Lesson video

Thank you for choosing to learn using this video.

My name is Miss Davis, and I'm gonna be helping you as you work your way through this lesson.

We're looking at a lot of our algebra skills, we're also looking at our fraction skills.

You are definitely gonna want to write things down as we go through.

You learn best in mathematics when you give things a go.

So make sure that you've got some paper and a pen to hand so that you can try things out.

Of course, I'm gonna help you as we go through.

So if there's bits that you're not sure about what the next step would be, have a think yourself, but then watch the video and I will show you how I would do these questions.

As with a lot of things in algebra, there are different ways of doing things.

Just because your way is not the same as my way doesn't mean that you are doing it wrong.

So bear that in mind as you're exploring some of these concepts.

Let's get started then.

Welcome to this lesson where we're gonna be checking and securing your understanding of solving equations with simple algebraic fractions.

Few key words that we're gonna use today.

We're gonna talk lots about what a solution is.

We're also gonna use this word reciprocal.

If you wanna have a quick reminder, pause the video and read through them now.

We're gonna start by solving simple equations with fractions.

Now you'll have seen equations like this before.

We're just gonna have a reminder of how we solve these so we can apply them to questions where we have fractions.

So we have a look at a bar model for 3a equals 21.

In order to solve for a, we can split the bar into three sections, and what we've done there is we've isolated the value of one positive a.

There's two ways you can think about what we've done.

You could say that we have divided the bottom bar by three, but this is the same as multiplying by the reciprocal of three, which is a third.

So you could say we've multiplied the bottom bar by a third.

Let's look at that when we have our equations, we've got 3a equals 21, the reciprocal of three is a third.

So we can multiply both sides of the equation by a third to maintain equality.

3a times a third is just a, 21 times a third is seven, so we get eight equals seven.

I want you to think carefully about that method of working out and our steps for that so we can apply it to some trickier questions.

Okay, well let's have a look at when the coefficient of the variable is a fraction.

So I show you this bar model.

We could write this equation as a half x equals four, or equally X over two equals four.

A half of x and x over two are the same thing.

Now what could we to work out the value of one positive x? Well, if I know that a half x is four, if I multiply that by two, I would get what one x is and that's because a half and two are reciprocals of each other.

Let's see what that looks like as an equation.

So a half x equals four, we multiply by the reciprocal.

So half x times two equals four times two to maintain a quality.

So x equals eight.

So how could we use that idea to solve the equation? A fifth x equals negative 1.

2.

Well the reciprocal of a fifth is five, so we multiply both sides of the equation by five.

A fifth x times five is just x, negative 1.

2 times five is negative six.

So we've got x equals negative six.

Right quick check then.

I'd like you to match the equations on the left to the equivalent equations on the right, which you're gonna show us the solution.

Give that a go.

Let's have a look.

We've got a third x equals nine.

So we need to multiply both sides of the equation by the reciprocal of a third, which is three.

So nine times three.

X over nine is the same as writing a ninth of x.

So we need to multiply both sides by nine.

For C, we need to multiply both sides by a third 'cause that's the reciprocal of three.

For D, we're gonna multiply both sides by one ninth.

And again, we're gonna multiply for the last one, both sides by one ninth, but this time we've got nine multiplied by one ninth.

Jun and Sofia are solving the equation a third x plus four equals seven.

Jun reckons the reciprocal of a third is three.

So we can multiply by three first.

Sofia thinks we should subtract four first.

What do you think? All right.

Both options are valid as long as the quality is maintained by performing the same operations to both sides of the equation.

We're gonna look at both options.

So whichever method you decided you would use is absolutely fine 'cause either method is going to work.

The one thing you've gotta be careful of is that we're doing the same thing to the whole expression on both sides of the equation.

So for Jun, that's gonna mean multiplying all the terms by three.

So it's just gonna have to be a bit careful about that.

Let's have a look how that works.

So if we're gonna multiply the whole expression on the left hand side by three, we need to multiply a third x by three and four by three.

We also need to multiply the right hand side by three.

So we get x plus 12 is 21.

Now we can subtract 12, and x is nine.

Right.

Sofia said we should subtract four first.

So let's look at that method.

So subtracting four from both sides gives us a third x equals three, and now we can multiply by the reciprocal of a third, which is three.

So a third x times three equals three times three, therefore x equals nine.

Just pause the video, and how I think about which method you preferred.

Make sure you can follow all of those steps.

So how could we solve the equation x plus four over three equals seven? Sofia says, I don't think subtracting four makes sense this time.

Well, let's have a look.

The equation could be written as x over three plus four over three equals seven.

If you think about when we add fractions, if they have a common denominator, we can add the numerator.

