Video: Solving a System of Linear and Quadratic Equations to Find the Set of Points of Intersection of Two Given Graphs

Find the set of points of intersection of the graphs of 𝑦 = 3𝑥 and 𝑥² + 𝑦² = 40.

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Video Transcript

Find the set of points of intersection of the graphs of 𝑦 equals three 𝑥 and 𝑥 squared plus 𝑦 squared equals 40.

Well, the first thing to look at is points of intersection. So what does this actually mean? But I think a little sketch is gonna help us understand. So what am I drawing is a rough sketch of our two graphs. So we got our 𝑦 equals three 𝑥 and our 𝑥 squared plus 𝑦 squared equals 40. Well, the points of intersection are in fact the places where these two graphs would meet. So with these graphs, we can see that there will be in fact two points of intersection. And these are where those two points are. So now we know that we’re gonna be looking for two points of intersection. And we know what points of intersection are.

Let’s go about solving this problem. So to do that, we’re gonna write down our two equations. And as you can see, I’ve actually labelled them. So we’ve got equation one and equation two, or graph one and graph two. So what we said before is there are points of intersection where the two graphs meet. So what we want to do is actually make our two equations equal to each other. And the way we can do that is by using the substitution method.

So to enable us to do that, what I’m gonna do is I’m gonna substitute our value for 𝑦 so that 𝑦 is equal to three 𝑥 into equation two. So we’re gonna get 𝑥 squared plus three 𝑥 all squared. And this is three 𝑥 all squared because we’ve substituted our three 𝑥 for our 𝑦. And this is all gonna be equal to 40. So then we’re gonna get 𝑥 squared plus nine 𝑥 squared is equal to 40. Being careful of the three 𝑥 all squared is three 𝑥 multiplied by three 𝑥, so you get nine 𝑥 squared not just three 𝑥 squared.

So then we can collect our like terms. We can collect our 𝑥 squared terms. So we’re gonna get 10𝑥 squared is equal to 40. And then we divide both sides by 10 which gives us that 𝑥 squared is equal to four. And then we’re gonna take the square root of both sides. So we’re gonna get that 𝑥 is equal to negative two or two. And the reason it’s negative two or two is cause if we look back at our graph in the top right-hand side, we know we’re gonna have two solutions. So we’re gonna have two different values of 𝑥. So there now are two values of 𝑥.

And now we’re gonna find our 𝑦-coordinates or 𝑦-values. And the way we’re gonna do that is by substituting our 𝑥-values back into equation one. You could actually do it into either equation. I’ve just chosen to do it into equation one. Looks that’ll be the simplest in this question. So when 𝑥 is equal two, we’re gonna get 𝑦 is equal to three multiplied by two cause we’ve substituted in two for our 𝑥-value. So we can get a 𝑦-value, a 𝑦-coordinate of six. Or if 𝑥 is equal to negative two, then we’re gonna substitute in negative two for our 𝑥-value. So we’re gonna get 𝑦 is equal to three multiplied by negative two. So we’re gonna get our 𝑦-value or 𝑦-coordinate of negative six.

So therefore, we can say that our set of points of intersection are: two, six and negative two, negative six. And I’ve shown that using our set notation to identify that.

So just a quick recap of what we’ve done. So first of all, we used the substitution method. So we substituted 𝑦 equals three 𝑥 into equation two. Then we’ve solved that to find our 𝑥-values which there were two of, cause we knew there’d be two values because of our little sketch. And then we substitute those 𝑥-values back into one of our equations in order to find our 𝑦-values, 𝑦-coordinates.

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