# Video: Solving Linear Equations With the Unknown on a Numerator

A detailed explanation of how to solve linear equations that involve the unknown in the numerator of a fraction. For example, 𝑥/7 = 4/5, (21𝑐)/4 = 6 + 3𝑐, 2/3 + 𝑦/9 = 4/5, or (𝑒 + 12)/20 = 11/15.

16:29

### Video Transcript

Before watching this video, you should make sure that you know how to solve basic linear equations. And we’ve got other videos that will show you how to do that. So what we’re gonna cover in this video is solving linear equations involving fractions with the unknown as the numerator. So stuff a bit like this where we’ve got the unknown, in this case the letter 𝑐, but it’s involved in a fraction and it’s on the top of the fraction. In this particular case, we’ve got a 𝑐 over here as well, but we’re gonna look at a range of different examples.

So let’s get on and do some examples. Now the first one is solve 𝑥 over three is equal to twelve. Now the general approach in these is to look for reciprocal operation. So here we’ve got 𝑥 divided by three. The reciprocal operation would be two multiplied by three. So we’re gonna use those sort of approaches to untangle these equations and solve to find the value of 𝑥. So as we’ve said with this one, 𝑥 divided by three is equal to twelve. The opposite of dividing by three is multiplying by three. And we’re gonna multiply both sides of our equation by three, so we’re doing the same thing to both sides of the equation means that it’s still equal. Now on the left-hand side, we’ve got a fraction times three. So I’m actually gonna express three as a fraction, three over one, because that makes it a little bit easier to see what’s going on. Now I’m going to cancel the three, so three divided by three is one and three divided by three is one. So that left-hand side of the equation just becomes 𝑥 times one over one times one, in other words just 𝑥. At the right-hand side, twelve times three is thirty-six. So that was only really a one stage process. We just had to multiply by three to get rid of that fraction, so our answer is 𝑥 equals thirty-six. And we can always check back with our original question; 𝑥 is thirty-six, thirty-six divided by three, yep that’s twelve. So yeah, we think we’ve got the right answer.

Now number two, little bit of a step-up. Solve 𝑥 over seven equals four over five. So now we’ve got fractions on both sides of our equation. Now looking at the 𝑥 term, we’ve got 𝑥 divided by seven, and the reciprocal operation for dividing by seven is multiplying by seven. So in order to get 𝑥 on its own, just one 𝑥, we’re gonna have to multiply by seven. And we need to balance the equation by doing the same thing to both sides. Now this time around, both sides of my equation are a fraction. So the version of seven that I’m gonna multiply by is seven over one. I’m gonna make that into a fraction as well, and then that enables us on the left-hand side to cancel out. Seven divided by seven is one, seven divided by seven is one, and I can’t cancel anything on the right-hand side. So on the left, I’ve got 𝑥; and on the right four times seven is twenty-eight and five times one is five. So 𝑥 is twenty-eight over five as a top heavy number or five and three-fifths as a mixed number. So our answer then is 𝑥 is five and three-fifths.

Now number three, solve five 𝑥 over eight is equal to a half. Well we’ve got fractions on both sides of the equation, and 𝑥 isn’t on its own on the numerator. This time, we’ve got five times 𝑥. So again we’re gonna need to use reciprocal operations to get rid of that denominator from the fraction involving 𝑥. So the opposite of dividing by eight is multiplying by eight. So again I’m gonna multiply both sides of my equation by eight. But the version of eight that I’m gonna multiply by is eight over one, because they’re both fractions. And then on the left-hand side, eight divided by eight is one, eight divided by eight is one, so I’ve got five 𝑥 times one over one. So that’s just five 𝑥 on the left-hand side. And on the right-hand side, I’ve got a two on the denominator, so that divided by two to make one. And I’ve got eight on the numerator, so that’ll divide by two to make four. So I’ve got four over one. That’s just four. So five times 𝑥 is equal to four, so we need to think about reciprocal operations again. The opposite of multiplying by five is dividing by five, so I’m gonna divide both sides of equation by five. So dividing the left by five, I’ve got five 𝑥 over five. And dividing the right by five, we’ve got four over five. So the whole point of doing that was so that I could say, oh look, the bottom is divides by five. Five divided by five is one. And the top divides by five, five divided by five is one. I’ve got one times 𝑥 over one, which is just 𝑥. And I can’t do any more cancelling at the right-hand side, so 𝑥 is equal to four-fifths. That’s our answer.

Now onto the next question. Solve twenty-one 𝑐 over four equals six plus three 𝑐. Now there’s a couple of differences here: one, we’ve got a 𝑐 on both sides of our equation; and two, we’ve got two different terms on the right-hand side. So it’s slightly a more complicated example than the previous one. And my first target is to get rid of that fraction involving 𝑐, so I ought to think about reciprocal operation. So the reciprocal operation to dividing by four is multiplying by four, so I’m gonna multiply both sides of my equation by four. And it’s important because I’ve got two terms over on the right-hand side that I multiply both of those by four. I have to multiply everything on both sides by four. And on the left-hand side, I’m gonna change that four into a four over one. We’re gonna put it in its fraction format because then that enables us to cancel out these two fours, so four divided by four to make one, four divided by four to make one. So on the left-hand side, I’ve just got twenty-one 𝑐. So I’ve successfully got rid of that fraction. And the other terms, six times four is twenty-four. And three 𝑐 times four, four times three 𝑐 is twelve 𝑐.