So x over three, add four over three is the same as x plus four over three.

Equally, if we think about taking out a factor of a third, we could write that as a third lot of x plus four and that would give us a third of x plus a third of four or four over three.

So those things are all equivalent.

So Sofia is perfectly correct, subtracting four is not gonna help us find the solution.

I'd like you to think about this one.

What could we do instead? Okay, well if we split it into two fractions like we have on the left hand side, we could subtract four thirds and then we'd know what x over three is and then we can multiply both by three.

If we looked at the second way of writing that, we can multiply both sides by three first 'cause it's the reciprocal of a third, then we'd know what x plus four is and we could subtract four.

Jun reckons that multiplying the original equation by three first is gonna be easier this time.

I wonder if you agree.

Well let's look at this original equation.

If we multiply both sides by three, the left hand side is gonna become x plus four, and the right hand side will be 21.

Now it's an easy one step equation to find out what x is.

I wonder if you agree with Jun that that was probably the quickest method.

So as with all equations, we could check by substituting.

So let's do a quick check.

Is 17 plus four all over three equal to seven? Well that would be 21 over three and that is seven.

So there are often choices of methods when solving equations.

So Andeep is trying to solve this equation.

I want you to have a read through his steps.

Can you explain what he has done at each step and then is there anything that you would've done differently or would you have made the same decisions as Andeep? Give it a go.

Let's look at what he's done then.

So he started by multiplying both sides by five because the fraction has a denominator of five.

We multiply both sides by five, we get an equation without any fractional terms. Then he's decided to divide both sides by two, and then he's added three and he's done that all in one step.

You might have felt that you would've wanted to write out the fact that three is equivalent to six over two and then done five over two, add six over two.

That would've been absolutely fine.

So I wonder if you said something like this, Sofia liked Andeep's method but she would've expanded the brackets first.

Now that's not necessarily easier, it's just another way to do it.

Right.

I'm gonna offer an alternative.

Two dots of x minus three over five can actually be written as two fifths times x minus three.

So what could we do now to efficiently solve this equation? You think about how we could use the fact that this bracket now has a coefficient of two fifths to solve our equation.

All right.

Well what we can do is we can multiply by the reciprocal of two fifths.

Of course the reciprocal of two fifths is five over two.

So if we multiply both sides of the equation by five over two, we get left with x minus three on the left because of course two fifths times five over two is one 'cause they're reciprocals.

So we get left with one lot of x minus three and then five over two on the right hand side.

Now we can add three to both sides.

Again, you might wanna write three as six over two.

Then we've got five over two plus six over two is 11 over two.

Now essentially what we've done is exactly the same as what Andeep did, but we did it in one step whereas he did it in two.

So Andeep multiplied by five but then divided by two.

What we've done is multiplied by five over two.

Okay, let's see if you can apply this then.

So which of these would be a sensible first step to solve the equation? Two ninths of y equals negative four.

What do you think? Alright, I'm happy if you went for either multiply by nine, or multiply by nine over two.

Let's look at the differences between the methods.

If you multiply both sides by nine, you would've got two y equals negative 36 and then you could do a second step of dividing by two or multiplying by half, to get y equals negative 18.

If you decided to multiply both sides by nine over two, because nine over two is the reciprocal of two nights, two nights y times nine over two gives us one y and then we just have to do negative four times nine over two, which is negative 18.

It's entirely up to you whether you want to do in the two individual steps or whether you are happy just multiplying by the reciprocal in one step.

Alright, time to have a practise then.

So you have got eight equations and eight solutions.

I'd like you to match the equations to their solutions.

See if you can solve using reciprocals first and then don't forget, you can check your answers by substituting them back into the equation.

Give that a go, come back when you're ready for the next bit.

Well done.

We focus lots now on these different methods and different choices you can make when solving an equation.

So I'd like you to have a look at these two different methods to solve this equation.

See if you can fill in those missing parts, give that a go, come back when you're ready for the next bit.

And finally, six equations, you are free to solve them in any way that you like.

When you think you've got those solutions, we'll have a look at the answers together.

Right.

Good effort guys.

And I'm hoping that you spent some time checking your answers back using substitution.

Just pause the video and check you've got the correct solution for each.

Well done.

Let's look at these different methods then.

So the first method we've multiplied by the reciprocal of five thirds, 'cause five thirds is the coefficient of the bracket.

The reciprocal of five thirds is three fifths.

Now 10 times three fifths.

So you can multiply 10 by three and divide by five or divide by five, then multiply by three.

Either way we get six.

Then we can add three to both sides and divide by two.

I'm just gonna leave that as x equals nine over two at the moment.

If you wrote it in a different format, that's absolutely fine.

Right.

