Lesson video

Well done for taking that first step and choosing to load this video to learn today.

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

Lots of skills that we're gonna be using, so feel free to pause the video and give things a go, give yourself time to think about what you are doing, and then I'll help you out by adding any hints, any tips as we work our way through.

Let's get started then.

Welcome to this lesson on solving equations with algebraic fractions.

We're gonna bring together lots of skills today.

We're gonna look at simplifying algebraic fractions, as well as, adding, subtracting, and multiplying algebraic fractions.

So some of our equations today are gonna produce quadratic equations.

Many will be able to solve by factorising.

However, if we can't solve by factorising, we might need to use the quadratic formula.

If you want a reminder of what the quadratic formula is, pause the video and have a read through now.

We're gonna start by looking at equations with unknowns on one side.

So let's have a look at this equation.

Aisha says, "It would be easier if the left-hand side "was written as a single fraction." Let's look at how to do that then.

So the lowest common multiple of 4x and 6x is 12x.

So we multiply the numerator and the denominator of the first fraction by three and then the second fraction by two, we get a common denominator of 12x.

We can now combine our fractions to get 17 over 12x equals 2.

Now, 17 over 12x is equivalent to 17 times 1 over 12x.

So that means we can solve this by multiplying by the reciprocal.

You might have seen equations like this before.

Now, it's gonna be easier to multiply by the reciprocal of 1 over 12x, rather than the reciprocal of 17 over 12x.

The reciprocal of 1 over 12x is 12x.

To maintain equality, we can multiply both sides of the equation by 12x.

The left-hand side becomes 17 and the right-hand side, 24x.

Dividing both sides by 24, we get our solution.

It's important that we use brackets to ensure correct priority of operations.

Let's look at this example.

What would you do first? The most efficient first step is gonna be to combine the fractions by multiplying.

So you get 6 lots of x + 4, there are three lots of x + 1 = 4.

What could we do now? Now, we've got options.

You might have said, we can simplify by dividing the numerator and the denominator by three.

You might have said, expand the brackets or multiply by the reciprocal of one over three lots of x plus one.

We're gonna look at that final method for now.

So we multiply both sides by that denominator, we get 6 lots of x plus four equals 12 lots of x plus 1.

Jun says, we don't need to expand the brackets, we can divide both sides by six.

Now people often get into a habit of expanding a bracket as soon as they see one.

This can sometimes cause us unnecessary calculations.

So actually looking for common factors can be more helpful.

Now had we divided by 3 earlier, we wouldn't have needed to divide at this stage, but dividing both sides by six gives us this equation.

And of course now we can expand the remaining brackets and solve like any other linear equation.

You might wanna pause the video and just check those steps to make sure you are happy.

Right.

Jun has a good point.

Because our answer is an integer, it's gonna be really easy to check with substitution.

Now we can always check our answers by substitution.

However, where our answers are fractions is potentially not as easy to substitute.

But in this case we've got a nice easy answer.

Let's substitute and check.

Got 2 plus 4 over 3 times 6 over 2 plus 1 equals 4.

That gives us 6 over 3 times 6 over 3 is 4 or 6 over 3 is 2.

So 2 times 2 is 4, and that is true.

So x equals 2 satisfies that equation.

Brilliant.

We didn't need to learn anything new.

We are just using skills we already have.

So what do you think would be the most efficient first step to solve this equation? Pause the video.

What would you do? Now as long as equality is maintained, all of these options are valid.

However, combining the fractions first is often the most efficient first step.

Let's just look at this option of multiplying both sides by 3x though.

We'd have to multiply every term by 3x.

The first term would become 15x over 6x minus 1 equals 6x.

That's actually quite efficient 'cause 15x over 6x simplifies to 5 over 2.

So we get 5 over 2 minus 1 equals 6x and then we can solve from there.

So that first option was also reasonably efficient.

Subtracting two from both sides is not going to be as helpful.

It doesn't simplify our equation in a way that's gonna help us find x.

If we found a common denominator of 6x and then subtracted, we get 3 over 6x equals 2 and that is going to be fairly efficient.

Adding 1 over 3x to both sides does not help us 'cause we've got two terms containing x now on different sides of the equation.

So sometimes combining algebraic fractions will produce a quadratic equation.