Now I’ve got twenty-one 𝑐 on the left-hand side. I’ve got twelve 𝑐 on the right-hand side. If I do the reciprocal of adding twelve 𝑐, if I take away twelve 𝑐 from both sides of that equation, I’ll end up with a positive number of 𝑐’s over here. And I’ll end up with no 𝑐’s over here. So I’ve taken away twelve 𝑐 from the left-hand side and twelve 𝑐 from the right-hand side. So on the right-hand side, twelve 𝑐 take away twelve 𝑐 is nothing, so those two things cancel out. And on the left-hand side, twenty-one 𝑐 take away twelve 𝑐 leaves me with nine 𝑐. So I’ve got down to nine 𝑐 is equal to twenty-four. So again think reciprocal operations. And nine times 𝑐, the opposite of timesing by nine is dividing by nine, so I’m gonna divide both sides of my equation by nine. On the left-hand side, obviously, they cancel out. Nine divided by nine is one, nine divided by nine is one, so I’m just left with 𝑐. And on the right-hand side, twenty-four and nine are both divisible by three. So twenty-four divided by three is eight and nine divided by three is just three, so I’ve got eight over three. And that’s a top heavy fraction. If I convert that to a mixed number, I get two and two-thirds. So that’s our answer: 𝑐 is equal to two and two-thirds.

Now number five, find the value of three 𝑦 given that two-thirds plus 𝑦 over nine is equal to four-fifths. Well there’s a couple of things we need to be wary of here. Obviously the equation’s a little bit more complicated. We’ve got three fractions spread over both sides of the equation and it’s asking us to find the value of three 𝑦. So we’re not just solving for 𝑦, we’re trying to then find the value of the expression three 𝑦. So it’s gonna be a two-stage process. Now I’ve got a bit of a choice as to what I do first. I can either attack this term here and do the reciprocal operation of dividing by nine and multiply through everything by nine to get that 𝑦 out of a fraction, or I could do the fraction calculations involving two-thirds and four-fifths. So take away two-thirds from both sides and just evaluate that fraction and then go on to multiply by nine afterwards. Now in fact that’s the way I’m gonna approach this. Because if I multiply three by nine to start off with, I’m gonna end up with bigger numbers and more complicated fractions to work with. So subtracting two-thirds from both sides of the equation, on the left-hand side I’ve got two-thirds plus 𝑦 over nine take away two-thirds. And two-thirds take away two-thirds is nothing. So that just leaves me with 𝑦 over nine.

Now over on the right-hand side, I’ve got four-fifths minus two-thirds. Now in order to do fraction subtraction, we do need to have a common denominator. So I need to look at- we’ve got five and we’ve got three. They haven’t got any common factors other than one, so I’m gonna have to use the lowest common multiple of fifteen. So I’m gonna have to multiply the second fraction by- I’m gonna call it five over five. So five over five is just one, so I’m just multiplying this number by one. So I’m not changing the magnitude of the number, so it’s still two-thirds. But now if I did two over five, two times five over three times five, that’s ten over fifteen. So it’s an equivalent fraction to two-thirds, but it’s got a denominator of fifteen. And if I multiply four-fifths by three over three, that now will also have a denominator of fifteen. Again, three over three is one, so it’s an equivalent fraction but it’d be twelve over fifteen. So twelve-fifteenths take away ten-fifteenths just leaves me with two-fifteenths. So I’m looking at 𝑦 over nine is equal to two-fifteenths. Now I don’t wanna know what 𝑦 over nine is, I want to know what 𝑦 is. So the reciprocal operation for dividing by nine is multiplying by nine. So I’m gonna multiply both sides of that equation by nine. And in fact because they’re both fractions, I’m gonna do nine over one rather than just nine, put nine in its fraction format. Now on the left-hand side, the nines cancel which is why I did that. So I’ve just got 𝑦 — and on the right-hand side, nine divided by three to make three and fifteen divided by three to make five — so I’ve got two times three over five times one. So I’ve got 𝑦 is equal to six over five. But that’s not the end of the question remember, because we’re looking for the value of three times 𝑦, so we need to multiply that by three. So three 𝑦 is equal to three times six over five. So let’s put that into its fraction format again. So three times six over five, that’s eighteen over five, which is a top heavy fraction. Putting that into a mixed number, three 𝑦 is equal to three and three-fifths.

