Video Transcript
We did an introduction to addition
and subtraction of rational expressions in a previous video. But in this one, we’re gonna be
looking at some slightly more difficult examples of adding and subtracting rational
expressions, and learn a few techniques to look out for in simplifying the
results.
So the first one here, we’re gonna
do a subtraction, three 𝑥 over 𝑥 minus one minus 𝑥 over 𝑥 plus three.
Well there are no common factors in
either of these denominators. So we’re gonna have to find
equivalent fractions for each of these, in order to get a common denominator. And the way that we’re gonna do
that is, we’re gonna take this first denominator here and we’re gonna multiply top
and bottom of the second fraction by that denominator. And then we’re gonna take the
second denominator and do the same for the first fraction. So now both of the denominators are
basically 𝑥 plus three times 𝑥 minus one. Now what you’ll remember is that 𝑥
plus three over 𝑥 plus three is just one. So we’ve taken that first fraction
and we’ve multiplied it by one. We’ve taken the second fraction and
multiplied it by 𝑥 minus one over 𝑥 minus one, which again of course is just
one. So multiplying something by one
doesn’t change it. So we’ve still got the same
question here, but we’ve now got it in a format where both of those fractions have
got the same denominators.
So again, you might remember from
before that it’s a really good idea to put those denominators and all these little
terms in brackets, when they’re supposed to be together. And then we combine the two terms
into one term with that common denominator. So that was three 𝑥 lots of 𝑥
plus three for our first term, and 𝑥 lots of 𝑥 minus one for our second term. And now we’ve gotta multiply those
out. So three 𝑥 lots of 𝑥, and three
𝑥 lots of positive three. But it’s negative 𝑥 lots of 𝑥,
and negative 𝑥 lots of minus one. So three 𝑥 lots of 𝑥 is three 𝑥
squared, and three 𝑥 lots of positive three is positive nine 𝑥. Negative 𝑥 times 𝑥 is negative 𝑥
squared, and negative 𝑥 times negative one is positive 𝑥. And now we’ve gotta simplify that
numerator. So we’ve got three 𝑥 squared and
we’re taking away 𝑥 squared, so that gives us two 𝑥 squared. And then we’ve got nine 𝑥 and
we’re adding another 𝑥, so that’s gonna give us plus ten 𝑥.
Well now we can factor the
numerator. So we’ve got a common factor of two
in each of those coefficients, and we’ve also got a common factor of 𝑥 which we can
factor out. So that gives us two 𝑥 lots of 𝑥
plus five all over 𝑥 plus three lots of 𝑥 minus one. Now if we’d have been a bit luckier
there, that factor there, 𝑥 plus five, would have been the same as one of the
brackets on the denominator. We’d have been able to cancel
something else out and make it simpler. But as it is, that’s gonna be our
answer. Nothing else is gonna cancel
down. There we go, two 𝑥 lots of 𝑥 plus
five over 𝑥 plus three times 𝑥 minus one.
So the process, just to summarise
this, that we’re gonna be using is first of all, we’re looking for a common
denominator. Now we’re looking to try and
simplify that as much as possible, as soon as possible. But essentially, we’ve got to look
for this technique here, to try and get a common denominator. Secondly, we’re looking to combine
all of this into a single rational expression, or a single fraction. And then we’re trying to evaluate
that numerator, perhaps do some factoring, simplify it down and see if we’re lucky
enough if any of the things cancel down in our finar answ- final answer, to simplify
that as much as possible.
Right then.
In the next question, we’ve got to
simplify three over 𝑥 plus two plus two 𝑥 over 𝑥 squared minus four.
