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Video: Solving Simultaneous Non-Linear Equations Using Algebraic Substitution

Tim Burnham

Learn how to use the method of algebraic substitution to solve two simultaneous equations, where one of them is nonlinear. We work through two examples in detail, explaining the tips and techniques and common pitfalls and misconceptions.

13:09

Video Transcript

Sometimes we have to solve a pair of simultaneous equations for which one of them is nonlinear. And in these cases, the method of algebraic substitution is usually the most appropriate. You can plot graphs and read off the coordinates of the points of intersection, but they can’t be guaranteed to give you a hundred percent accurate answer. Algebraic methods give us much more control over our accuracy, although they can sometimes take quite a lot of working out.

So the first example we’re gonna look at is this. Use algebraic substitution to find the coordinates of the points of the intersection of the line 𝑦 equals four and the circle 𝑥 minus one all squared plus 𝑦 minus two all squared equals sixteen. So this is an algebraic substitution exercise. But just for the sake of interest, I’m just gonna show you what the graph looks like, so you can see what we’re headed for. So the line 𝑦 equals four is a line on which the 𝑦-coordinates are always four. So that’s a horizontal line, as we can see here, the blue line going through four on the 𝑦-axis. And the circle, 𝑥 minus one all squared plus 𝑦 minus two all squared equals sixteen, has, for its centre, it’s got an 𝑥-coordinate of one and it’s got a 𝑦-coordinate of two. So there’s the centre, and the radius is the square root of sixteen, that’s four.

Now clearly, that line cuts this circle in two places, when we look at the graph. But that’s not always gonna be the case. If we happen to have moved it up a little bit, perhaps to 𝑦 equals six, we would’ve had a tangent just running across here, which would just touch that circle in one place. So there would only be one answer. If we’d have moved the line even further up, okay, it wouldn’t have cut through the circle at all, and we wouldn’t have any points of intersection. But as it is, this particular question, we’re gonna get two different answers. So let’s just get rid of the diagram and go back to the algebra and see how we get on.

So the first thing I’m gonna do is just write out the equations, put my little brace in to say that they’re simultaneously true, and I’m gonna label them numbers one and two. Now equation one tells us that 𝑦 is four; 𝑦 is always four. So in fact, we already know the 𝑦-coordinates of whatever points of intersection there are; they’re both gonna be four. And the second equation here has got two unknowns in it. So it’s got lots and lots of different possible solutions, so a different combinations of 𝑥 and 𝑦 that will match that. So, what we’re gonna do is substitute the 𝑦 equals four from the first equation into the second equation. So that means wherever we see 𝑦 in that second equation, we’re gonna replace it by four, because the first equation told us that 𝑦 was equal to four. Now we’re gonna multiply this out. So the second term here, four minus two is two; two squared is four. And 𝑥 minus one all squared, well just to make sure you don’t make any mistakes, write it out properly; it means 𝑥 minus one times 𝑥 minus one. So we need to multiply each term in the first bracket by each term in the second bracket. And when we do that, we’ve got 𝑥 squared minus one 𝑥 minus another 𝑥 plus one plus the four equals sixteen. So obviously, minus one 𝑥 minus another 𝑥 is minus two 𝑥, one plus four is five. And on the right-hand side, we’ve still got sixteen. So our quadratic equation is 𝑥 squared minus two 𝑥 plus five equals sixteen.

So now what I’m gonna do is take away sixteen from both sides, to leave myself zero on the right-hand side. And then I’ve got a formula, the quadratic formula, that I can use to solve that quadratic equation. Well, so the quadratic formula tells us when 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 is equal to zero, when 𝑎 is not zero. Then 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎. So I need to find the values of 𝑎, 𝑏, and 𝑐 from my quadratic equation. And that is, 𝑎 is one, 𝑏 is negative two, and 𝑐 is negative eleven. And then we need to substitute them in here to find out our values of 𝑥. So 𝑥 is gonna be the negative of negative two plus or minus the square root of 𝑏, negative two, all squared minus four times 𝑎, is one, times 𝑐, is negative eleven, all over two times 𝑎, two times one. So the negative of negative two is two. And that’s gonna be plus or minus, negative two times negative two is four, and four times one is four, and four times negative eleven is negative forty-four. So we’re taking away negative forty-four, which means we’re adding forty-four. And on the denominator, two times one equals two. Now in the square root there, we’ve got four plus forty-four, so that’s forty-eight. But forty-eight is sixteen times three, and sixteen is a square number. So I’m just gonna rearrange that slightly. So root forty-eight is the same as root sixteen times three, and root sixteen times three is the same as root sixteen times root three, and the square root of sixteen is four. So this little expression here for 𝑥 is two plus or minus four root three all over two. Well that numerator will factorise; I can take two out as a common factor. So the top becomes two lots of one plus or minus two root three, and the two will cancel on the top and the bottom. So this is just one plus or minus two root three. So 𝑥 is equal to one plus two root three, or one minus two root three.

So that’s our two 𝑥-coordinates. So remember, the two 𝑦-coordinates were both four. So the exact form of our two pairs of coordinates are one minus two root three, four, and one plus two root three, four. Or if I rounded them to two decimal places, that’ll be minus two point four six, four and four point four six, four. The question didn’t tell us to round them. In fact, it didn’t say what level of accuracy is wanted. Generally speaking, it’s a good idea to keep the answers as accurate as you can, for as long as you can in the question. So working in their surd form, in one minus two root three form and one plus two root three form is a really good policy.

