Systems of equations with graphing khan academy answers

Video transcript

We're told to graph this system of equations and identify the number of solutions that it has. And they have the system of equations here. So they want us to graph each of these equations and think a little bit about the solutions. So the first equation here-- I'll rewrite it, so I'll graph it in the same color that I write it. This first equation's already in slope-intercept form, y is equal to 3x plus 1. We see that the slope, or m, is equal to 3, and we see that the y-intercept here is equal to 1. So let me be clear, that is also the slope. I just called it m because a lot of times people say y is equal to mx plus b. So we can graph it. We can look at its y-intercept, the point 0, 1 must be on this graph. So that's the point 0, 1 right there. This is the y-axis, that is the x-axis. And the slope is 3. That means if we move 1 in the positive x-direction, we're going to move up 3 in the positive y-direction. So we move 1 in the x-direction, we move up 3. If we moved 2 in the x-direction, we would move up 6. Just like that. Because 6 over 2 is still 3. Likewise, if we moved down 1, if we went negative 1 in x, we would go negative 3 in y. So negative 1, negative 3. Because negative 3 divided by negative 1 is still 3. If we went negative 2 in x, we would go negative 6-- 1, 2, 3, 4, 5, 6 in y. So these are all points along the line, and I can connect the dots now. So let me do that. So let me connect the dots as best as I can. This should be a line, not a curve. My hand isn't 100% steady, but I think you get-- let me do it a little bit better than that. I think I can do a better job than that. Let me draw-- that's even worse. All right. Last attempt. That's throwing me off. So last attempt right here. There you go. So that's that first line right there. y is equal to 3x plus 1. So let me do the second one now. So it's written in standard form right now, 2y plus 4 is equal to 6x. We want to get this in slope-intercept form, y is equal to mx plus b. So a good place to start could be to subtract this 4 from both sides. So it goes on the other side. So let's subtract 4 from both sides of this equation. The left-hand side, we're left with just a 2y, and then the right-hand side becomes 6x minus 4. So 2y is equal to 6x minus 4. And then to get everything in terms to solve for y, we just have to divide everything by 2. So let's divide everything by 2, and we get y is equal to 3x minus 2. So that's the second equation in slope-intercept form. So same drill here. The y-intercept is negative 2. So we go-- that's negative 1, negative 2 right there, and its slope is 3. And notice its slope is the same as the other line. So it's going to have the same inclination. If we move 1 in the x-direction, we move up 3 in the y-direction. 1, up 3. Just like that. If we go back 1 in x, we go down 3. Back 1 in x, we go down 3. Just like that. So if we connect the dots here, it'll look something like this. I'll do my best to draw a straight line. So the second graph, 2y plus 4 equals 6, we put it into slope-intercept form and we graphed it. Now, the whole point of this question was to identify the number of solutions that it has, the system. A solution to a system of equations is an x and y value that satisfy both of these equations. Now, if there were such an x and y value that satisfied both of these equations, then that x and y value would have to lie on both of these graphs. Because this blue line is all of the pairs of x and y's that satisfy the first equation. The red line is all of the pairs of x's and y's that satisfy the second equation. So if something's going to satisfy both, it's got to be on both lines. When you look here, are there any points that are on both lines? Well, no. These two lines never intersect. A point of intersection is a point that is common to both of these lines. No, they don't intersect. No intersection. So there is not a solution to this system of equations. There is no solution. We know that because these two lines don't intersect. And you didn't even have to graph it. The kind of giveaway was that these are two different lines. They have different y-intercepts, but their slopes are identical. So if you have two different y-intercepts and your slopes are identical, then you have two different lines that will never intersect. And if they represent a system, or if they're the graphs of a system of equations, that system has no solution.

