### Video Transcript

In this video, we’re gonna look at how to solve equations with algebraic fractions. We’re gonna go through the different steps one by one, and we’re also going to check our answers before giving them at the end of the question.

Well our first question is, solve the equation two 𝑥 plus three over four plus 𝑥 minus five over three is equal to three over two. Well in these questions, it’s always a good idea first to put parentheses around any numerators or denominators that contain algebraic expressions, just to make it clear that those terms all live together on those numerators or denominators.

Right, now the next thing we need to do is find the lowest common multiple of the denominators. Well one way of tackling this is to express each number as the product of its prime factors, so two and two are both prime numbers; two times two is four, and since two is a prime number, I can write it down here as well.

Now two isn’t a prime factor of three. In fact, the only prime factor of three is three, so I’m gonna write that separately over here. Now grouping our prime factors like that means that we can say, well look, we’ve got two is a common prime factor between two and four, so we can write that down here. Then we’ve just got two, which is a prime factor of four, so we write that down here. And then we’ve just got three, which is a prime factor of three, so we can write that down here.

Now two times two times three is equal to 12, so 12 is the least common multiple of four, three, and two. So I need to multiply through each of the terms in my original equation by 12. In fact, I’m gonna write 12 as 12 over one because each term is a fraction, so I’ve got 12 over one times two 𝑥 plus three over four plus 12 over one times 𝑥 minus five over three plus 12 over one times three over two.

Okay, but why did we do that? Well because we can now simplify each of those terms and get rid of the denominators. That first term, 12 over one times two 𝑥 plus three over four, if I divide the bottom by four, I get one; if I divide the top by four, I get three. So I’ve got three times two 𝑥 plus three all over one, or just three times two 𝑥 plus three.

Doing the same for the second term, this time I’m gonna divide three by three. And over on the right-hand side, I can divide three by two. So we’re no longer dealing with algebraic fractions; that’s great.

Well now let’s simplify this; let’s multiply out those terms. We can use the distributive property of multiplication. So we’ve got three times two 𝑥 is six 𝑥 and three times positive three is positive nine. Then I’ve got four times 𝑥 is four 𝑥 and four times negative five is negative 20, and six times three over on the right-hand side is equal to 18.

Now I can gather up my like terms. Six 𝑥 plus four 𝑥 is 10𝑥; nine take away 20 is negative 11, so 10𝑥 minus 11 is equal to 18. Now if I add 11 to both sides of my equation, negative 11 plus 11 is nothing, so on the left-hand side, I’ve just got 10𝑥, and 18 plus 11 is 29, so 10𝑥 is equal to 29. If I divide both sides by 10, I’ll be able to see what 𝑥 is equal to, because 10 divided by 10 is just one, so that just leaves me with one 𝑥 on the left-hand side, and that means that 𝑥 is equal to 29 over 10 or 2.9. So we’ve got two equivalent versions of our answer there: 𝑥 is equal to 29 over 10 or 𝑥 is equal to 2.9.

Now before we move on, we’re just gonna substitute that answer back into the original question just to check whether it’s right or not. So the left-hand side of the original equation is two 𝑥 plus three all over four plus 𝑥 minus five over three. So let’s substitute in 𝑥 is equal to 29 over 10, and that means two times 29 over 10 plus three all over four plus 29 over 10 minus five all over three.

Well this two here really means two over one, and when I write it like that, I can see that the two cancels with the 10 to make five on the denominator there, so that becomes 29 over five, and this three here is the same as 15 over five, and writing it that way means I’ve got 29 over five plus 15 over five; that’s a simple fractional addition.

Now looking at this five here, I could rewrite that as 50 over 10, so I get a common denominator with this 10 here. Then 29 over five plus 15 over five is 44 over five, so I’ve got 44 over five all over four or 44 over five divided by four for that first term there, and 29 over 10 minus 50 over 10 is negative 21 over 10, and that’s negative 21 over 10 all over three, or negative 21 over 10 divided by three.

Now of course, four is the same as four over one, and three is the same as three over one. Now to divide fractions, all I do is flip the second fraction and turn it into a multiplication. So 44 over five divided by four over one is the same as 44 over five times one over four, and likewise 21 over 10 divided by three over one is the same as 21 over 10 times one over three.

