Video Transcript
In this video, we’re gonna look at solving simultaneous equations using a graph.
Now this method is all about drawing graphs. Or perhaps, you’ll be given the graphs in the question, so sometimes you have to plot the curves and plot the lines. And you’re looking for the points at which they intersect, so where does the line cut through the curve. And then you’re looking at the points, looking at the 𝑥-coordinate, looking at the 𝑦-coordinate; these are your solutions. Okay. Let’s go ahead and have a look at some examples.
Okay. So the first example, use the graphs to solve the simultaneous equations, 𝑦 equals two 𝑥 plus three and 𝑦 equals 𝑥 squared minus four 𝑥 plus three. And we’ve been given a pair of axes and we’ve had got the two graphs already plotted on there for us. So that red curve is 𝑦 equals 𝑥 squared minus four 𝑥 plus three, and the blue line is 𝑦 equals two 𝑥 plus three.
So when we are solving simultaneous equations, what we’re trying to do is find the points where those two equations simultaneously hold true. So that means if I put a particular 𝑥-coordinate into one of these equations, I’ll get the same 𝑦-coordinate out, whether I put it into the first equation or the second equation. So what it’s basically saying is, read off the 𝑥- and the 𝑦-coordinates of every point where these two graphs intersect. So the first point is here, and the coordinates of that point are zero, three. So that means the 𝑥-coordinate is zero and the 𝑦-coordinate is three. So that’s our first solution. 𝑥 equals zero, 𝑦 equals three.
So let’s call our equations number one and number two, and let’s just check this answer. So taking the first equation and plugging in a value of zero for 𝑥, we’ve got 𝑦 is equal to two times zero plus three. So two times zero is zero, so zero plus three is three. So yes, when we plug zero into that first equation, we get an answer of three out. So let’s try that same thing in the second equation. Plugging in a value of zero for 𝑥, it gives us 𝑦 equals zero squared minus four times zero plus three. So that’s zero minus zero plus three, which is three. So yeah, when I put in 𝑥 to the second equation, I get an answer of 𝑦 equals three out.
Now you’re not expected to do that when you’re doing the questions. You’re just expected to read off the coordinates and present them in the right way. But I’m just going through these checks, just to demonstrate how the thing’s working in this example. And our second point of intersection then, is this one here. And the coordinates of that point are six, fifteen. So the 𝑥-coordinate is six and the 𝑦-coordinate is fifteen. And what that’s saying is, if I put an 𝑥 value of six into either of those equations, I’m gonna get an answer of fifteen out for 𝑦. So let’s just check that.
So in the first equation, 𝑦 is gonna be two times six plus three. So that’s twelve plus three, which is fifteen. So yep, that works. And in the second equation, we’re doing 𝑦 is six squared minus four times six plus three. So that works out to be thirty-six take away twenty-four plus three, which is indeed fifteen. So yep, that looks like they’re the right answers. So that was our answer. If we wanted to be a little bit more kind of careful, we would make sure that it’s clear which 𝑥 value goes with which 𝑦 value. So when 𝑥 equals zero, 𝑦 is equal to three and when 𝑥 equals six, 𝑦 is equal to fifteen.
So in question two, using the graphs, solve the simultaneous equations 𝑦 equals two 𝑥 plus three and 𝑦 equals a half 𝑥 cubed minus two 𝑥 squared minus four 𝑥 plus three. And again, we’ve been given those graphs. So we haven’t actually got any plotting to do. That’s nice! So we’re looking for which points do the graphs intersect. And in this case, we’ve got three points. First one has an 𝑥-coordinate of minus two, or negative two, and a 𝑦-coordinate of negative one. The second one has got an 𝑥-coordinate of zero and a 𝑦-coordinate of three. And the third one has got an 𝑥-coordinate of six and a 𝑦-coordinate of fifteen. So our answer is those three points. When 𝑥 is negative two, 𝑦 is negative one. When 𝑥 is zero, 𝑦 is three. And when 𝑥 is six, 𝑦 is fifteen. And we could, if we wanted to check our answer, plug those 𝑥-values into equation one and equation two here, and then check that we get the same 𝑦-value in each equation for the same 𝑥-value.
And for number three, using the graphs, solve the simultaneous equations 𝑦 equals a half 𝑥 minus two, so that’s the red line, and 𝑥 minus three all squared plus 𝑦 minus two all squared equals twenty-five. So in this case, we’ve got the equation of a circle.
Now if you know anything about the equation of a circle, which you don’t have to, to answer this question, you’ll know that the 𝑥-coordinate of the center is three, the 𝑦-coordinate of the center is two, and the radius is the square root of twenty-five, so that’s five, which is all very interesting, but it’s not actually relevant to the question. So we’ve got to solve the simultaneous equations. Where do those two graphs intersect? Well here where the 𝑥-coordinate is zero and the 𝑦-coordinate is negative two, and here where the 𝑥-coordinate is eight and the 𝑦-coordinate is two. So we’ve got two solutions here: When 𝑥 is zero, 𝑦 is negative 2 and when 𝑥 is eight, 𝑦 is two.
So if I put the 𝑥-value of zero into either of those equations and worked out the corresponding 𝑦-coordinate, it would be negative two for each equation. And if I put eight into each of those equations, I would also get an answer of 𝑦 equals two for both equations.
And for our last example, again, we’ve gotta use the graphs to solve these simultaneous equations. This time, we’ve got the curve 𝑦 equals 𝑥 cubed minus two 𝑥 squared minus three 𝑥 plus two, so it’s a cubic graph. And the blue one, the line, is 𝑦 equals negative a half 𝑥 plus one.
So what we mean by simultaneous equations is, we’re looking for pairs of 𝑥𝑦 coordinates that will work the same in either equation. So where do those graphs intersect, so we’ve got three points here, here, and here. So there’s three solutions to this. And we need to work out the 𝑥- and the 𝑦-coordinates of each. Now in this case, they don’t seem to line up neatly with exact whole numbers. So our life’s a little bit more tricky, so we have to think very carefully about what these solutions might be. So for that first point on the left, it looks like the 𝑥-coordinate is just to the left of negative one, and there are five ticks in between each of our numbers on the 𝑥-axis. So each one of those is nought point two. So this looks like it’s gonna be between minus one point zero and minus one point two. So let’s say that’s minus one point one. And the 𝑦-coordinate looks like it’s the third tick up after one, so that looks like it’s gonna be one point six.
And for our second solution here, it looks like the 𝑥-coordinate is somewhere between nought point two and nought point four, let’s say that’s nought point three. And the 𝑦-coordinate looks like it’s about the fourth tick up after zero, so that’s nought point eight.
Now for the third solution, it looks like we’re around about two point two four six eight, two point eight for our 𝑥-coordinate. And the 𝑦-coordinate, it’s negative and it looks about two ticks down, so that’s negative nought point four.
So two things to note there: One, it’s actually quite tricky when you start getting these noninteger solutions. You can’t be quite as accurate as you could if you did this algebraically. And also, you know, you have to be quite careful. You have to look at the scale, look at what the ticks represent, and think very carefully about your solutions.
So there we have it, using graphs to solve simultaneous equations. It’s just a matter of looking at the graphs, see where they cross over, see the intersections between the lines and the curves, and then read off the 𝑥-coordinates and read off the corresponding 𝑦-coordinates. If you want to be really thorough, you can plug the 𝑥-coordinates to each equation and see that you get the correct 𝑦-coordinates in both cases, to make sure that they are, in fact, simultaneously true.
