Lesson Video: Adding and Subtracting Algebraic Fractions Mathematics

Video Transcript

In this video, we will learn how to add and subtract algebraic fractions, which are simply fractions where either the numerator, denominator, or both involve algebraic expressions rather than only numbers. For example, two over 𝑥 plus four and three over 𝑥 minus one, these are each examples of algebraic fractions.

Before we dive into working with algebraic fractions though, let’s recall how we can add or subtract numeric fractions. And we’ll consider the example of two-fifths plus one-quarter. We know that in order to add or subtract fractions, we need a common denominator. And we usually look to make this the lowest common multiple of the two original denominators. In this case, that’s the lowest common multiple of five and four, which is 20. Notice that in this instance 20 is also equal to the product of five and four. Although that won’t always be the case.

We then convert each of our fractions to an equivalent fraction with the denominator of 20. In the case of our first fraction two-fifths, this means multiplying both the numerator and denominator by four. And in the case of our second fraction, we need to multiply both the numerator and denominator by five. That gives eight over 20 plus five over 20. And as the two fractions now have the same denominator of 20, we add them together by simply adding their numerators, giving an answer to this numeric problem of 13 over 20.

Now adding or subtracting fractions like this is probably something you’ve been doing for many years. The rules for adding and subtracting algebraic fractions are exactly the same. Before we can add or subtract two algebraic fractions, we must first convert them to equivalent fractions with a common denominator. Our denominators, and perhaps also the numerators, will be algebraic expressions, but the process that we follow is exactly the same. So let’s look at an example.

Given that 𝑎 over 𝑏 and 𝑐 over 𝑑 are algebraic fractions, write 𝑎 over 𝑏 plus 𝑐 over 𝑑 in the form 𝑥 over 𝑦.

In this question, we’re being asked to find the sum of two algebraic fractions. If we imagine instead that 𝑎, 𝑏, 𝑐, and 𝑑 are whole numbers instead of algebraic terms, then this problem looks very similar to a question such as two-fifths plus one-quarter. We know that in order to add these two numeric fractions together, we need to find a common denominator. And we can find this by multiplying the two original denominators together. In our numeric example, that gives a common denominator of 20. In the algebraic example though, the common denominator will be the product of 𝑏 and 𝑑, which we can write as 𝑏𝑑.

We also need to convert the numerators so that the fractions we’re adding are equivalent to the ones we started with. In our numeric example, we need to multiply both the numerator and denominator of two-fifths by four and the numerator and denominator of one-quarter by five. In each case, we’re multiplying by the other denominator. In the algebraic example, in the case of 𝑎 over 𝑏, we need to multiply both the numerator and denominator by 𝑑. And in the case of 𝑐 over 𝑑, we need to multiply both the numerator and denominator by 𝑏.

That gives 𝑎𝑑 over 𝑏𝑑 plus 𝑏𝑐 over 𝑏𝑑. And we now have two algebraic fractions with a common denominator. We can therefore add them by adding their numerators, which gives 𝑎𝑑 plus 𝑏𝑐 all over 𝑏𝑑. This fraction can’t be simplified any further. So we have our answer to the problem. We’ve written 𝑎 over 𝑏 plus 𝑐 over 𝑑 in the form 𝑥 over 𝑦. 𝑥 is equal to the algebraic expression 𝑎𝑑 plus 𝑏𝑐 and 𝑦 is the algebraic expression 𝑏𝑑.

So whilst questions involving algebraic fractions may look a little scarier than numeric questions, we just follow exactly the same process. Let’s now look at a slightly more complicated example where the denominators of the two fractions we’re adding are algebraic expressions rather than single terms.

Write four over 𝑥 plus two plus two over 𝑥 minus one as a single fraction in its simplest form.

We’re being asked to add these two algebraic fractions together. So we need to recall the methods by which we can do this. The process by which we add algebraic fractions is exactly the same as when we add numeric fractions. So first, we must find a common denominator for our two fractions. We can find this by multiplying the two original denominators together, giving 𝑥 plus two multiplied by 𝑥 minus one. We then need to convert each fraction to an equivalent fraction with this denominator.

For the first fraction, if we’re multiplying the denominator by 𝑥 minus one, we must also multiply the numerator by 𝑥 minus one. And for the second fraction, if we’re multiplying the denominator by 𝑥 plus two, we must also multiply the numerator by 𝑥 plus two. We therefore have four multiplied by 𝑥 minus one over 𝑥 plus two 𝑥 minus one plus two multiplied by 𝑥 plus two also over 𝑥 plus two 𝑥 minus one.