So we're just gonna remind ourselves how to deal with quadratic equations.

Let's have a look at this one.

The lowest common multiple of 4 and x minus 4 is 4 lots of x minus 4.

So we multiply the numerator and the denominator of the first fraction by x minus 4 and the numerator and denominator of the second fraction by four.

We can then combine those by adding.

Now at this stage you might want to expand the brackets.

I'm gonna multiply by the denominator first.

It doesn't make a lot of difference here.

So multiplying both sides by 4, lots of x minus 4, gives me x minus 3 times x minus 4 plus 8 equals 16, lots of x minus 4.

Right, what would you do next? We are gonna have to expand the brackets now.

There's nothing more we can do at this stage and then we'll need to rearrange to equal zero.

So the product of two binomials, don't forget we've got plus 8 equals 16 x minus 64.

Rearranging gives us x squared minus 17 x plus 60 equals zero.

Right, what would you do next? Why is this helpful to us? We could factorised to find the solutions.

If factorising is not possible, then we could use another method to solve a quadratic.

For example, the quadratic formula.

In this case it will factorised.

If you want to factorised it yourself, pause the video now.

And we get x minus 5 x minus 12 equals zero.

That means x minus 5 has to equal zero or x minus 12 has to equal zero.

Therefore x is five or 12.

And there we go, we've got our solutions to that equation.

Again, there's lots of steps to that.

So we need to make sure we're presenting our working really clearly and we're checking for common mistakes.

Right, Alex is solving this equation.

I'd like you to read through his working and see if you can identify any mistakes.

Off you go.

Now Alex did a really good job here.

His first mistake was when he got to the point where he needed to factorised.

So that is the line of working that is incorrect.

Everything up to there he did really well.

Let's just have a look at what he did.

So first he picked a common denominator, x plus 1 x minus 3 and he converted both fractions so they had that common denominator.

So multiplied the numerator and the denominator at the first fraction by x minus 3 and the second fraction by x plus 1.

Then he combined the fraction and he remembered he had to subtract both terms, so he subtracted x and subtracted 1.

He's done a couple of things in this next step, he's simplified the numerator to 2x minus 10 and he's multiplied by x plus 1 x minus 3.

He's possibly done a few too many steps in one go here, but we'll let him off.

He's also expanded those binomials and written it as x squared minus 2 x minus 3.

Right here he has expanded the brackets further by multiplying every term by 3 and then he's rearranged to equal zero.

Now this is where he went wrong.

3x squared minus 8x plus 1 does not factorised.

Negative 1 and negative 1 do have a product of positive 1 but negative 1 times x is negative 1x.

3x times negative 1 is negative 3x and they don't sum to negative 8x.

In fact, this does not factorised.

So Alex is gonna need to use another method, for example, the quadratic formula to solve this equation.

His algebraic manipulation skills up to that point though were really good.

Right, your turn.

For question one, you need to fill in the gaps to solve the equation.

So follow along with the method that's been used.

Can you work out those missing terms? Then for question two, you've got one to try yourself.

You can use any method you like, but you might want to look at what was done on the left hand side.

Give those two a go, come back when you're ready for the next two.

Good start.

So for this second one again we're doing exactly the same we're solving to find x.

You'll notice that we end up with two solutions for question three.

So see if you can fill in the missing lines of working.

And then have a go at question four.

Off you go.

And finally, two for you to have a go at yourself.

Think about the methods you were using before.

Which elements of those can you apply to these as well? If you want to use technology to check your answers.

For example, if you have a calculator that can solve quadratic equations for you, feel free to do that.

You might wanna challenge yourself though to factorised them and solve them first if possible.

When you're happy with your answers, come back and we'll look at them together.

Let's have a look.

So we've picked a common denominator of 20x.

So you should have 10 over 20x and 28 over 20x.

That gives us negative 18 over 20x.

We can see then that I've multiplied the right hand side by 20x, so I need to do the same with the left hand side.

That leaves me with negative 18.

And a final answer of negative 18 over 20 or negative nine tenths.

Let's see if you could apply that to your own example.

So the easiest common denominator here is gonna be 6x.

So the second fraction can stay the same.

The first fraction, if we multiply the numerator and denominator by 2, we get 2 lots of x plus 1 over 6x and then we can combine them by adding, so that's 2x plus 2 plus 5 or 2x plus 7 over 6x equals 2.

