WEBVTT 00:21.600 --> 00:30.965 Solving a system of linear equations in two variables. We're going to begin by determining whether an ordered pair is an actual 00:30.965 --> 00:44.000 solution to a specific system of linear equations. And before we do that, let's talk about what a solution to a system of equations actually is. So if you 00:44.001 --> 00:56.557 recall in solving a linear equation, your solution is the ordered pair that makes the equation true. When we incorporate this into a system of linear equations, 00:56.690 --> 01:10.738 meaning more than one linear equation, then the solution or the ordered pair of this system of linear equations must satisfy both equations in the system. 01:10.738 --> 01:24.552 We're going to begin by taking a look at this example here. Our job is to determine whether the ordered pair (five, four) is a solution to the system: 01:24.552 --> 01:40.634 two x minus five y equals negative ten and two x plus four y equals six. In order for this ordered pair (five, four) to be a solution, it will have to satisfy  01:40.634 --> 01:54.849 both equations in the system. We'll begin with the first equation: two x minus five y equals negative ten. Now my ordered pair is 01:54.849 --> 02:08.000 (five, four), so by substituting five for x and four for y, we're trying to figure out if what we have on the left side of 02:08.000 --> 02:20.500 the equation is equal to what we have on the right, which is negative ten. Well two times five is ten, minus twenty, 02:20.501 --> 02:29.882 and that is negative ten. So because I get a true statement, negative ten equals negative ten, 02:29.882 --> 02:42.263 then that tells us that (five, four) is a solution to this equation. But in order for it to be a solution to the system of equations, 02:42.263 --> 02:50.000 it must also satisfy the second equation in the system. So similar to what we did with the first equation, 02:50.000 --> 02:56.977 we're going to use our second equation: two x plus four y equals six, 02:56.977 --> 03:19.833 and we're also going to substitute five for x and four for y. Two times five is ten, plus sixteen. As you notice here, ten plus sixteen is twenty-six, and twenty-six 03:19.833 --> 03:32.212 definitely does not equal six. Therefore the ordered pair (five, four) does not satisfy the second equation in our system, and because (five, four) 03:32.240 --> 03:56.070 does not satisfy both equations, then what we say is that the ordered pair (five, four) is not a solution to our system of equations. 03:56.080 --> 04:03.500 Now we'll solve a system of linear equations by graphing. To solve a system of equations by graphing, 04:03.500 --> 04:12.300 it's necessary to graph each of the lines represented in the system of equations. Once the lines are graphed, 04:12.300 --> 04:22.529 we're looking for any point of intersection between the lines. That point of intersection, the x-value and the y-value, 04:22.529 --> 04:36.810 are the values that satisfy or make true each of the equations in the system. Let's take a look at this example. We want to solve this system by graphing. 04:36.810 --> 04:51.120 Negative two x plus y equals negative five, and two x minus four y is equal to eight. Notice both of these equations are in standard form: 04:51.120 --> 05:00.120 A x plus B y equals C. To find the slope of an equation in standard form, 05:00.120 --> 05:14.148 we're going to take negative A and divide it by B. To find the y-intercept of an equation in standard form, we're going to take C 05:14.148 --> 05:29.196 and divide it by B. So let's look at equation number one here and figure out the slope and the y-intercept. 05:29.200 --> 05:48.185 My A value is negative two, so the negative of negative two is positive two, and B is one. For the y-intercept our value of C is negative five, 05:48.200 --> 05:57.162 and again B is one, but it's not necessary to write it here. The reason I left the one here is because with slope 05:57.162 --> 06:07.162 we're looking at rise over run, or a change in y over a change in x. Now let's take this information and actually graph the first equation. 06:07.162 --> 06:16.376 My y-intercept is negative five, so along the y-axis I'm going to negative five and I'm putting my point. 06:16.376 --> 06:27.125 With the slope of positive two over positive one, from my y-intercept I'm going up two and over one to the right. 06:27.125 --> 06:44.292 Now I'm going to use slope again to get another point on this graph. So up two, over one. And now I can draw the line. 06:44.292 --> 07:05.292 So let's look at the second graph in our system: two x minus four y is equal to eight. 