### 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.