Multiplying both sides by 6x.

And we can rearrange and we get x as seven tenths.

Well done for clear working out there.

Right, we're upping the level of challenge a little bit this time.

So our denominator is gonna be x lots of x minus one.

Then we've gotta be careful when we're expanding our brackets.

So 8 times x is 8x.

8 times negative 1 is negative 8.

But remember we're subtracting both of those.

So we're subtracting 8x and we're subtracting negative 8.

And of course subtracting negative 8 is the same as adding eight.

Denominator stays the same.

Then we can see that we've multiplied the right hand side by x squared minus x.

We do that for the left hand side, we get x plus 8.

If you left that as 9x subtract 8x plus 8, that's fine.

It's gonna help us to simplify.

Rearranging to equal zero is what we need to do in order to solve.

Our brackets are x minus four x plus two and we get x is 4 and x is negative 2.

Right, we've got really similar method on the right hand side.

We've got a common denominator.

Making sure we subtract both terms. That's helpful to us because x subtract x is zero.

Here I've decided to divide both sides by two just to be a little bit efficient.

Rearranged to equal zero and factorised.

Nice one to factorised that one and we get one answer, x is negative two.

Let's have a look at five then.

So same idea as before.

We've got a common denominator.

Add together.

Multiply by the denominator.

I've done a bit of expanding as well.

Expand all my brackets and rearrange.

What's really helpful here is we've got no constant term so we can factorised into a single bracket.

So we get x is zero or x is 4.

For question six, exactly the same idea.

Unfortunately this time I am gonna have a little bit of a trickier quadratic.

What you might have chosen to do at this stage is factor out two and have it as two lots of 6x squared minus x minus 2, you might have found that easy to factorised.

I didn't I factorised as it was.

Either way you get an answer of negative a half or two thirds.

You might wanna pause and just check that final one, there was quite a lot of elements to it.

Well done and we're getting really good now at solving equations with algebraic fractions.

This time we're gonna have a look at equations with unknowns on both sides.

Now this isn't actually any more tricky than what we've just done.

So let's have a look at something like this.

Alex says we can rearrange this so the fractions are both on the same side and then it's gonna look exactly like the equations we've just looked at.

So if we subtract x plus 1 over 5x from both sides.

Alex is right, we need a common denominator.

I've gone for 20x.

And then simplified 11 minus 4x over 20x equals zero, just like we did in the first part of the lesson.

Now we can multiply both sides by 20x.

We get 4x is 11 and x is 11 over 4.

Now, there are often choices for how to solve equations.

Let's look at a different equation.

So we've got 5 over x plus 3 equals 4 over x.

Alex says we could rearrange so the fractions are both on the same side again.

Whoa, Jun says, we could convert to a common denominator first without rearranging.

What would you do? Of course, either method is absolutely fine.

However, let's look at what happens with Jun's method.

So let's convert to a common denominator first.

So in this case it's gonna be x lots of x plus 3.

So multiply the numerator denominator of the first equation by x and the second equation by x plus 3.

Now the denominators are the same, that means the numerator must be equal.

So we can write the equation 5x equals 4 lots of x plus 3.

Or 5x equals 4x plus 12.

Like 'cause this x equals 12.

So actually Jun's method is a little bit more efficient.

'Cause with Alex's method we had to subtract a fraction from both sides first, find a common denominator and then manipulate again.

Whereas here finding a common denominator allows us just to equate the numerators.

So there's a little bit less rearranging to do.

I just had an idea.

Would multiplying by the reciprocal work as well? Well let's have a look.

We multiply by the reciprocal of 1 over x plus 3.

That's x plus 3.

Remember we need to do it to both sides.

So we get five equals 4x plus 12 over x and then if we multiply by the reciprocal of one over x, which is x, which is the same as just multiplying by the other denominator, isn't it? We get 5x equals 4x plus 12, we get x equals 12.

Now in this case, Aisha's method is the same as Jun's.

She just presented it in a slightly different way.

That's because the common denominator that Jun went for was x plus 3 times x, he just multiplied the 2 denominators together.

So essentially what Aisha's done is exactly the same thing, but she's just written it as a multiplication rather than a finding a common denominator.

We're gonna have a look at this one.

