that our final solution comes out as--
00:43:21.500 --> 00:43:31.500
this .50h + .10l, divided by 100, equals .22; or we can think of the number of points that are being put in to make our final number of points.
00:43:31.500 --> 00:43:35.600
It is kind of abstract; it doesn't really make as much sense; but I think it is a whole lot easier to understand.
00:43:35.600 --> 00:43:37.500
So, that is why I am telling it to you.
00:43:37.500 --> 00:43:47.300
50h + 10l = 2200; I also want you to notice that what we have here and what we have here are actually equivalent equations.
00:43:47.300 --> 00:43:55.900
If you multiply the top one, the red one, by 10000 (by 100 and then by 100 again),
00:43:55.900 --> 00:43:59.800
the first 100 would cancel out the fraction on the left and bring us to 22 on the right.
00:43:59.800 --> 00:44:07.700
And then, the next 100 would bring the .50 to 50, the .10 to 10, and the now-22 to 2200.
00:44:07.700 --> 00:44:10.400
We see that we have the same thing; they are actually equivalent.
00:44:10.400 --> 00:44:16.500
It is just this multiplication; so this point method and the percentage method are going to end up giving us the same thing.
00:44:16.500 --> 00:44:22.200
But I think this point method is a lot easier to understand, in terms of the mechanics of creating it mathematically.
00:44:22.200 --> 00:44:27.300
And it is an easy, fast way to create it on a test, when you have a low amount of time on your hands.
00:44:27.300 --> 00:44:30.100
I recommend thinking about it; but if you don't like it, don't use it.
00:44:30.100 --> 00:44:31.900
All right, at this point, we are ready to solve it.
00:44:31.900 --> 00:44:34.300
Solving it actually won't end up being that hard.
00:44:34.300 --> 00:44:39.300
I am in the mood for using elimination, because we have these nice h's and l's that are ready to go.
00:44:39.300 --> 00:44:51.400
We can multiply them by -10 pretty easily, so we know that -10h - 10l =...multiplied by -10, so -1000.
00:44:51.400 --> 00:45:01.900
We add that to 50h + 10l = 2200; so -10h - 10l = -1000 gets added together here.
00:45:01.900 --> 00:45:09.800
These cancel out entirely; we are left with 40h = 2200 - 1000, or 1200.
00:45:09.800 --> 00:45:18.400
40h...divide by 40 on both sides; we have 1200/40, or 120/4, which is 30.
00:45:18.400 --> 00:45:33.000
So, we know that h = 30, and then from h + l = 100, if h + l = 100, then we know that 30 + l = 100, so l = 70.
00:45:33.000 --> 00:45:54.600
Great; so in the end, these two pieces of knowledge here and here mean that we want to add 30 milliliters of 50% solution and 70 milliliters of 10% solution.
00:45:54.600 --> 00:46:00.600
And that is how we will build our 22% acid solution with 100 milliliters in the end.
00:46:00.600 --> 00:46:12.400
Great; Example 5: Solve the system of equations, if possible: 3x + y - 2z = -6, 4x + 1/2y + 2z = -5, and -3x + 2y - z - 9.
00:46:12.400 --> 00:46:19.200
All right, this is a three-variable system of equations; but ultimately, we are just going to use substitution and elimination.
00:46:19.200 --> 00:46:23.600
We can either stick just to substitution, stick just to elimination, or use a combination of them.
00:46:23.600 --> 00:46:26.800
I am going to use a combination of them, because I feel like it.
00:46:26.800 --> 00:46:31.400
But I want you to know that you can really use any method, as long as you are careful with your work--
00:46:31.400 --> 00:46:33.200
you are making sure that you don't make mistakes.
00:46:33.200 --> 00:46:35.100
Either method will work, or a combination of them.
00:46:35.100 --> 00:46:38.200
You can switch over; sometimes the easiest thing won't be obvious,
00:46:38.200 --> 00:46:42.200
and you will just have to try playing with it for a while, and then you will figure out what it is.
00:46:42.200 --> 00:46:47.200
Just play with it, even if you are not quite sure what will be easiest; just get started on it, and things will work out in the end.
00:46:47.200 --> 00:46:55.600
All right, at this point, I notice that we have 3x here and -3x here; so I am going to add those two equations, and I will get cancellation.
00:46:55.600 --> 00:47:07.700
3x + y - 2z = -6 and -3x + 2y - z = 9: we add those two together (I'll put a line under it):
00:47:07.700 --> 00:47:14.700
3x and -3x cancel out, so we have 3y - 3z = +3.
00:47:14.700 --> 00:47:24.700
Let's divide everything by 3 to make it a little bit simpler: y - z =...sorry, not 3; but we divided by 3, so y - z = 1.
00:47:24.700 --> 00:47:28.600
OK, now we will come back to that in a few moments.
00:47:28.600 --> 00:47:34.600
And now, we also might see that there is a -2z here; there is a positive 2z here; we can add those ones together, as well.
