Video: Introduction to Addition and Subtraction of Rational Expressions

Understand adding and subtracting numeric rational expressions, and then we move on to simple linear expressions in numerators or denominators.


Video Transcript

In this video, we’re gonna be looking at how to add and subtract rational expressions. These are fractions with polynomials either on the numerator or denominator or both and some who call them algebraic fractions. So we’ve called it the introduction because we’re gonna be looking at some simpler examples and just introducing some of the ideas to help you down the road.

But first, let’s just remember how to add simple fractions that’ve got different denominators. And we’re gonna be paying close attention to how we discuss our working out and how we set it all down on the page. So these are different denominators: a sixth is smaller than a half, so we’ve got a small chunk of something plus a slightly bigger chunk of something. The only way we can sensibly work with this is to get a common denominator. So what we’re looking for is equivalent fractions for a sixth and a half, but they’ve got actual different denominators on them. So the quickest way that we can find to do this is we can take this denominator here, and we can multiply top and bottom of the other fraction by that denominator. So what we’re doing is we’re taking a sixth and we’re multiplying it by, well two divided by two is one. And when you multiply anything by one, you’re keeping it the same. So this bit of calculation here is still a sixth, but it’s just a sixth but written slightly differently. And we’ll also take the denominator on the first fraction six, and we’ll multiply top and bottom of the other fraction by that. So again a half times six divided by six is one a half times one it’s just a half. Now if you’re holding your hands up in horror already cause you can see a quicker easier way of doing this, don’t worry. I’ll come back and talk about that in a second. Let’s just follow this example through. So we’ve got two times one over twelve plus one times six over twelve. And we can combine that into one quite big fraction, so two times one is two and one times six is six. So we’ve got two plus six over twelve, which is eight over twelve. Now we can divide the top and the bottom there by four. Eight divided by four is two. Twelve divided by four is three. So this gives us an answer of two-thirds.

So let’s just recap what we did there. First of all, we found a common denominator for those fractions. And this was the method that we did that; we took this denominator here and multiplied top and bottom of the other fraction by. We took this denominator here and multiplied top and bottom of the other fraction by it. Then we managed to combine those into one fraction. And then finally, we added the numerators together. And we were able to cancel down in this case by dividing by four, so we’re simplifying our final answer. Now this is basically the technique that we use. Even if we’ve got polynomials, we’ve got algebraic fractions. Now as I said, we’re just gonna go through this once again and do it slightly differently this time. The numbers that we had here were very easy to work with. So it wasn’t a great problem, but there is another slightly different way of looking at the problem and which will help us when we get into the more complicated algebraic rational expressions. So what we could’ve done to start off with is just factorise that first denominator. So six is equal to three times two. Now when we’re looking for a common denominator, this denominator is a multiple of two and this denominator is also a multiple of two. So all we need to do is to take this three from here, and we’re gonna multiply the top and the bottom of the other fraction by three. So three over three is one again. So that same principle applies before.

So what we’ve done in that first step is we’ve effectively sort of reduced the amount of work that we’ve got to do with not having to multiply that first term by anything. We’re just taking the-the missing factor if you like from the common denominator, and we’re creating another version of the second term, which does have that same denominator. So of course one times three is three, so we’ve got one over three times two plus three over three times two or two times three. And we can combine that into one fraction, because they both’ve got the same denominator, and that gives us four over three times two. But I can divide the top by two, so four divided by two is two. I can do that at the bottom by two, which cancels down to make one. So I’ve got the same answer as we had before: two-thirds. But all the way through, we’ve dealt with slightly smaller numbers than we did last time. So again, those were easy numbers, not a big problem. But on the more complicated questions, this might actually turn out to be a real advantage for us. So this is the basic technique. Let’s look at some algebraic fractions or some rational expressions now.

So in this first one, we’ve got to simplify 𝑥 over three plus 𝑥 over five. Now three is a prime number, five is a prime number, so we’re not gonna be able to factorise any of the denominators. So let’s go ahead and write this out. To get our common denominator, I’m going to multiply both of these fractions by one. But the version of one that I’m gonna multiply the first one by is five over five, because that was the denominator of the other fraction. And the version that I’m gonna multiply the second fraction by is three over three. So combining those into one big fraction, because we have the common denominator here five times three, we’ve got five times 𝑥 is five 𝑥, 𝑥 times three is three 𝑥. So we’re adding those two together all over five times three on the bottom, which is gonna be fifteen. And five 𝑥 plus three 𝑥 is eight 𝑥. Now eight and fifteen don’t have any common factors. So in fact eight 𝑥 over fifteen is gonna be a final answer in this case. So it’s just like working with fractions as we did before. But in this case, we have some letters in the mix, which just make the whole thing a little bit more tricky.