Alex says we can rearrange the fractions of both on the same side.

Or we can convert to a common denominator first or multiply both sides by one denominator than the other.

Have a look at this example.

What would you do? Well, Alex's and Jun's methods are very similar.

Jun's gonna have less steps.

It's unnecessary to rearrange the equation so the fractions are on the same side before finding a common denominator and then having to rearrange later.

Aisha is actually gonna be a little more complicated this time.

It will work, but it's gonna involve an unnecessary quadratic expression.

We'll have a look at that in a moment.

So let's look at Jun's method.

The good thing here is that the easy common denominator is gonna be 6x.

So we only need to change the first fraction.

So it's 3 lots of x minus 3 all over 6x equals 2x plus 1 over 6x.

So we can equate the numerators and solve.

Nice and efficient this time.

Right, so Aisha's method is to multiply by one denominator then by the other to remove any fractional terms. So start by multiplying both sides by 6x.

So we get 6x lots of x minus 3 over 2x equals 2x plus 1.

So we've got 6x lots of x minus 3 equals 2x lots of 2x plus 1.

We've multiplied by both denominators.

Then we can expand the brackets.

Rearranged to equals zero, factorised, we get X equals zero or x equals 10.

Okay, something weird has happened here.

We've got slightly different answers.

Now one of these answers is not valid.

I want you to pause the video.

Can you explain why? Okay, well, if you have a look at our second line of working where we had 6x lots of x minus 3 equals 2x, lots of 2x plus 1.

Now at that stage we could have simplified by dividing both sides by 2x.

Now often we say that that is not a valid step when manipulating equations.

'Cause if you divide both sides by x, you essentially remove a solution.

That is because it is possible for x to be zero and obviously dividing by zero is undefined.

Now in this case x cannot be zero.

The reason being is our fraction is x minus 3 over 2x dividing by zero is undefined.

So it's not possible for 2x to be zero.

Therefore it's not possible for x to be zero.

That's a really subtle idea.

So don't worry if you're not a hundred percent sure about that one.

If you try to be as efficient as possible and use the lowest common multiple where possible, you're not gonna come across this problem that often.

So x equals 10 is the only valid solution.

Alright, I'm gonna have a go at one and then you are gonna try one yourself.

So in this example you could use Aisha's method and multiply both fractions by 5 minus 2x and both fractions by x minus 10.

Or you could use a common denominator.

They're gonna be exactly the same in this case.

I'm gonna equate my numerators.

I need to find the product of the binomials on the left and expand the bracket on the right and then rearranged to equal zero.

We've got a nice expression to factorised.

So x is 7 or x is negative 5.

Right, have a look at what I did on the left hand side, see if you can do the same on the right.

So again, you could use Aisha's method and multiply both sides by 2 minus x and then both sides by x plus 2.

Or you could find a common denominator, which is just gonna be 2 minus x times x plus 2.

Either way, you end up with this equation.

and then if we expand and simplify, rearrange to equal zero and we've got a nice one to factorised again and we've got X minus 3 x plus 10, so x is 3 or negative 10.

Where possible, factorising can help spot common factors.

Just make things a little easier.

So this looks like quite a complicated equation at the moment.

If I factorised those denominators might help me spot a nice common denominator, and it does.

So what would you do now? Any ideas? Now two of the fractions have a denominator with a factor of x plus 4.

So that means those fractions are gonna be easier to combine.

So what I'm gonna do is simplify those first.

So if I rearrange to get this equation, I can now find a common denominator of the fractions on the right hand side.

I'm gonna go with 6 lots of x plus 4.

Then I can simplify.

That just makes my equation a little bit easier.

I've lucked out in this case because my numerators are equal.

I know my denominators must be equal.

So I end up with this equation.

Divide in both sides by three and then expanding and solving.

So lots of steps to that problem.

However, because I've presented them really neatly, I've done my working out down the page and I've aligned everything, it's really easy for me to go back through and check.

I could also substitute x equals negative one into my original equation and check it does satisfy that equation.

Right, I'm gonna try one on the left and then you are gonna give one a go.

So factorising denominators helps me see a common factor.

The common denominator I'm gonna go with then is 2x lots of x minus 3.

If I multiply the numerator and denominator at the first fraction by x and the second fraction by 2, then I can put the numerator equal to each other.