Okay. Let’s move on to our next question then. Solve the simultaneous equations 𝑦 equals negative two 𝑥 plus four, 𝑦 equals 𝑥 squared minus four 𝑥 plus three. So we’ve got a linear equation, which if we graphed it, it will be a line. And we’ve got a quadratic equation, which we graphed it, it will be a parabola. So these could intersect in one place, two places, or no places at all. But we’re just gonna use the algebraic method, rather than drawing the graphs. And we’re gonna use the method of substitution to actually try and solve this problem.

So in the first question, we knew that the 𝑦-coordinate was always gonna be four, because the-the straight line told us that. In this one, we’ve got a slopey straight line, so we don’t actually know what the 𝑦-coordinate is gonna be. But we do have a nice simple expression that tells us what 𝑦 is gonna be, minus two 𝑥 plus four. So we can use that instead of 𝑦 in the second equation, and then we’ll have one big equation just involving 𝑥s. So we’ve written out the second equation but instead of 𝑦 at the beginning, we’ve written what we know 𝑦 is from the first equation, negative two 𝑥 plus four. So now I’m gonna rearrange that to come up with a simple quadratic equation, like we did in the first question. So I’m gonna add two 𝑥 to both sides and then I’m gonna subtract four from both sides, to give myself a quadratic equal to zero. And then I can use the quadratic formula to solve that. So adding two 𝑥 to both sides, I’ve just got four left on the left-hand side. And negative four 𝑥 plus two 𝑥 gives me negative two 𝑥 on the right-hand side. So now I’ll add four, or rather I’ll subtract four from both sides of my equations, which will leave me with zero on the left-hand side, and 𝑥 squared minus two 𝑥 minus one on the right-hand side. And we’ve got 𝑎 is equal to one, 𝑏 is equal to negative two, and 𝑐 is equal to negative one. So lots of negative signs which we need to be really careful of. So there’s our quadratic formula again. When 𝑎𝑥 squared plus 𝑏𝑥 plus 𝑐 equals zero, provided 𝑎 is not equal to zero, then 𝑥 is equal to negative 𝑏 plus or minus the square root of 𝑏 squared minus four 𝑎𝑐 all over two 𝑎.

So the next step is just to plug all of those values for 𝑎, 𝑏, and 𝑐 into the quadratic formula, as we have done there. Really good idea to put negative numbers in brackets, just to make sure they don’t make any mistakes. And so the negative of negative two is two. So we’ve got two plus or minus the square root of negative two times negative two there, so 𝑏 squared, will be four, and four times one is four, times negative one is negative four. So we’re taking away negative four, which means we’re adding four. So we’ve got four plus four up there, and two times one on the bottom is two.

So we’re gonna do the same trick we did last time to simplify that root eight. So we’ve got two plus or minus root eight over two. Eight is the same as four times two, so I can rewrite it like that. And remember, four is a square number, so the square root of four times two is the same as the square root of four times the square root of two. And the square root of four is two. So root eight is the same as two root two. And same is the two terms on the numerator are both two times something, I can factor out that two. And then the two on the numerator and denominator will cancel down, so I’m just left with one plus or minus root two.

So the two 𝑥-coordinates that we’re looking for are one plus root two and one minus root two. So what we’re gonna do now is substitute back into the other equation, or we can substitute into either equation in fact. But I’m gonna use the straight line, because the numbers are gonna be a lot easier. So having at least two 𝑥-coordinates, which of the two corresponding coordinates on the 𝑦 equals minus two 𝑥 plus four line match those 𝑥-coordinates. So as we’ve said, we’re gonna take the 𝑥-coordinate one minus root two, plug it into the equation number one there 𝑦 equals minus two 𝑥 plus four, and work out what the corresponding 𝑦-coordinate is. So this is the value here, the 𝑥-value minus two times the 𝑥-value plus four. So let’s use the distributive rule to multiply out those parentheses. And negative two times one is negative two, negative two times negative root two is positive two root two, and then we’ve got plus four on the end. So here we’ve got negative two plus four, so that’s just two. So that’s our first coordinate sorted. So we’ve got 𝑥 is equal to one minus root two, 𝑦 is equal to two plus root two [two root two].

So let’s make a note of that solution, and then go on and find the other one. So when 𝑥 is equal to one plus root two, we’ll substitute that again and find the corresponding 𝑦-coordinate for that 𝑥-coordinate. So now we’ve substituted this value for 𝑥 in the equation, and got 𝑦 equals minus two times one plus root two plus four. And using the distributive rule to multiply out the parentheses, negative two times one is negative two, and negative two times positive root two is negative two root two, and then we’ve got the plus four on the end. So negative two plus four is two, which means that the 𝑦-coordinate that goes with 𝑥 is equal to one plus root two, is two minus two root two. So we’ve got our other solution.

So here are our pairs of solutions, 𝑥 is one minus root two when 𝑦 is two plus two root two, and 𝑥 is one plus root two when 𝑦 is two minus two root two. If we’d have been asked to give our solutions as coordinate pairs, that’s what they would’ve looked like. And if we’ve been asked to give our answers to two decimal places say, then that’s what our answer would’ve looked like.

So with these algebraic substitutions, the top tip is to keep your answer as accurate as possible as long as possible, and to just be aware that you need to be very careful. There can be quite long and drawn-out calculations, and as lots of positives and negatives and roots and tricky things going on. So you’ve really gotta be careful to keep track of those. But overall, I think you’ll agree that we would never have got that level of accuracy out of our answers, if we were just trying to read the coordinates of the graph, especially at this scale. So overall, algebraic substitution can be a bit long and drawn-out, a bit tricky, but it does give you a completely accurate answer.