Okay let’s turn the notch up just one little bit. This is very slightly harder than the last one, but not massively difficult.

Simplify two 𝑥 over five plus three 𝑥 over four. So again those denominators don’t have any common factors other than one, so I’m gonna have to multiply both of the fractions by some version of one in order to get equivalent fractions. And I’m gonna multiply the second fraction by five over five, and I’m gonna multiply the first fraction by four over four. And now I’ve got common denominators of four over four times five in each case. And because I’ve got those common denominators, I can merge those two fractions into one. So I’ve got four lots of two 𝑥 plus five lots of three 𝑥 all over four times five, so let’s evaluate some of those. Well four times two 𝑥 is eight 𝑥, five times three 𝑥 is fifteen 𝑥, so we’re adding fifteen 𝑥, and four times five on the bottom is twenty. And eight 𝑥 plus fifteen 𝑥 is twenty-three 𝑥. Nothing cancels down, so that’s our answer.

Now for this example, I’m gonna do this in two different ways. One, I’m just gonna blindly go in and just gonna follow the method that we’ve been talking about so far, and then we’re gonna rerun it and see if we can spot any slightly more efficient ways of doing it.

So we’re gonna simplify one over three 𝑥 plus two over eight 𝑥. Well I’m gonna take the first denominator and I’m going to multiply the top and the bottom of the other fraction by that. And I’m gonna take the second denominator and I’m gonna multiply the top and the bottom of the first fraction by that. Then if I combine them into one fraction, I’ve got one lot of eight 𝑥 plus two lots of three 𝑥 on the numerator and the common denominator of eight 𝑥 times three 𝑥 on the bottom. So this gives me eight 𝑥 plus six 𝑥 on the top and then three times eight is twenty-four. And 𝑥 times 𝑥 is 𝑥 squared. Now eight 𝑥 plus six 𝑥 is fourteen 𝑥. So I’ve got fourteen 𝑥 over twenty-four 𝑥 squared, but now I can do a bit of cancelling. Fourteen and twenty-four can both be divided by two. So fourteen divided by two is seven, twenty-four divided by two is twelve. 𝑥 can be divided by 𝑥 to just make one, and 𝑥 squared can be divided by 𝑥 just to make 𝑥. So on the top, I’ve got seven times one, which is seven. And on the bottom, I’ve got twelve times 𝑥, which is just twelve 𝑥. So that’s the answer that we’ve got.

Okay let’s rerun that again then. But now we might spot that in the question, this second term, two over eight 𝑥, I could’ve divided the top by two and I could’ve divided the bottom by two as well to make four. So I’ve got one over three 𝑥 plus one over four 𝑥. Now looking at those denominators, this has got a factor of 𝑥 and this’s also got a factor of 𝑥. So in fact, what I’m gonna multiply each of these fractions by is not four 𝑥 and three 𝑥, but I’m only gonna multiply by the-the missing factor in that common denominator. So for the second fraction here. I’m going to multiply by three on the top and three on the bottom. And for the first fraction here, I’m only gonna multiply by four on the top and four on the bottom. So that first term over here was four times one, the second term over here is one times three. So when I combine them to make one fraction, the denominator is four times three times 𝑥. And when I evaluate those, four times one is four, one times three is three, and four times three times 𝑥 is twelve 𝑥. Well four plus three is seven. So seven over twelve 𝑥 is our answer, because none of those cancel down. So by looking at the question a little bit more carefully in the first place and by being a little bit more careful about how we created a common denominator, we can see this second calculation here was actually a lot easier to do than our first calculation there.

Okay moving on.

Simplify four over 𝑥 plus three plus two over 𝑥 minus two. Well it looks a little bit more complicated, but the process we’re gonna follow is exactly the same as what we’ve just been doing. Now the one thing that I would recommend is putting brackets around your denominators, cause that’s gonna make life a little bit easier as we go through here and make sure we don’t make any mistakes. So we’re gonna take our first denominator and multiply the top and the bottom of the other fraction by that value. And we’ll take the second denominator and multiply the top and the bottom of the first fraction by that value. So the principle still sticks, 𝑥 minus two divided by 𝑥 minus two is just one. So we’ve still got one times the first fraction, so it’s still the first fraction. And 𝑥 plus three divided by 𝑥 plus three is still one. So we’ve got one times the second fraction. So it’s still just two over 𝑥 minus two. Okay let’s multiply those out. Well we combine them into one fraction here. We’ve got four lots of 𝑥 minus two plus two lots of 𝑥 plus three all over the common denominator of 𝑥 minus two times 𝑥 plus three. So now we’re going to do four lots of 𝑥 and four lots of negative two and we’re gonna add two lots of 𝑥 and two lots of positive three. So four lots of 𝑥 is four 𝑥 and four lots of negative two is negative eight. Two lots of 𝑥 is positive two 𝑥 and two lots of three is positive six. Now I’m not multiplying out the denominator at this stage, because we’re not finished tidying up the numerator. Who knows? Things might factorise and something might cancel out. If I multiply the denominator out, I wouldn’t be able to spot that.