So x must be 8 over 3.

See if you can use that to help you solve this one.

Let's have a look.

Factorising helps us see a common factor.

So the lowest common multiple is gonna be 3x lots of x plus 1.

Then we can put those numerator equal to each other.

We get 2x of 6 and x of 3.

Well done if you did that one by yourself.

Right, time for a practise then.

I'd like you to think about all those skills that we have used to help you solve each equation.

You wanna think about sensible common denominators.

You wanna think about factorising to spot any common factors to help you with those denominators.

Give these three a go.

Well done.

Couple more for you to have a look at.

Fantastic.

Have a go at these two.

Thinking carefully about how you're going to solve your quadratic equations.

You could use the equation function on your calculator to check your answers if you wish.

And finally, see if you can have a go at this challenge.

When you think you've got an answer, come back and we'll look at them together.

Fantastic effort.

They were getting quite challenging at the end there.

Let's look at these together.

So the first one you get x equals negative 5.

For B, we get x equals 25 over 4.

The C, factorising helped me see that the most sensible common denominator was gonna be 4x lots of 2x minus one and you get x equals 20.

You need to pause the video and just check through those three before moving on.

For 2A you've got choices.

It might have been slightly quicker to use Aisha's method here and multiplied both sides of the equation by x plus 8.

That would've just missed out that first step.

I've chosen the common denominator of x plus 8 and written x plus 5 as x plus 5 x plus 8 over x plus 8.

Either way, you end up with the same quadratic, you can factorised that as x plus 10, x plus 3.

So x is negative 10 or negative three.

B was similar.

We just had slightly more rearranging to do Factorising, we got x minus 7 x plus 2 equals zero, so x is 7 and x is negative 2.

Again, just pause if you need to read through a couple of those steps.

For 3A, the method here hasn't changed.

We're still finding a common denominator or multiplying by both denominators then rearranging to get a quadratic.

Now the helpful thing about this quadratic is all the terms are divisible by 2.

So we can write it as 2 lots of x squared minus 9x plus 8 equals zero and that's a bit easier to factorised.

So you get two lots of x minus 8 x minus 1.

So x is 8 or x is 1.

For B, there wasn't much in the way of common factors to help us here.

We end up with a quadratic of 3x squared minus 16x plus 16 equals zero.

And then factorising, we get 3x minus 4 x minus 4 equals zero.

Then we get X as four thirds or x as 4.

And finally, so factorising is gonna help us here.

Hopefully you factorised the denominators like I did and you spotted that there were some common factors.

So what I've chosen to do then is I've rearranged them.

So the two fractions, which has a denominator with a factor of x plus 4 on one side and the ones with a denominator with a factor of x plus 2 on the other side.

That gives us that third line down.

Now we can convert them over common denominator.

The left hand side, I've gone with 3x lots of x plus 4.

The right hand side I've gone with 4 lots of x plus 2.

You might wanna pause the video and just check that stage there.

What that allows me to do is simplify my equation to X minus 6 over three x, lots of x plus 4 equals 7 over 4 lots of x plus 2.

Now unfortunately at this stage there's not much I can do to make this more efficient.

So if I convert 'em over a common denominator, it's gonna be 12 x lots of x plus 2 x plus 4.

From there I can equate my numerators and I'm just expanding my binomials and them rearranging.

You get the quadratic of 17x squared plus a 100x plus 48 equals zero.

There was a lot of manipulating to get to that stage, but it was all things you've seen before.

So just look back through your really neat working and see if you can spot where you've gone wrong if you haven't got that quadratic.

Now, even if that quadratic was factorizable, it's not gonna be easy to do.

I've got a coefficient of x word of 17, so I'm gonna use a quadratic formula.

I get two answers.

X is negative naught 0.

5 to one decimal place or x equals negative 5.

4 to one decimal place.

Now I can use tech to check here because I have a calculator which has an equations function on it.

I can type in the value for A, B and C and it will solve that for me.

Of course, that requires me to have the correct quadratic in the first place.

Right, that was really tough, that last one.

So well done if you gave that a good go.

Well done for all your hard work today.

We've looked at lots of different parts of algebraic fractions.

You might wanna pause and just have a read through of that summary to see how much you've achieved.

Thank you for joining me and I look forward to working with you again.