Now I can do a bit of canceling. Four divided by four is one, 44 divided by four is 11, three divided by three is one, and 21 divided by three is seven. So I’ve got 11 over five minus seven over 10, but to subtract fractions, I need to have common denominators. So 11 over five is the same as 22 over 10, and 22 over 10 minus seven over 10 is just 15 over 10.

Now with 15 over 10, I can divide the bottom by five to give me two and the top by five to give me three, so I’ve got three over two. Now three over two was what it was supposed to equal, the right-hand side. So we’re saying that the left-hand side with the particular solution that we put in there does equal the right-hand side, so we’re pretty confident we’ve got the right answer.

So if you remember the steps we went through there, first of all, we put these parentheses in just to make it nice and clear what was going on, which terms had to be kept together. Then we found the least common multiple of our denominators. Then we multiplied through each term by those denominators in order to get rid of the fractions. And finally, we just had to rearrange the equation that we got and solve it. And then finally finally, we just decided to check our answer as well.

Okay, let’s move on to another question. Solve four over three 𝑥 minus one plus two over 𝑥 plus one equals three. Now the first thing we notice about this question is that, this time round, our algebraic expressions are on the denominators, but let’s not worry about that for now. Let’s put our parentheses around the numerators or denominators to make it clear which terms have to stay together.

Now the second part of our process was to find the least common multiple of our denominators and then multiply through by that and simplify. Now the least common multiple of three 𝑥 minus one and 𝑥 plus one is just three 𝑥 minus one times 𝑥 plus one, so let’s multiply each term by that combination, three 𝑥 minus one times 𝑥 plus one.

And remember, on the left-hand side, they’re both fractions, so I’m actually gonna multiply by three 𝑥 minus one times 𝑥 plus one over one. And our first term then becomes three 𝑥 minus one times 𝑥 plus one times four all over one times three 𝑥 minus one.

Well dividing the bottom by three 𝑥 minus one and the top by three 𝑥 minus one cancels those two things out. The second term, we’re adding three 𝑥 minus one times 𝑥 plus one times two over one times 𝑥 plus one. Well dividing the bottom by 𝑥 plus one gives us one; dividing the top by 𝑥 plus one cancels that term there.

So just rewriting that out neatly, I’ve got four times 𝑥 plus one plus two times three 𝑥 minus one is equal to three times three 𝑥 minus one times 𝑥 plus one. Okay, let’s use the distributive property to multiply out these parentheses. Four times 𝑥 is four 𝑥, and four times positive one is plus four, two times three 𝑥 is six 𝑥, and two times negative one is negative two.

Now I’m gonna multiply the two parentheses together, so I’m gonna do three 𝑥 times 𝑥 to give me three 𝑥 squared, three 𝑥 times positive one to give me positive three 𝑥, negative one times 𝑥 to give me negative 𝑥, and negative one times positive one to give me negative one.

Then tidying up these two terms here, three 𝑥 take away one 𝑥 is two 𝑥, and now I can use the distributive property to multiply out these parentheses here. Well three times three 𝑥 squared is nine 𝑥 squared, three times two 𝑥 is positive six 𝑥, and three times negative one is negative three. Then over on the left-hand side, I’ve got four 𝑥 add six 𝑥, which gives me 10𝑥, and four take away two, which is just positive two.

Now I need to rearrange this into a quadratic that I can solve. So if I subtract 10𝑥 from both sides on the left-hand side, I’ve got 10𝑥 take away 10𝑥, which is nothing, just leaving me with two, and on the right-hand side, six 𝑥 take away 10𝑥 is negative four 𝑥.

Now I can just subtract two from both sides, giving me nine 𝑥 squared minus four 𝑥 minus three minus two on the right-hand side, and minus three minus another two is minus five, and over on the left-hand side, two take away two is zero. So we’ve got nine 𝑥 squared minus four 𝑥 minus five is equal to zero.

So our original problem, four over three 𝑥 minus one plus two over 𝑥 plus one equals three, after a bit of rearranging, is equivalent to nine 𝑥 squared minus four 𝑥 minus five equals zero, so that’s a quadratic equation that we’ve got to solve, and that’s gonna be a little bit easier than working with all those fractions.

Alright then, let’s solve this quadratic equation. Well first of all, let’s do nine times negative five, which is negative 45. Now what we need to do is find all the factors of negative 45. Well obviously, negative one times 45 is negative 45, or one times negative 45 is negative 45, likewise negative three times 15 or three times negative 15 and lastly negative five times nine or five times negative nine.