As the two fractions now have a common denominator, we combine them by adding their numerators, giving four multiplied by 𝑥 minus one plus two multiplied by 𝑥 plus two all over 𝑥 plus two multiplied by 𝑥 minus one. To simplify, we distribute each set of parentheses in the numerator, giving four 𝑥 minus four plus two 𝑥 plus four all over 𝑥 plus two 𝑥 minus one and then see if anything can cancel. We have negative four plus four, so these two terms will directly cancel one another out. And then four 𝑥 plus two 𝑥 simplifies to six 𝑥. So we have our final answer of six 𝑥 over 𝑥 plus two multiplied by 𝑥 minus one.

This is a single fraction, and it’s in its simplest form as there are no shared factors that can be canceled from the numerator and denominator. Our answer to the problem then is six 𝑥 over 𝑥 plus two multiplied by 𝑥 minus one.

In our next example, we’ll see how to add or subtract algebraic fractions where the denominator of one fraction is a factor of the denominator of the other.

Write two over 𝑥 minus three squared plus seven over 𝑥 minus three as a single fraction in its simplest form.

In this question then, we’re finding the sum of two algebraic fractions. We recall that the first step in any problem involving adding or subtracting fractions, whether they’re numeric or algebraic, is to find a common denominator. But we need to be a little careful here because if we look at our denominators carefully, we see that they already have something in common. One of our denominators is 𝑥 minus three, and the other is simply 𝑥 minus three all squared.

One denominator is therefore a factor of the other. So to find a common denominator, we aren’t simply going to multiply the two denominators together. Let’s think about the numeric example one-quarter plus three-eighths. Now the product of the denominators four and eight is 32, but this isn’t the lowest common multiple of the two numbers. As four is a factor of eight, the lowest common multiple of the two denominators is simply eight. We would therefore only need to convert one of our two fractions to an equivalent fraction in order to be able to add them. We would convert the fraction one-quarter to two-eighths and then add this to three-eighths, giving a final answer of five-eighths.

Let’s now apply the same logic to the algebraic problem. 𝑥 minus three is a factor of 𝑥 minus three squared. So the common denominator we’ll use is 𝑥 minus three squared. And we’ll only need to convert our second fraction. Our first fraction is unchanged. And in order to create a denominator of 𝑥 minus three squared in the second, we have to multiply by 𝑥 minus three. So we do the same to the numerator, giving two over 𝑥 minus three squared plus seven multiplied by 𝑥 minus three over 𝑥 minus three multiplied by 𝑥 minus three. Of course, we can write the denominator of our second fraction as 𝑥 minus three squared so that we can see that they are common.

As the two fractions we’re adding now have the same denominator, we combine them by adding the numerators. We can then distribute the parentheses in the numerator only to give two plus seven 𝑥 minus 21 all over 𝑥 minus three squared and finally simplify the expression in the numerator to give negative 19 plus seven 𝑥 over 𝑥 minus three squared. So we found the sum of these two algebraic fractions as a single fraction in its simplest form. Our answer is negative 19 plus seven 𝑥 over 𝑥 minus three squared.

In our next example, we’ll see how to subtract an algebraic fraction from a term which is simply an integer.

Write two minus two over 𝑥 as a single fraction in its simplest form.

The first step in any problem involving adding or subtracting fractions whether they’re numeric or algebraic is to find a common denominator. Now in this example, we’re subtracting a fraction, two over 𝑥, from an integer, two. A comparable question using numbers only would be something like two minus three-quarters.

Now in a simple question like this, we may be able to work out the answer in our heads, but when we were first learning how to add or subtract fractions, we would probably have followed a process something like this. We may first have thought of the integer two as the fraction two over one. We may then have written the two fractions with a common denominator of four. So two over one would become two multiplied by four over four, which is eight over four. As the two fractions now had the same denominator of four, we could subtract three-quarters from eight-quarters by subtracting the numerators, giving five over four or five-quarters. And in the case of a numeric example, we may then convert this to a mixed number of one and a quarter.

Let’s now apply the same logic then to this algebraic problem. The denominator of the fraction we’re subtracting is 𝑥. So this is the common denominator we want to use for the two fractions. We can also think of this as one multiplied by 𝑥 if we wish. To express the integer two as a fraction with the denominator of 𝑥, we’d also need to multiply the numerator by 𝑥. So the integer two is equivalent to the fraction two 𝑥 over 𝑥. We therefore have two 𝑥 over 𝑥 minus two over 𝑥. And as the denominators of these two fractions are the same, we can combine them by subtracting the numerators, giving two 𝑥 minus two all over 𝑥.

