Lesson video

Hello, Mr. Robson here.

Well done.

Maths is always a good choice, and don't you love it when the lesson title has the word complex in it? That means it's going to be really exciting.

Let's get learning.

Our learning outcome is that we'll be able to solve complex linear equations, including those involving reciprocals.

Some keywords for today, the first, equation.

An equation is used to show two expressions that are equal to each other.

I imagine you're familiar with the word equation and familiar with that definition.

Rational equation might be new, but it's nothing to be afraid of.

A rational equation is an equation that contains fractions with unknowns in the numerator, denominator, or both.

I can't wait to see what they look like.

Three parts of today's lesson and we're gonna begin with an introduction to complex linear equations.

Alex and Izzy are in a lesson on solving equations and come across this one, 3x plus 4 over 5 equals 11.

Alex says, "I'm worried.

That one looks complex!" And I tend to agree; it does look complex.

But Izzy is a little more confident.

She says, "It's actually not that bad.

Just use this equivalent fraction.

11 over 1 equals 55 over 5." I wonder what she means.

Can you see how Izzy's idea might be useful? Pause this video and have a little think.

It is a useful idea.

Let's explore why.

It enables us to rewrite the equation like so.

Nothing changes on the left-hand side, but the right-hand side has changed, not in value just in appearance, because 55 over 5 is equivalent to 11.

We importantly haven't changed the value, just the way we write that number.

Why is that useful? It's because of this: 3x plus 4 over 5 equals 55 over 5, we've got common denominators.

That is a really powerful position to be in.

The denominators are now equal, therefore the numerators must be equal.

We just solve 3x plus 4 equals 55.

That was a brilliant bit of maths by Izzy to turn that complex equation into a far more simple one.

Once we're in the position to solve 3x plus 4 equals 55, we'll just remove four from both sides, then divide three by three, we have a solution, x equals 17.

Izzy's idea made a complex equation far more simple.

Let's see if you can spot which equivalent fraction is going to be useful this time for the equation 25 minus 2x over 3 is equal to 5.

Which equivalent fraction is going to be useful in helping us to solve this one? Will it be 5 over 1 is equal to 25 over 5? Will it be 1/3 is equal to 5/15? Or will it be 5 over 1 is equal to 15 over 3? Which one do you think will be the most useful equivalent fraction to use? Pause and have a little think about this one.

I hope you said option C, 5 over 1 is equal to 15 over 3.

It's the only one which gives us equal denominators, and we need equal denominators to have equal numerators to create a simple equation to solve.

1/3 equals 5/15, so that is an equivalent fraction, but it's not a useful thing to know when solving this equation.

It doesn't manipulate the value five.

For the top one, 5 over 1 being equal to 25 over 5, it didn't give us equal denominators.

What do I mean by that? If we used 5 over 1 or turned 5 into 25 over 5 and rewrote the equation, so notice nothing's changed on the left hand side, on the right hand side instead of five, we're expressing it's 25 over 5, we haven't got equal denominators.

If we haven't got equal denominators, we won't have equal numerators.

It's no easier to solve from this position.

The correct thing, 5 over 1 equals 15 over 3.

When we change the five into 15 over 3, you'll notice equal denominators, which means we have equal numerators.

25 minus 2x equals 15 is now in a form that we can solve far easier.

You'll get to a solution, x equals five.

Alex and Izzy now tried this equation which looks different again, 3x over 2 plus 4 equals 10.

Izzy's confident again.

"I can use equivalent fractions," she says.

Okay, nothing's changed in the 3x over 2 term, four, we're expressing that as eight over two.

Good.

10, we're expressing as 20 over 2.

Good.

Aha, equal denominators, so we must have equal numerators.

3x plus 8 equals 20.

We've turned a complex-looking equation into a far simpler one by using our knowledge of equivalent fractions.

Well done, Izzy.

3x plus 8 equals 20.

3x must equal 12, x must equal 4, right? Remember that solution, x equals four.

Because Alex says, "That works! But, isn't it quicker to start like this?" Hmm, what's Alex done? Well done.

You spotted it.