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Video transcript

- The following two equations form a linear system. This is one equation; it has X and Y so it's gonna define a line. And then I have another equation that involves X and Y, so it's gonna define another line. It says: "Graph the system of equations "and find its solution." So we're gonna try to find it visually. So let's graph this first one. To graph this line, I have the little graphing tool here. Notice if I can figure out two points, I can move those points around and it's going to define our line for us. I'm gonna pick two X values and figure out the corresponding Y values and then graph the line. So let's see how I could do this. So let's see; an easy one is what happens when X is equal to zero? Well if X is equal to zero, everything I just shaded goes away and we're left with -3y is equal to nine. So -3y equals nine. Y would be negative three. So when X is equal to zero, Y would be negative three. So let me graph that. When X is equal to zero, X is zero, Y is negative three. Now another easy point actually instead of trying another X value, let's think about when Y is equal to zero 'cause these equations are in a standard form so it's easy to just test. Well what are the X and Y intercepts? So when Y is equal to zero, this term goes away, and you have negative X is equal to nine, or X would be equal to negative nine. So when Y is zero, X is negative nine. So when Y is zero, X is negative nine, or when X is negative nine, Y is zero. So I've just plotted this first equation. So now let's do the second one. We'll do the same thing. What happens when X is equal to zero? When X is equal to zero, so this is going to be our Y intercept now. When X is equal to zero, -6y is equal to negative six. Well Y would have to be equal to one. So when X is zero, Y is equal to one. So when X is zero, Y is equal to one. Get one more point here. When Y is zero, when this term is zero, Y being zero would make this entire term zero, then 6x is equal to negative six or X is equal to negative one. So when Y is zero, X is negative one or when X is negative one, Y is zero. When X is negative one, Y is zero. And so just like that, I've plotted the two lines. And the solution to the system are the X and Y values that satisfy both equations; and if they satisfy both equations, that means they sit on both lines. And so in order to be on both lines, they're going to be at the point of intersection. And I see this point of intersection right over here, it looks pretty clear that this is the point X is equal to negative three and Y is equal to negative two. So it's the point negative three comma negative two. So let me write that down. Negative three comma negative two. And then I could check my answer; got it right. Let's do another. Let's do another one of these. Maybe of a different type. So over here it says: "A system of two linear equations "is graphed below. "Approximate the solution of the system." Alright so here I just have to just look at this carefully and think about where this point is. So let's think about first its X value. So its X value, it's about right there in terms of its X value. It looks like, so this is negative one. This is negative two, so negative 1.5 is gonna be right over here. It's a little bit to the left of negative 1.5, so it's even more negative, I would say negative 1.6. And I'm approximating it, negative 1.6. Hopefully it has a little leeway in how it checks the answer. What about the Y value? So if I look at the Y value here, it looks like it's a little less than one and a half. One and a half would be halfway between one and two. It looks like it's a little less than halfway between one and two, so I'd give it 1.4, positive 1.4. And let's check the answer, see how we're doing. Yep, we got it right. Let's actually just do one more for good measure. So this is another system. They've just written the equations in more of our slope intercept form. So let's see, Y is equal to negative seven, X plus three. When X is equal to zero, we have our Y intercept. Y is equal to three. So when X is equal to zero, Y is equal to three. And then we see that our slope is negative seven. When you increase X by one, you decrease Y by seven. So when you increase X by one, you decrease Y by one, two, three, four, five, six, and seven. When X goes from zero to one, Y went from three to negative four, it went down by seven, so that's that first one. Now the second one: our Y intercept. When X is equal to zero, Y is negative three, so let me graph that. When X is zero, Y is equal to negative three. And then its slope is negative one. When X increases by one, Y decreases by one. So the slope here is negative one. So when X increases by one, Y decreases by one. And there you have it. You have your point of intersection. You have the X-Y pair that satisfies both equations. That is the point of intersection. It's gonna sit on both lines which is why it's the point of intersection. And that's the point X equals one, Y is equal to negative four. So you have X equals one and Y is equal to negative four. And I can check my answer and we got it right.

How do you solve systems of equations by graphing?

Which graph shows a system of equations with no solutions?