Now before we do anything else with
that, let’s have a look at this denominator here, 𝑥 squared minus four. We should recognise, that’s the
difference of two squares. So we can factor that into 𝑥 minus
two lots of 𝑥 plus two. So let’s write out our question
with that expressed that way first, and see if that gives us any clues as to how we
can move forward. So when we do that, we can see
that, look, the denominator here is 𝑥 plus two, but this denominator has also got a
factor of 𝑥 plus two. So if we want a common denominator,
we can just take this bit here, this factor, 𝑥 minus two, and multiply top and
bottom of that first fraction to get a- an equivalent fraction with that same
denominator. So remember, 𝑥 minus two over 𝑥
minus two is just one. So we’ve just multiplied that first
fraction by one. So we haven’t changed the size of
it, the magnitude of it, we’ve just got an equivalent fraction. But now they’ve both got a
denominator of 𝑥 minus two times 𝑥 plus two. So let’s combine those into one
fraction.
And the first term was three lots
of 𝑥 minus two. I’ve just turned those around and
called it three lots of 𝑥 minus two. It doesn’t matter which way you
multiply those. And then we’ve got two 𝑥, for our
second term, on the numerator. So we’re just gonna multiply out
that numerator and see if we can simplify it at all. And three lots of 𝑥 is three 𝑥,
and three lots of negative two is negative six. So we’ve got three 𝑥 minus six
plus two 𝑥 on the top, and we can now say three 𝑥 plus two 𝑥 is five 𝑥. And that leaves us with five 𝑥
minus six over 𝑥 minus two 𝑥 plus two. Looking at that numerator, we
haven’t got any common factors. We can’t simplify that down. Nothing cancels top and bottom. So in fact, that is gonna be our
final answer.
So the next example we’re gonna
look at is simplifying one over 𝑥 squared plus five 𝑥 plus six plus one over 𝑥
squared plus seven 𝑥 plus twelve. Now it’s not gonna be any good just
to, sort of, multiply the top and bottom of the second fraction by this denominator
here, because we’re gonna end up with incredibly complicated expressions to deal
with. But what we can do, first of all,
is factorise these two denominators and then pick out if they’ve got a common part,
we will multiply by that. If they’ve got other bits, then
we’ll multiply by those. So let’s js- let’s just factorise
those two denominators first.
Well they’re both quadratics, so we
know we’re gonna have 𝑥 here and 𝑥 here because they’ve already got one 𝑥 squared
and one 𝑥 squared. So they’re gonna have 𝑥 here and
𝑥 here. And we’ve gotta find numbers in the
first case that add to make positive five, and multiply to make positive six. So that’s gonna be plus two and
plus three. And in the second case, we’re
looking for numbers which add to make seven and multiply to make twelve. So we’re looking at plus four and
plus three.
Now before we carry on, let’s just
do a quick check. 𝑥 times 𝑥 is 𝑥 squared. 𝑥 times three is three 𝑥, two
times 𝑥 is two 𝑥, so that’s two 𝑥 and three 𝑥 is five 𝑥. And then two times three is
six. So that one works. And 𝑥 times 𝑥 is 𝑥 squared. 𝑥 times three is three 𝑥, four
times 𝑥 is four 𝑥, three 𝑥 and four 𝑥 is seven 𝑥. And four times three is twelve. So that one works as well.
Okay. So both of those denominators have
got 𝑥 plus three as a factor. So I’m gonna have to take this
missing factor and multiply top and bottom of the second fraction by that. And I’m gonna have to take this
missing factor and multiply top and bottom of the first fraction by that. And when we do that, we’ve got a
common denominator with three brackets multiplied together, three sets of
parentheses multiplied together there. So I’m gonna combine those into a
single fraction. I’ve got one lot of 𝑥 plus four
plus one lot of 𝑥 plus two. Well of course, one lot of those
things is just one lot. So that’s 𝑥 plus four plus 𝑥 plus
two. And 𝑥 plus 𝑥 is two 𝑥, and
positive four add another two is plus six. And looking at that, on the
numerator here, we’ve got a common factor two and six. So two is a common factor; we can
factor that to give us two lots of 𝑥 plus three all over 𝑥 plus two times 𝑥 plus
three times 𝑥 plus four. Now if we look at that very
carefully, we’ve got two times 𝑥 plus three. 𝑥 plus three is a factor on the
numerator, but 𝑥 plus three is also a factor on the denominator. So if I divide both of them by 𝑥
plus three, they cancel out, which gives us our final answer of two over 𝑥 plus two
times 𝑥 plus four.