So onto number six, find the value of 𝑒 given that 𝑒 plus twelve over twenty is equal to eleven over fifteen. So now the 𝑒 is in the numerator, but it’s got another term in that numerator. So the numerator is 𝑒 plus twelve, so we’ve got a slightly more complicated situation than we’ve seen. Now my first top tip here is that we should always put parentheses around these numerators, because what I’m gonna do is use the reciprocal operations. We’re gonna multiply everything through by twenty. But it’s important when I do that, that I don’t just multiply the twelve by twenty or just multiply the 𝑒 by twenty, but I multiply the whole of that numerator by twenty. And again because they’re both fractions, I’m actually gonna change that to twenty over one to make life a little bit easier. So on the left-hand side, I can cancel the twenties down. Remember, that was the whole point of multiplying through by twenty, so that I could simplify the left-hand side and get rid of that fraction. So that just leaves me with one lot of 𝑒 plus twelve. So I can now get rid of the parentheses cause there aren’t any other terms on that left-hand side. So I’ve just got 𝑒 plus twelve, and then on the right-hand side I can do a bit of cancelling. Five is going to fifteen three times and five is going to twenty four times. So I’ve got eleven times four over three, which is forty-four over three.

Now remember I want to get 𝑒 on its own, so I’ve gotta do the reciprocal operation of adding twelve. That’s subtracting twelve from both sides of my equation. So on the left-hand side, I’ve got 𝑒 plus twelve take away twelve. They’re gonna cancel each other out. That just leaves me with 𝑒. And on the right-hand side, given that I’m doing fractions subtraction, I need to turn that twelve into a fraction. And to do fractions subtraction, I need to have common denominators. And the denominator of forty-four over three is three, so I’m going to multiply the twelve by one. And the version of one that I’m gonna multiply it by is three over three. So it’s still twelve, but it’s now a version of twelve, an equivalent fraction for twelve which has got three as the denominator. So I’ve got forty-four over three minus thirty-six over three, which means that 𝑒 is equal to eight over three. And as mixed number, that’s two and two-thirds. So our answer then is 𝑒 is equal to two and two-thirds.

Now in our final example, we’ve got find the value of 𝑚 given that m over three plus four-fifths is equal to two 𝑚 over seven. Now in this question, we’ve got a few terms. We’ve got 𝑚 on both sides of the equation, so we’ve got quite a bit of work to do. Now we’ve gotta do a couple of different approaches. We could either try to gather the 𝑚 terms together in one place. Or we could try to get rid of the fractions first and sort of undo these three and the seven on the denominators here to try to- and then gather the 𝑚 terms together afterwards. And in fact, that’s what I’m gonna do. So my first move then is gonna be to multiply through by seven in order to try to get rid of this particular denominator. And given that every single term is a fraction, I’m gonna multiply three by seven over one. That’s seven in its fraction form just to make things a little bit easier.

Now on the right-hand side, I’ve got seven, which can cancel with the seven. So on the right side, I’ve just got two 𝑚. And it doesn’t look like anything else cancels in any of the other fractions, so those first two terms are seven 𝑚 over three plus twenty-eight over five. So now I’m going to attack this fraction here, trying to get rid of this denominator, and I’m gonna multiply everything through by three. And again, because I’ve got fractions involved on the left-hand side, I’m gonna change that to three over one in each case. Still three, but it’s gonna make the work a bit easier. Now the first time, I can divide top and bottom by three, so they cancel out. So I’ve just got seven 𝑚. Second term can’t do any cancelling. So twenty-eight times three over five times one, but no cancelling goes on there, so that just leaves me with eighty-four over five. And on the right-hand side, two 𝑚 times three is six 𝑚.

So now I’ve got seven 𝑚 on the left, six 𝑚 on the right. So if I take away six 𝑚 from both sides of my equation, I’ll end up with all the 𝑚’s on one side. So on the right-hand side, I’ve got six 𝑚 take away six 𝑚, which just leaves me with zero. On the left-hand side, I’ve got seven 𝑚 takeaway six 𝑚, so that’s just one 𝑚. But I’ve still got my plus eighty-four over five term. So that leaves me with 𝑚 plus eighty-four over five is equal to zero. Well I want to know what 𝑚 is equal to, so I’m gonna need to do the reciprocal operation, and the opposite of adding eighty-four over five is taking it away. So I’m gonna have to subtract that from both sides of my equation. And when I do that, I’ve got eighty-four over five take away eighty-four over five on the left-hand side, so they cancel out. So I’ve just got 𝑚. And zero take away eighty-four over five is negative eighty-four over five, so 𝑚 is equal to negative eighty-four over five. Now negative eighty-four over five, perfectly good answer, but it’s a top heavy fraction, so I’m gonna convert that to a mixed number; 𝑚 is equal to negative sixteen and four-fifths. And that’s our answer.

Now to summarise what we’ve just been doing, we were using reciprocal operations to eliminate fractions from our equations. We were gathering terms involving the unknown variable. We were simplifying fractions where we could to make our calculations easier. And lastly, remember to always check that you’ve answered the actual question. So for example, we had a question that asked us to find the value of three 𝑦, and if we have only found the value of 𝑦, then we wouldn’t have got full marks for our question.