00:47:34.600 --> 00:47:50.300
3x + y - 2z = -6, and 4x + 1/2 + 2z = -5; now, if you wanted to, you could have previously just gone to substitution once you had y - z = 1.
00:47:50.300 --> 00:47:51.900
You could have gone to substitution from the beginning.
00:47:51.900 --> 00:47:55.700
But I decided to start with elimination, because I saw these things that could cancel out easily.
00:47:55.700 --> 00:48:04.900
3x + 4x is 7x, plus 1/2 + 1 becomes 3/2y; -6 + -5 becomes equal to -11.
00:48:04.900 --> 00:48:10.600
Great; so at this point, we have stuff involving x and y and stuff involving y and z.
00:48:10.600 --> 00:48:21.300
So, if that is the case, we want to overlap for the y; we want to be able to figure out what y is, so let's get y as the one that is not going to be substituted out.
00:48:21.300 --> 00:48:31.000
So, y - z = 1; we will change that into z = y - 1; we subtract y, and then multiply by -1 on both sides.
00:48:31.000 --> 00:48:46.400
z = y - 1; and over here, we can get what x is equal to: 7x = -3/2y - 11, or x = -3...we divide that by 7,
00:48:46.400 --> 00:48:51.300
so that would become 14, minus 11, over 7; great.
00:48:51.300 --> 00:48:55.500
Now, that is not that friendly to substitute; but that is what we have gotten to at this point.
00:48:55.500 --> 00:49:00.300
It might have been easier if we had gone with a slightly different method, but we have something; we can work it out.
00:49:00.300 --> 00:49:02.800
It might be a little bit more math than we were hoping to have to do.
00:49:02.800 --> 00:49:07.800
It might be a little bit more "number-crunching"; but it is not going to be that hard to work through.
00:49:07.800 --> 00:49:10.900
At this point, let's figure out which one we want to plug it into.
00:49:10.900 --> 00:49:21.200
Let's choose 3x + y - 2z = -6; that will be fine: 3x + y - 2z = -6.
00:49:21.200 --> 00:49:26.100
Now, we know that x is equal to this stuff over here, so we have 3 times -3...
00:49:26.100 --> 00:49:30.500
Oh, oops, that was one big mistake here; I forgot to put that y down.
00:49:30.500 --> 00:49:41.200
I am sorry about that; so -3 over 14 times y minus 11 over 7, plus...
00:49:41.200 --> 00:49:45.500
we don't have anything substituting for y, because we want y, because we are going to solve for y at the end of this...
00:49:45.500 --> 00:49:51.900
minus 2 times...and over here, z is equal to y - 1, equals -6.
00:49:51.900 --> 00:50:06.500
We start working this thing out: 3 times -3/14 will become -9/14y; minus 33/7, plus y, minus 2y, plus 2, equals -6.
00:50:06.500 --> 00:50:11.000
Let's compact some stuff together: we will compact our y's to what we can...
00:50:11.000 --> 00:50:19.800
-9/14 times y...and we have the y here and the -2y here, so that would become -y.
00:50:19.800 --> 00:50:24.800
And we can't really combine 33/7 and 2 very easily, without bringing in more fractions.
00:50:24.800 --> 00:50:28.800
So, let's just subtract them to the other side, because we know that eventually we are going to have to subtract on the other side.
00:50:28.800 --> 00:50:37.400
So, we subtract by 2 on both sides; -6 - 2 becomes -8; we add 33/7 on both sides, so we get + 33/7.
00:50:37.400 --> 00:50:42.500
OK, now, at this point, we can...I don't like dealing with the fractions here;
00:50:42.500 --> 00:50:45.300
so at this point, I am going to just say, "Let's get rid of the fractions."
00:50:45.300 --> 00:50:48.900
As opposed to trying to put things over common denominators, we are going to get rid of those denominators eventually.
00:50:48.900 --> 00:50:55.800
So, we will just multiply the whole thing by 14, 14 times this whole thing.
00:50:55.800 --> 00:51:08.000
That will get us -9y - 14y =...-8 times 14 becomes -112, and 33/7 times 14...
00:51:08.000 --> 00:51:20.600
well, that will cancel the 14 to just 2 times 33; the 14/7 becomes 2, so 14 times 33/7 is the same thing as 2 times 33, so that is + 66.
00:51:20.600 --> 00:51:31.700
That is a negative; and so, -9y - 14y becomes -23y; -112 + 66 becomes -46.
00:51:31.700 --> 00:51:36.500
Divide by -23 on both sides, and we get y = +2.
00:51:36.500 --> 00:51:40.300
Great; so there is our first thing--at this point, we have these two equations.
00:51:40.300 --> 00:51:43.800
We have our z equation and our x equation, so it is just a matter of plugging into those.
00:51:43.800 --> 00:51:51.800
z =...plug in our 2 for y...2 - 1; so z is equal to positive 1--that one is pretty easy.
00:51:51.800 --> 00:52:02.200
This one is a little bit more difficult; we have x = -3/14 times 2 minus 11/7.