Well let’s look at some subtraction
ones then. So 𝑥 plus three over 𝑥 plus four
minus 𝑥 minus one over 𝑥 plus one. Well there are no common factors in
the denominators there. So let’s just write this out and
try and get a common denominator.
So first, as we said before, make
sure you put those numerators and denominators, cause they’re slightly more
complicated numerators now, in brackets to keep those terms together, that need to
be kept together. So we’re going to use this
denominator here to multiply top and bottom of this fraction. And we’re gonna take this
denominator and multiply top and bottom of this fraction by that. So now we have 𝑥 plus one times 𝑥
plus four as our common denominator. We can combine those two terms
together into one single fraction. And now we’re gonna have to
multiply out these parentheses and these parentheses. Now this is where it gets quite
tricky. Because we’ve got a minus sign in
here, we’re taking away the whole of that second expression here, from the first
expression here. So we’ve gotta be quite careful how
we lay that out. So I’m actually gonna multiply out
the parentheses and leave them in brackets with a minus in front, and then will
evaluate it afterwards.
So multiplying out the first
bracket then, we’ve got 𝑥 times 𝑥 is 𝑥 squared. And we’ve got 𝑥 times three is
positive three 𝑥, one times 𝑥 is positive one 𝑥. So three 𝑥 and one 𝑥 is four
𝑥. And then we’ve got one times three,
both positive, so that’s three. And for the second term here, we’ve
got 𝑥 times 𝑥 is 𝑥 squared. 𝑥 times four is plus four 𝑥. Negative one times 𝑥. We’re taking away an 𝑥, so four 𝑥
take away one 𝑥 is three 𝑥. And negative one times positive
four is negative four. So we’re gonna be taking away 𝑥
squared. We’re gonna be taking away positive
three 𝑥. So take away three 𝑥, negative
three 𝑥. And we’re taking away negative
four, which is like adding four. So let’s rewrite our numerator.
Well the first few terms here are
gonna remain as they are. And then as we said, we’re taking
away 𝑥 squared. We’re taking away positive three
𝑥, or taking away three 𝑥. And we’re taking away negative
four, which is the same as adding four. So now we can simplify that. We’ve got 𝑥 squared take away 𝑥
squared. Well that’s nothing, so they’re
gonna cancel out. We’ve got positive four 𝑥 and
we’re taking away three 𝑥, so that’s just one 𝑥. And we’ve got positive three and
we’re adding on four, so that’s seven, which leaves us with 𝑥 plus seven all over
𝑥 plus one times 𝑥 plus four. Well we could put brackets around
the top if we wanted to, or we could even multiply out the denominator. But nothing else is gonna
cancel. That is our final answer.
And the most difficult thing there
was the fact that we had quite a lot of working out to do. And taking care of the negative
signs here was really the crucial and-and most difficult step. That’s where most people go wrong,
if they’re gonna go wrong with that kind of question.
So in our next subtraction of
rational expressions, we’ve got three over 𝑥 minus two take away two 𝑥 minus two
over two 𝑥 squared minus 𝑥 minus six. And as we’ve seen in some of these
questions before, the first thing to try and do is to factor anything that we
can. And it looks like the numerator of
the second term will factor. And it looks like the denominator
will also factor here. So let’s try and do that.
So we’ve put the denominator of the
first term in a bracket, because that’s always a good idea. And the numerator on the second
term is quite easy to factor. So two is the common factor, so
we’ve got two lots of 𝑥 minus one. Now the denominator, slightly
trickier, so two 𝑥 squared, so we- we’re gonna have two brackets; it’s a
quadratic. One of them is gonna have to be two
𝑥, and one of them is gonna have to be 𝑥, because two 𝑥 times 𝑥 is two 𝑥
squared. And now we’ve gotta sort of play
around a bit, to try and work out what the other terms are gonna be.