07:05.292 --> 07:19.143 To graph the second line, we're going to use x- and y-intercepts. So recall to find an x-intercept, we're going to substitute zero for y. 07:19.143 --> 07:33.593 So substituting zero for y here, leaves me with two x is equal to eight, which makes my x-intercept four. To find the y-intercept, 07:33.600 --> 07:47.000 we will substitute zero for x, and in substituting zero for x, I'm left with negative four y is equal to eight. Therefore, y is equal to negative two. 07:47.000 --> 07:54.875 Now let's go to the graph and plot these intercepts. So an x-intercept of four means 07:54.875 --> 08:06.992 I'm going on the x-axis to four, and I'm plotting the point. A y-intercept of negative two means on the y-axis I'm going 08:06.992 --> 08:18.526 down to negative two and plotting the point there. Now we can draw the line. 08:18.992 --> 08:46.526 What we recognize now is that this point right here (two, negative one) is the point of intersection. That means two for x 08:46.526 --> 09:06.013 and negative one for y makes each of these equations in the system true. So we say that (two, negative one) is the solution to this system of equations. 09:06.013 --> 09:18.198 We're going to continue graphing systems of linear equations. In graphing systems of linear equations, we can have either one solution, 09:18.200 --> 09:30.000 no solution, or infinitely many solutions. So this first system we're going to look at is two x minus y equals six 09:30.000 --> 09:44.000 and six x minus three y equals six. So to begin graphing the first equation, we're going to find the intercepts. To find the x-intercept, 09:44.000 --> 09:56.040 we're going to make the y-value zero leaving me with two x is equal to six. Therefore, I know that x is equal to three. My x-intercept is three. 09:56.040 --> 10:07.040 To find the y-intercept, we'll make x zero leaving us with negative y is equal to six 10:07.040 --> 10:28.128 which makes y equal to negative six. In our second equation, we can also find intercepts to graph this line easily. The x-intercept: six x is equal to six, 10:28.160 --> 10:42.315 which makes the x-intercept one. And for the y-intercept, negative three y is equal to six, making our y-intercept negative two. 10:42.320 --> 10:55.563 So let's begin by graphing the first equation. With an x-intercept of three, I'm going along my x-axis to three, and I'm plotting the point. 10:55.563 --> 11:06.363 With the y-intercept being negative six, down on the y-axis to negative six, and there's the y-intercept. 11:06.363 --> 11:31.591 I can now connect the two points. And there I have the line. Now doing the same for the second equation, an x-intercept of one 11:31.600 --> 11:51.186 and a y-intercept of negative two. Connecting the points gives me this line. 11:51.186 --> 11:59.186 Notice here that it looks like we have two parallel lines. We could only be sure of that 11:59.200 --> 12:07.530 by actually finding the slopes of each of these lines and comparing them, as well as comparing the y-intercepts. 12:07.530 --> 12:13.343 If you were to find the slope of line number one and line number two, you would realize that it's the same. 12:13.530 --> 12:23.343 And by looking at the graphs here, they obviously have different y-intercepts. Therefore making the two lines parallel. 12:23.343 --> 12:43.096 With parallel lines there is no point of intersection. Therefore, the solution to this system of equations is that there is no solution. 12:43.120 --> 13:14.193 Now if we move over to our second system of equations, in my first equation I notice that my x-intercept is three and my y-intercept is negative six. 13:14.200 --> 13:45.425 For equation number two, finding the x-intercept gives me three, and finding the y-intercept, negative six. So let's go ahead and get these lines on the graph. 13:45.440 --> 14:07.007 Starting with equation number one, I have an x-intercept of three and a y-intercept of negative six. Connecting these points and drawing the line, 14:07.007 --> 14:25.944 gives me that. Now for our second equation, I also have an x-intercept of three, and I also have a y-intercept of negative six. 14:25.944 --> 14:44.000 When I pick up the ruler to draw this line, what I notice is that it's the exact same line as equation number one. 14:44.000 --> 14:51.524 It's very hard for me to pick out only one point of intersection for both of these lines, 14:51.524 --> 15:18.151 and that's because they intersect at every point on the lines. Therefore, there is not only one solution to this system. There are infinitely many solutions. 15:18.160 --> 15:29.496 So here is an example of a system that has no solution, or no point of intersection. This is an example of a system that has infinitely many solutions, 15:29.520 --> 15:35.602 or infinitely many points of intersection.