Let's look at this right hand side.

So for this right hand side, we've rewritten that as five lots of two x minus three all over three.

We've got to multiply two x minus three by five and divide it by three.

In this method we've decided to expand the brackets, so we should have negative 15 and still all divided by three.

Now the easiest next step would be to multiply both sides by three.

So we'd have 10x minus 15 equals 30.

We could then add 15 to both sides.

And if 10x is 45, x is 4.

5.

Now of course that's the same as our answer for our first step, 'cause nine over two is the same as 4.

5.

You might wanna pause and just have a think about which method you preferred and what you would've done if you had been faced with that question.

And finally, you could have solved this in any way you liked.

You might wanna pause the video and read through my steps.

I have tried to come up with the most efficient method possible.

You might want to just look and see if maybe you could have been more efficient in some of your steps.

I want to just draw attention to F, I've decided to subtract 10 from both sides first.

Then because I've got a coefficient of negative three fifths, if I multiply both sides by negative five over three, then I'm gonna get what x minus five equals and that just shortens some of those steps.

That being said, it's absolutely fine if you want to add steps and do things in a way that suits you, pause the video when you're happy, we'll move on to the next part of the lesson.

Well done.

We're starting to gain a lot of confidence now with working with fractions.

So now we're gonna have a look at fractions where the variable term is in the denominator.

So let's have a look at example.

So we've got 12 over a equals four.

Now you might already know what a is gonna be in this case, but we want to think about methods that work so we can apply this to other equations in the future.

So Izzy says that reciprocal of 12 over eight is a over 12.

Can we use that? Well, we know, a value and its reciprocal have a product of one.

So if I do 12 over eight multiplied by a over 12, that's gonna give me one.

Of course to maintain a quality we need to multiply the other side by a over 12 as well.

So it gives us one equals 4a over 12.

Now this looks similar to some of the equations we looked at a moment ago.

So we could multiply both sides by 12 and then divide by four.

Of course we could simplify that fraction 'cause 4a over 12 does simplify whatever method you like.

I'm gonna do it by multiplying both sides by 12 and then dividing both sides by four.

So we get a equals three.

Alright, we're thinking about efficiency as always.

Alex says, I think it would've been more efficient to just multiply both sides by a.

Pause the video, do you think Alex's idea will work? So let's think about what 12 over a means.

12 over a is the same as 12 lots of one over a or 12 times one over a.

So yes by A is a perfectly reasonable step because a is the reciprocal of one over a.

So we multiply both sides by a one over a times a is one.

So we just have 12 on the left hand side and four A on the right hand side.

We got to this step in the last method as well.

Now we can divide both sides by four and that gets us a equals three.

And Alex is perfectly correct, of course that makes sense.

12 divided by three is four.

We can substitute into check as always.

Okay, we're gonna have a go at doing these together.

So I'm gonna show you one on the left hand side and you are gonna try one on the right hand side.

We're gonna bring in some of our negative number skills here as well.

So 14 over y equals negative two.

If I multiply both sides by y because it's the reciprocal of one over y, I get 14 equals negative two y, and then I just want to divide both sides by negative two.

14 divided by negative two is negative seven.

And let's check by substituting 14 divided by negative seven.

Is that negative two? Yes it is.

Time for you to have a go at the one on the right hand side and then can you check your answer by substituting.

Give it a go.

Let's have a look.

We multiply both sides by b, we get negative 15 equals 3b, and then divide both sides by three.

So b is negative 15 over three, or negative five.

Let's check by substitution, negative 15 divided by negative five, is that three? Yes it is.

All right, we're gonna build on that a little bit more now.

So Alex is trying to solve two over 5x equals four.

Ooh is two over 5x the same as two fifths of x? Now Izzy's got a good suggestion.

I find trying things out with numerical values can help me see equivalent expressions.

Yeah, that's a really good tip.

If ever you're not sure how to manipulate something with algebra, or you're not sure if two things are equivalent, you could try substituting in a number and seeing if you get the same value.

So let's see if x was two, then two over 5x would be the same as two over 10, or a fifth.

If x is two, then two fifths of x will be the same as two fifths of two, or two fifths times two.

Of course that's four fifths.

So no they are not the same thing and we'll see why in a moment.

So two over 5x is actually equivalent to two lots of one over 5x or two times one over 5x.

Right, Alex is thinking again, now the reciprocal of one over 5x is 5x.

So let's see how we can use this.

We multiply both sides by 5x, we get two equals 20x and x equals two over 20 or a 10th.

Izzy's gonna do a bit of substitution for us.

So two over five, lots of a 10th is the same as two over a half, and of course two divided by a half is two times two, which equals four.

So we have checked that that works.