So we must have two 𝑥 and 𝑥,
because two 𝑥 times 𝑥 is 𝑥- two 𝑥 squared. But one of the other terms is gonna
be, well they have to multiply together to give negative six and they’re gonna
combine together to give this whole expression. And with a bit of trial and error,
we see that they’re two 𝑥 plus three and 𝑥 minus two. Well that’s handy because look,
we’ve got 𝑥 minus two and we’ve got 𝑥 minus two. So we’ve got a common factor
here. So the only thing that would be
missing from this denominator, for the first term, would be this term here, two 𝑥
plus three. So I’m gonna multiply top and
bottom by two 𝑥 plus three. And that’s an equivalent fraction
to the first term. And we’ve now got two terms both
with a common denominator of two 𝑥 plus three times 𝑥 minus two. So we can combine those into a
single fraction, which is three lots of two 𝑥 plus three for the first term on the
numerator, and we’re taking away two lots of 𝑥 minus one for the second term. So we’re gonna have to multiply out
these brackets on the numerator. So negative two lots of 𝑥 and
negative two lots of negative one. Well three lots of two 𝑥 is six
𝑥, three lots of positive three is positive nine, negative two lots of 𝑥 is
negative two 𝑥, and negative two lots of negative one is plus two. So we can now simplify this. We’ve got six 𝑥 take away two 𝑥
is four 𝑥, and nine add two is eleven, positive eleven.
So again, looking at the numerator,
there’s no common factors. We can’t factor that down
anymore. Nothing cancels on the numerator
and the denominator. So there we have it; that would be
our answer to this question. So again, once more, we just have
to watch what we’re doing in terms of negative signs, when we’re subtracting
these. And in this case, we had to do
quite a lot of work at the beginning to factor our numerator and denominator, which
just made the whole multiplication thing, finding the equivalent fraction with the
common denominator, a little bit easier.
Right. Let’s round it off with this quite
tricky question here.
So in this case, we’ve got to
subtract two rational expressions. We’re gonna subtract one from the
other. But we’ve gotta show that it equals
a particular expression. Now sometimes, when you do your
calculation, you might end up with an answer in one format. But in this case, they’re asking
for you to present in a very specific format. So let’s go ahead and see what
we’re gonna do with this. First of all, it looks like we’re
gonna have to do a bit of factorisation on the denominator here. But if we look at that first term
and we put a bracket around the numerator, we can see that par- that the numerator,
if I divide that by 𝑥 plus five, I get one. If I divide the denominator by 𝑥
plus five, that also becomes one here. So we’ve got one over two minus 𝑥
times three plus 𝑥. So we’ve got a slightly simpler
first term. So let’s have a look at what we can
do next.
Well in our working out, we can
show that cancelling going on and we can show that the first term is equal to one
over two minus 𝑥 times three plus 𝑥. And the second term, we can rewrite
by factoring the denominator. And to get negative 𝑥 squared,
I’ve gotta have a negative 𝑥 times a positive 𝑥. And we’ve just gotta work out what
the numbers are, that multiply together to make six and add together to make
negative one, which is the coefficient of that 𝑥 term here. So that’s gonna be two minus 𝑥 and
three plus 𝑥. So we’ve basically rearranged or
reexpressed those first two terms. And it turns out that if we look at
the denominators for each of those, they’ve got common denominators. So we’re gonna be able to add them
together. So let’s just reexpress that
left-hand side. And remember, the first term can be
reexpressed as one over two minus 𝑥 times three plus 𝑥. And we’re taking away that second
term, which we’ve just reexpressed as 𝑥 over two minus 𝑥 into three plus 𝑥. So we’ve got our common denominator
here. So we can combine these two things
together into one single fraction.
And that looks incredibly close to
what we’ve actually got to show, but it’s not quite the same. Instead of one- instead of 𝑥 minus
one, we’ve got one minus 𝑥. And instead of 𝑥 minus two, we’ve
got two minus 𝑥. But we’ve got 𝑥 plus three, three
plus 𝑥, they’re the same thing. Three plus 𝑥 is the same as 𝑥
plus three. So we’ve got to be a bit creative
here about what we’ll do. Well what we could do is multiply
the top and the bottom by negative one. And that gives us these
expressions, which when I reorder them, give us the answer we were looking for.
