Question Video: Solving Equations Involving Rational Functions

Video Transcript

Solve 𝑦 equals two 𝑥 plus one over three 𝑥 plus four with an expression for 𝑥 in terms of 𝑦.

And the equation we were given in the question gives us 𝑦 in terms of 𝑥: 𝑦 on its own on the left-hand side and some expressions involving 𝑥 on the right-hand side. So the question is asking us to take that equation and rearrange it so that we got 𝑥 on its own on the left-hand side and some 𝑦 terms on the right-hand side.

Now, the first thing we notice is that the 𝑥 variable appears in this denominator. And we should also remember that all the terms on the numerator need to be grouped together and all the terms on the denominator need to be grouped together. Now, I want to eliminate that denominator from the right-hand side. And I can do that by multiplying the right-hand side by the multiplicative inverse of one over three 𝑥 plus four because one over three 𝑥 plus four times three 𝑥 plus four over one is equal to one.

Basically, dividing the top by three 𝑥 plus four gives us one. Dividing the bottom by three 𝑥 plus four plus four gives us one. So we’ve got one times one over one times one, which is just one. So these two terms cancel each other out. But the problem is we’ve now unbalanced our equation. We’ve multiplied this side by three 𝑥 plus four over one, so it’s no longer equal to this side. We need to use the multiplication property of equality. And that means that if we multiply one side of the equation by something, we need to multiply the other side of the equation by the same thing if that remained equal.

Now, that means we’ve got to multiply the left-hand side by three 𝑥 plus four over one as well. Now, when we do that, we’ll notice that 𝑦 doesn’t have a denominator and that three 𝑥 plus four over one is just the same as three 𝑥 plus four. And this means on the left-hand side, we’ve got 𝑦 times three 𝑥 plus four. Now, we can use the distributive property on the left-hand side of the equation to give us 𝑦 times three 𝑥 and 𝑦 times four.

Now, 𝑦 times three 𝑥 can be written as three 𝑥𝑦. And 𝑦 times four can be written as plus four 𝑦. And of course that is just equal to two 𝑥 plus one on the right-hand side. Now, let’s just remember what we’re trying to do. We’re trying to isolate the 𝑥 on the left-hand side of our equation. A good next step will be to get rid of the two 𝑥 on the right-hand side. And I can do that using the subtraction property of equality; I can subtract two 𝑥 from both sides.

And then on the right-hand side, two 𝑥 plus one take away two 𝑥. Well, two 𝑥 take away two 𝑥 is just nothing. So that just leaves us with positive one. And if we look at the left-hand side, there weren’t any simple 𝑥 terms that we could subtract two 𝑥 from. So we’ve got three 𝑥𝑦 plus four 𝑦 minus two 𝑥.

Now, we need to leave ourselves with only terms involving 𝑥 on the left-hand side. So we’re gonna need to try and eliminate this 𝑦 term from the left-hand side. And we can do that again using the subtraction property of equality and subtracting four 𝑦 from both sides. And that means on the left-hand side, three 𝑥𝑦 plus four 𝑦 minus two 𝑥 minus four 𝑦. Well, the four 𝑦 take away four 𝑦 gives us nothing. So we’re just left with three 𝑥 𝑦 minus two 𝑥. And on the right-hand side, we’ve just got one minus four 𝑦.

So now we’ve got all of the 𝑥 terms on the left-hand side and no 𝑥 terms on the right-hand side. But we only want 𝑥 terms on the left-hand side, and there’s an 𝑥𝑦 term there at the moment. But remember three 𝑥𝑦 means three times 𝑥 times 𝑦 and two 𝑥 means two times 𝑥. So we can see we’ve got a common factor of 𝑥 in both of those terms. So we can use the distributive property to factor 𝑥 out of both of those terms. So that means we’ve got 𝑥 times three 𝑦 minus two is equal to one minus four 𝑦.

Now remember one over three 𝑦 minus two is the multiplicative inverse of three 𝑦 minus two. So if I multiply both sides by one over three 𝑦 minus two, then these two things cancel out to give us one. So the left-hand side just becomes 𝑥. And on the right-hand side, I just got one minus four 𝑦 times one. So that’s just one minus four 𝑦 over three 𝑦 minus two. And in fact, I don’t even need those parentheses there. And there we have our answer: an expression for 𝑥 in terms of 𝑦. 𝑥 is equal to one minus four 𝑦 over three 𝑦 minus two.

Now just before we go, I just wanna mention if we’d made some slightly different choices along the way, we’d have ended up with a different version of our answer. For example, in this stage here, if we’d have subtracted three 𝑥𝑦 from both sides rather than two 𝑥 from both sides, we’d have gathered all of the 𝑥 terms over on the right-hand side of that equation. Then we could have subtracted one from both sides to eliminate this term, then factored out this 𝑥 again using the distributive property, and finally multiplied through by the multiplicative inverse of two minus three 𝑦 to give us an answer of 𝑥 is equal to four 𝑦 minus one over two minus three 𝑦.

Now, this looks very similar to the answer that we got over here. But we’ve got four 𝑦 minus one on the numerator instead of one minus four 𝑦 and we’ve got two minus three 𝑦 on the denominator instead of three 𝑦 minus two. So which one of our two answers is correct? Well, they’re both right; they’re both equivalent versions of the same answer.

Look if I multiplied both of those answers by one, I think you’ll agree I’m not changing either of them. But in fact, if I multiplied this one by a different version of one, negative one divided by negative one. Negative one divided by negative one is just one. And then I use the distributive property to multiply the terms on the numerator by negative one and likewise the terms in the denominator. Then negative one times one is negative one; negative one times negative four 𝑦 is positive four 𝑦. So the denominator is negative one plus four 𝑦. And negative one times three 𝑦 is negative three 𝑦, and negative one times negative two is positive two. So the denominator is negative three 𝑦 plus two.

Then, if we remember the addition is commutative, it doesn’t matter which order I add these things together in, so negative one plus four 𝑦 is the same as four 𝑦 plus negative one or four 𝑦 take away one. And on the denominator, negative three 𝑦 plus two is the same as two add negative three 𝑦 or two take away three 𝑦. So whether you got this answer or this answer, we can see they amount to the same thing.

Now, it’s well worth remembering this trick of multiplying your answer by one — well, the version of one that is negative one over negative one, if you need to rearrange it into a slightly different format. Sometimes, you’ll be given different types of questions. For example, “show that 𝑦 equals two 𝑥 plus one over three 𝑥 plus four” can be expressed as 𝑥 equals four 𝑦 minus one over two minus three 𝑦.

So if your working out gives you the answer in this format, you might need to use this technique in order to rearrange that answer into the format that they’ve asked for in the question. So even if it seems a little bit strange at first, I would practice using this technique on some questions of your own.