We may also spot in this instance that the terms in the numerator have a shared factor of two. So we can give our answer in a factored form of two multiplied by 𝑥 minus one all over 𝑥, although this isn’t entirely necessary. So by following the exact same processes as when we subtract a fraction from an integer, we found that two minus two over 𝑥 as a single fraction in its simplest form is two 𝑥 minus two over 𝑥.

So we’ve now seen a variety of examples, but there are a couple of common mistakes that I’d like to highlight.

Suppose we add two algebraic fractions and through simplifying arrive at the result 𝑎 over 𝑎 plus 𝑏. A common mistake is to think that we can cancel a factor of 𝑎 in the numerator and denominator to give one over one plus 𝑏 or sometimes one over 𝑏. This is incorrect. 𝑎 isn’t a factor of the denominator. It’s not 𝑎 multiplied by something; it’s 𝑎 plus something. So we can’t divide through by it.

However, if in another problem, we arrived at an answer of two 𝑎 squared over 𝑎 multiplied by 𝑎 plus four, we could cancel by a factor of 𝑎 here as the numerator is two 𝑎 squared, which is two 𝑎 multiplied by 𝑎, and the denominator is 𝑎 multiplied by 𝑎 plus four. So we see that 𝑎 is a multiplicative factor of both the numerator and denominator. In this case then, it would be perfectly acceptable and in fact best practice to cancel that shared factor of 𝑎, leading to a final answer of two 𝑎 over 𝑎 plus four.

Another thing we need to be careful of is when we are subtracting algebraic fractions, we must make sure that we subtract all of the numerator in the second fraction. In the case of the example on screen, we would use a common denominator of 𝑥 plus two multiplied by 𝑥 plus one, giving four multiplied by 𝑥 plus one minus three multiplied by 𝑥 plus two in the numerator. All good so far.

The stage that often trips people up is when distributing the parentheses in the numerator. The first set distributes to four 𝑥 plus four. That’s fine. But in the second set, we must remember it’s negative three multiplied by the entire of 𝑥 plus two. The common mistake is to get the negative three 𝑥 correct but then write positive rather than negative six. In this case, if we were to proceed with positive six rather than negative six, we’d get the coefficient of 𝑥 in the numerator correct, but we’d get the constant term wrong.

So watch out for each of these common mistakes and make sure you don’t fall into either of these traps.

Before we finish, let’s just look at one final example which is slightly more complicated because it involves an algebraic expression in the numerator of one fraction.

Write 𝑥 over 𝑥 plus one plus 𝑥 plus four over 𝑥 minus three as a single fraction in its simplest form.

The first step is to find a common denominator for the two fractions, which we do by multiplying them together, giving 𝑥 plus one multiplied by 𝑥 minus three. We multiply the numerator and denominator of the first fraction by 𝑥 minus three and the numerator and denominator of the second by 𝑥 plus one. As the two fractions now share a common denominator, we combine them by adding the numerators.

We then distribute the parentheses in the numerator. And this is where this example is a little more complicated than the others we’ve seen so far because we end up with some quadratic terms. 𝑥 multiplied by 𝑥 gives 𝑥 squared. We can simplify the numerator by collecting like terms. 𝑥 squared plus 𝑥 squared is two 𝑥 squared. Negative three 𝑥 plus 𝑥 plus four 𝑥 is positive two 𝑥. And then we have a constant term of four.

We could in this instance also factor the numerator by two, giving a final answer of two multiplied by 𝑥 squared plus 𝑥 plus two all over 𝑥 plus one multiplied by 𝑥 minus three.

In this problem then, the methods are exactly the same. We first find a common denominator by multiplying the two individual denominators together. We then find equivalent fractions with this denominator and combine them by adding their numerators. It’s slightly more complicated though because we end up with a quadratic expression in the numerator as well as one in the denominator.

Let’s now review some of the key points that we’ve seen in this video. Firstly, to add or subtract algebraic fractions, we follow the exact same rules as when we’re working with numeric fractions. We first find a common denominator, usually by multiplying the two denominators together. Although as we’ve seen, if one denominator is a factor of the other, then we’ll only need to convert one.

We then convert each fraction to an equivalent fraction with this common denominator and combine by adding or subtracting the numerators. We can then distribute the parentheses in the numerator, simplify the resulting expression, and cancel if there are indeed any true shared factors between the numerator and denominator. Although as we saw in our common mistakes, we must be very careful when we’re doing this.

By following these rules then, we can extend our knowledge of adding fractions to questions where the denominator, and possibly numerator, of one or both fractions is an algebraic expression.