00:52:02.200 --> 00:52:09.600
So, that gets us to -6/14, minus 11/7; actually, it is easier to just cancel out.
00:52:09.600 --> 00:52:17.900
2/14 becomes 7, so that will cancel the denominator to just a 7; minus 11/7.
00:52:17.900 --> 00:52:20.600
That gets us -6 - 11...
00:52:20.600 --> 00:52:26.800
Oops, let's go back a few moments to before I made that mistake.
00:52:26.800 --> 00:52:37.600
-3/14, times 2...well, we can break this into 7 times 2, so the 2's cancel out, and we have -3/7 - 11/7.
00:52:37.600 --> 00:52:49.800
-3/7 - 11/7 gets us -14/7, because they combine (they have the same denominator); -14/7 simplifies to x = -2.
00:52:49.800 --> 00:52:52.400
Great; at this point, we have found each of our values.
00:52:52.400 --> 00:52:55.300
We have found our x-value, our y-value, and our z-value.
00:52:55.300 --> 00:53:04.200
So, in the end, we know that the point that works in all three of these equations is (-2,2,1).
00:53:04.200 --> 00:53:10.900
Great; now, once again, I am just going to say: you could have gone through and done this in other ways.
00:53:10.900 --> 00:53:13.900
You could have used just substitution; you could have used just elimination.
00:53:13.900 --> 00:53:17.000
You could have combined them in a different way than I did here.
00:53:17.000 --> 00:53:20.700
There might have been an easier way; there probably was definitely going to be a harder way, as well.
00:53:20.700 --> 00:53:26.600
But you end up just choosing one way, and you work through it, and eventually it will end up working out.
00:53:26.600 --> 00:53:31.300
As you do more of these problems, you will figure out what works better, what is easier for you,
00:53:31.300 --> 00:53:33.600
a general sense of how to get these things done fastest.
00:53:33.600 --> 00:53:37.900
But really, it is just a matter of practice and just slogging through it sometimes.
00:53:37.900 --> 00:53:46.800
All right, the final example: what if we wanted to solve this system of equations? u + 2v + 7w - 3x + 4y + 2z = 41,
00:53:46.800 --> 00:53:50.500
and then another one, and then another one, and then another one, and then another one, and then another one?
00:53:50.500 --> 00:54:03.100
Well, notice: we have 1, 2, 3, 4, 5, 6 variables total, and we have 1, 2, 3, 4, 5, 6 equations total.
00:54:03.100 --> 00:54:07.200
So we know that this is possible; so we could solve this system of equations.
00:54:07.200 --> 00:54:13.000
Or at least, we could figure out if it has no solutions, if it has one solutions, or if it has infinitely many solutions.
00:54:13.000 --> 00:54:16.600
This is horrifying; I don't want to do this--yuck!
00:54:16.600 --> 00:54:20.400
Solving this problem with the methods we know now would be possible.
00:54:20.400 --> 00:54:24.700
We could work through this with elimination; we could work through this with substitution.
00:54:24.700 --> 00:54:27.400
But it would take us forever to be able to work through this thing.
00:54:27.400 --> 00:54:31.100
It would be awful; it would be this huge task; it would just take us so much paper.
00:54:31.100 --> 00:54:36.700
We could get it done, but I don't want to; and it turns out that there are some great ways to do this.
00:54:36.700 --> 00:54:43.200
We have some tricks ready; later on, when we understand vectors and matrices,
00:54:43.200 --> 00:54:50.700
we will be able to see how to solve a monster like this, in the lesson using matrices, to solve systems of linear equations.
00:54:50.700 --> 00:54:58.200
You will be able to use what you know about matrices, matrix multiplication, matrix inverses...things that we will all learn about in the future; don't worry.
00:54:58.200 --> 00:55:04.300
You will have to learn all of this stuff, and you will be able to see that there is a really, really easy way to be able to just solve this stuff.
00:55:04.300 --> 00:55:07.800
You have access to a calculator, so you can compute the inverses easily.
00:55:07.800 --> 00:55:10.800
You will be able to just finish this thing really, really quickly.
00:55:10.800 --> 00:55:14.600
Solving this thing will be really easy once we talk through this stuff.
00:55:14.600 --> 00:55:21.200
So, we will actually come back to this many lessons in the future, when we get to using matrices to solve systems of linear equations.
00:55:21.200 --> 00:55:23.500
We will take this thing, and we will be able to solve it, just like that.
00:55:23.500 --> 00:55:26.900
It will be really, really easy to work through, which is pretty cool.
00:55:26.900 --> 00:55:31.300
All right, I hope you have a good idea of how systems of linear equations work.
00:55:31.300 --> 00:55:34.900
It is all about figuring out where the thing is intersecting in terms of that.
00:55:34.900 --> 00:55:38.100
That is really the idea of when these two things are going to be true.
00:55:38.100 --> 00:55:40.000
All right, we will see you at Educator.com later--goodbye!