Well four 𝑥 plus two 𝑥 is six 𝑥 and negative eight plus six is negative two. So we’ve got six 𝑥 minus two over 𝑥 minus two times 𝑥 plus three. But in fact, if we look carefully at the numerator, we can see that that will factorise. Six and two got common factor of two. So what do I have to multiply two by to get six 𝑥? Well that would be three 𝑥. And what do I have to multiply two by to get negative two? Well that would be minus one. So there we go. Now what we’ve ended up with inside the bracket isn’t the same as any of the brackets of the bottom. So nothing is gonna cancel, but this is a nice factored format of our answer. Now if I had multiplied out the denominator and I hadn’t factored the numerator, that’s the solution I would have got and still be a perfectly correct answer. But people tend to leave it in this factored format rather than this multiplied out format.

Right, so and next example is a subtraction example.

Simplify two over three 𝑥 minus one over five 𝑥. So we’re gonna approach this in just the same way we did before. But instead of adding the two fractions, we’re gonna be subtracting the second one from the first. So we can see that they both got an 𝑥 in common on the denominators for both of those fractions. So the missing factors are the things that we’re gonna be using to find equivalent fractions, so three. We’re gonna multiply the top and the bottom of the second fraction by three and the five, we’re gonna multiply the top and bottom of the first fraction by five. So that leaves us with five times two minus one times three all over five times three times 𝑥 when we combine that into one fraction. So that gives us ten minus three all over fifteen 𝑥, which is seven over fifteen 𝑥. And that won’t do any more cancelling, so there’s our answer. So subtraction just like addition, but you do have to take things away rather than adding them.

Okay one last one then.

We’re going to simplify three over three minus 𝑥 minus six over six minus 𝑥, so let’s do what we did before. And that is to put brackets around the denominators and then we’ve got to find equivalent fractions, so that we end up with a common denominator. So we’re gonna take this three minus 𝑥 and multiply the second fraction top and bottom by that value. And we’re gonna take the six minus 𝑥 and we’re gonna multiply the top and bottom of the first fraction by that value. So we’ve got a three lots of six minus 𝑥 as the numerator for the first fraction and six lots of three minus 𝑥 for the second numerator. So we’ll bring those terms together to make one single fraction. This is what we’ve got. and we’ve gotta be very very careful when we multiply out these brackets, because that second term on the numerator is minus the whole thing. So let’s have a look at how that works. So we’re gonna have three lots of six and three lots of negative 𝑥. Now three lots of six is eighteen and three lots of negative 𝑥 is negative three 𝑥, but now we are taking away six lots of three. And we’re taking away six lots of negative 𝑥, so six lots of negative 𝑥 is negative six 𝑥. If we’re taking away negative six 𝑥, that means that we’re adding six 𝑥. Now looking at that numerator, we’ve got eighteen take away eighteen, which is nothing. And we’ve got negative three 𝑥 add six 𝑥, which is positive three 𝑥. And none of those terms will cancel down, so we’re left with our answer of three 𝑥 all over six minus 𝑥 times three minus 𝑥.

So the thing really to pay attention to there was this step here. When we were dealing with taking away things. Remember we had to take away the whole of this expression here, which meant being very careful about the signs that we ended up with here. That’s the most common place that people go wrong when they do these questions. So really do watch out for that.

So let’s just recap again the whole process. Because whether we’re doing adding or subtracting of these fractions, these rational expressions, the process is just the same. First of all, we had to try and find common denominators, which meant finding equivalent fractions for each of them using these denominators very carefully to multiply the other fractions. I use this denominator here to multiply this fraction. Sometimes we had to multiply by the whole denominator, sometimes we could just multiply by one of the factors. We’ve got a way to get common denominators. Once we’ve got the common denominators, we can combine those two fractions into one big fraction, one big expression here, all over the common denominator. And then lastly, we’re evaluating that numerator, and sometimes we get something which can be factored. Sometimes it can’t be. Sometimes it will cancel with something on the denominator down here, but we’ve gotta look at all those processes and try to simplify that expression as much as possible when we get to the end. That one, two, three is the same for all of these questions.

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