Now we need to find which of those factor pairs sum to negative four, the coefficient of the 𝑥 term there. Well negative one plus 45 is 44, so that’s not it. Likewise, one add negative 45 doesn’t sum to negative four; negative three plus 15 doesn’t sum to negative four; minus five plus nine is positive four. But five plus negative nine is negative four, so I’m gonna split that negative four 𝑥 into two terms: five 𝑥 and negative nine 𝑥.

So all I’m doing is rewriting this middle term here, this 𝑥 term, as five 𝑥 take away nine 𝑥. Five 𝑥 take away nine 𝑥 is negative four 𝑥. Now I’ve got a choice here. I could either write five 𝑥 take away nine 𝑥 or I could take negative nine 𝑥 add five 𝑥. In fact, that’s the way round I’m gonna do it.

In fact, it doesn’t actually matter which order you write those two things in; you’re still gonna get the same answer at the end of the day, but when we look at the next stage of the process, we can see why I’ve written them round this way. I’m gonna treat this expression here as two separate expressions, which I’m gonna factor separately, so I’m gonna factor the nine 𝑥 squared minus nine 𝑥 and the five 𝑥 minus five separately.

So nine 𝑥 squared minus nine 𝑥, they’ve got a common factor of nine and they’ve also got a common factor of 𝑥. So I’m gonna factor out nine 𝑥. Now nine 𝑥 times 𝑥 would give me nine 𝑥 squared, and nine 𝑥 times negative one would give me negative nine 𝑥.

Now factoring the second half of that, I’ve got five 𝑥 minus five. Well they’ve both got a factor of five so- but positive five I’m gonna factor out, so positive five, and I’d need to multiply that by 𝑥 to get positive five 𝑥 and negative one to get negative five. And of course all of that is equal to zero.

It’s important when you factor those two terms that you aim to get the same contents of these parentheses here because that’s gonna be the common factor between those two whole terms and that’s what you’re gonna factor out in the next stage.

So 𝑥 minus one is a common factor of nine 𝑥 times 𝑥 minus one and positive five times 𝑥 minus one, and so 𝑥 minus one times nine 𝑥 gives us nine 𝑥 times 𝑥 minus one and 𝑥 minus one times positive five gives us positive five times 𝑥 minus one. And of course that’s equal to zero.

Now let’s just take a little diversion to see what would have happened if we had have written the nine 𝑥 and the five 𝑥 terms the other way round. We would have had nine 𝑥 squared plus five 𝑥 minus nine 𝑥 minus five equals zero.

Now when we factored that first half of the expression there, we’d- the only common factor we get is 𝑥, so we’re gonna have 𝑥 times nine 𝑥 plus five. Now it gets a little bit more tricky here because we said we wanted to make sure we had a common factor of nine 𝑥 plus five. So we’d write nine 𝑥 plus five and then we’d have to work out what factor do I need to take out of negative nine 𝑥, negative five to get nine 𝑥 plus five as a factor there.

Well I’d have to factor out negative one. So with negative one lots of nine 𝑥 plus five gives us negative nine 𝑥 minus five. Now I’ve got nine 𝑥 plus five as the common factor, and in the first instance I’m multiplying that by 𝑥 and in the second instance I’m multiplying it by negative one.

So we’ve got the same terms as we’ve got here, but just the other way round. And of course, those things are still equal to zero. Now the reason I chose to write it this way round instead of this way round in the first place was because I could see that I can factor out a nine and an 𝑥. I was doing as much factoring as early as possible, which makes the rest of the algebra a little bit easier to see on that left-hand side. It doesn’t matter which version you go for; they both come up with the same answer.

Now just to finish off, in order for 𝑥 minus one times nine 𝑥 plus five to equal to zero, either 𝑥 minus one is equal to zero or nine 𝑥 plus five is equal to zero, and that means either 𝑥 is equal to one or 𝑥 is equal to minus five-ninths. So our answer is either 𝑥 equals one or 𝑥 equals negative five-ninths, two possible different answers.

And again, if you had time, you could go back and substitute both of those values for 𝑥 in that original equation; you just check that the left-hand side does equal the right-hand side. For now, just take it from me that, in this case, they do.