You may want to just pause the video and follow through each of those steps.

So which of these are equivalent to three quarters of a? Think carefully and then we'll check your answer.

Well done if you said it was 3a over four, that's the same as three quarters of a.

How about this one? Which of these are equivalent to five over 3y? There was two correct answers.

Well done if you spotted both, we could have five lots of one over 3y, would give us five over 3y, or five thirds times one over y.

Remember we multiply fractions, we can multiply the numerator and the denominators.

So Jun has noticed that for all these examples we multiplied by the denominator of the fraction.

Now that is always going to work because that is multiplying by the reciprocal of the unit fraction.

I'll show you what I mean now.

So for any fraction, a over x, but a and x could be any terms you like.

A over x times x is gonna give you a.

Having said that for some equations it can be more efficient to add or subtract a term before multiply.

So what do you think to this one? Eight over 3x minus five equals 11.

What would be a sensible first step? Now we absolutely could multiply by 3x as long as we do that to every term.

So we'd end up with eight subtract 15x equals 33x.

That'd be absolutely fine to do, but it might not be the most efficient first step.

Well done if you suggested we could add five to both sides.

I'm gonna show you how this works and then you are gonna give it a go.

So if I add five to both sides, that just reduces the number of terms I have in my equation.

So we've got eight over 3x equals 16, and now we can multiply both sides by 3x.

I get my final answer as x equals a six.

Have a go at this one.

So if we subtract nine, multiply both sides by 2x and divide by negative four, get a final answer of x equals negative three.

Our final challenge then is to look at what would happen if there were multiple terms in the denominator.

For example, 20x minus three equals five.

But we can multiply by the denominator just like the other examples.

The method we are using does not change.

Jun's getting really confident with his algebra manipulation.

Of course we must remember the brackets because we need to make sure we multiply by the whole denominator.

And the whole denominator is the expression x minus three.

So we can write that as 20 equals five lots of x minus three.

The most efficient method would be to divide both sides by five and then add three.

Which of these is a correct first step to solve 24 over p plus six equals three? What do you reckon? Hopefully you're thinking about what Jun just reminded us of.

We need to multiply three by the whole expression p plus six.

So 24 equals 3p plus 18.

Time for you to have a go.

'Cause we're working with our algebraic fraction skills, we must have a really solid understanding of what these terms mean.

So if this first activity, I want you to match up any equivalent expressions.

There's a grid on the right hand side for you to record your answers.

It may be possible for multiple expressions to be equivalent.

So they don't just match up in pairs, there might be three, there might be four that match up.

You might find there's some that don't match with any.

That's absolutely fine.

Once you're happy with that, move on to the next bit.

Well done.

This time I'd like you to have a go at solving these six equations.

Give it a go.

Well done.

I'd like you to pause the video and check that you've got these all matched up correctly.

Does make a difference as to whether the variable is part of the numerator or the denominator.

They're gonna be very different expressions.

So check that you're happy with those and then we'll look at the answers to the next bit.

Well done.

So for the first one, I've multiplied both sides by a, and then divided both sides by three.

So a equals seven, and then I could check that works.

21 divided by seven is three.

For the second one, I've chosen to do this in two steps.

So if I multiply by 2a, that gives me 21 equals 6a.

So a equals 21 over six, which simplifies to seven over two.

You might have spotted there that we've doubled the denominator of our fraction, which is gonna half our value for a.

For C, if I multiply both sides by a, and divide both sides by negative five.

And of course 10 divided by negative five is negative two.

Well done for remembering your negative number skills.

For D, I've subtracted six from both sides, multiplied by the denominator, which is a, so three equals 9a, so a is three over nine or a third.

For E, again, I've subtracted 10 from both sides so that I've got less terms. So three over 4a is negative three, multiplying by 4a gives me three equals negative 12a, and then a is negative a quarter if I simplify.

Now of course, other methods that you can use, you might have done steps in slightly different order.

You could have simplified that second step by dividing both sides by three if you want it.

And finally, I want to multiply by that entire denominator.

So I've got three lots of a plus six in bracket, which is 3a plus 18.

I've decided to expand my brackets because eight isn't divisible by three and I'd have to write that as a fraction to expand in my brackets first and then subtracting 18.

So 3a is negative 10, and a is negative 10 over three.

Good work on those ones guys.

We've looked at a range of different types of questions now and we've shown that just because an equation has a fraction in it, it doesn't make it trickier, we just have to think about what steps we need to do to solve.

When you are working with fractures in the future, have a think about how you can use a reciprocal, and of course, when we're solving any equation, we've reminded ourselves that we need to maintain equality by doing the same thing to both sides.

Thank you for joining me today and I look forward to seeing you again.