He's added negative four to both sides of the equation, which is going to make it immediately more simple.

By adding negative four to both sides, on the left hand side, we've got 3x over 2, and on the right hand side, six, we've made that equation look a little more simple with a very simple step.

Well done, Alex.

"Great idea, Alex! That is really efficient." So let's have a look at where we go from here.

Alex has transformed the equation to 3x over 2 equals 6, and Izzy and Alex are gonna solve the same equation in slightly different ways.

Izzy's using equivalent fractions again, that's awesome.

6 is 12 over 2.

We're in a position where we've got equal denominators of two, therefore 3x must equal 12, x equals 4, the exact same solution, but that method's a little bit quicker.

I wonder what Alex is going to do differently.

3x over 2 is equal to 6, and Alex starts with that.

Can you see what he's done? He's multiplied both sides of the equation by two.

Why? Because 3x over 2 effectively means 3x divided by 2, and what's the inverse of dividing by two? Multiplying by two.

By multiplying the left hand side of the equation by two, we effectively eliminate the presence of the fraction, and we're left with just 3x on the left hand side.

Six multiplied by two on the right hand side equals 12, and you'll notice at the bottom of the screen, we've got different methods, same result.

Whichever method we used, we arrived at 3x is equal to 12, therefore x is equal to four.

You'll see this done in two different ways just like so, and it's fine to do either.

Let's check you've got that.

Which would be a useful next step in solving this equation? 7x over 3 plus 1 equals 15.

Would it be to turn it into 7x over 3 plus 3 over 3 equals 45 over 3? Or to turn it into 7x over 3 plus 1 equals 45 over 3? Or to turn it into 7x over 3 equals 14? What would be a useful next step? Pause and have a think about that.

Option A was a good start.

To turn one into the fraction 3 over 3 turn 15 into the fraction 45 over 3, it gives us common denominators, which leaves us with the equation to solve 7x plus 3 equals 45.

That's pretty efficient.

B was less efficient.

We can solve from there but it's a bit messier, and it could potentially lead to confusion because some people will read that and solve 7x plus 1 equals 45, which is not correct.

You need to do something with that one.

I turn it into the fraction, three over three.

Option C is a very good one.

That's adding negative one to both sides of the equation to leave us with just 7x over 3 on the left hand side and 14 on the right hand side.

Where would we go from there? We could multiply both sides by three to effectively remove the fraction from the equation.

We'll be left with 7x equals 42, so x must be six.

Alternatively, you might see it written like this.

Turn the 14 into the fraction, 42 over 3, and we've got equal denominators, therefore we've got equal numerators, 7x equals 42, so x must be six again.

You'll see that written in both those ways and either is fine.

Practise time now.

Question one, Alex solved this complex linear equation.

Fill in the missing blanks in his work.

Pause and fill in those spaces now.

For question two, some straight up solving of equations for you.

Equation A is 2x plus 10, all over 3, equals 14.

Equation B looks similar but slightly different, 2x over 3 plus 10 equals 14.

Solve both of those, pause and do that now.

Feedback time.

Filling in the missing blanks in Alex's work.

So a useful start.

The number two is the equivalent fraction to 16 over 8.

When we put that into the right hand side of the equation, we get 7x minus 5 over 8 is equal to 16 over 8.

We've got equal denominators, therefore we've got equal numerators.

We're just solving 7x minus 5 equals 16; from there, 7x must be 21 and x must be equal to three.

For question two, lots of ways that we could have written our working out for this, I'll show you one way.

I can start by using an equivalent fraction on the right hand side, turning 14 into 42 over 3.

Why over three? Because I need those denominators to be equal so the numerators are equal, those numerators being equal means 2x plus 10 equals 42, so 2x must be equal to 32, so x must be equal to 16.

You might have gone a slightly different route around that equation.

As long as you arrived x equals 16, there's a good chance your working out is correct.

For question B, the simplest way is to add negative 10 to both sides of that equation, to have it read 2x over 3 equals 4.

From there, lots of ways we can consider this.

You might use equivalent fraction, four being equal to 12 over 3, therefore, 2x must be equal to 12, x is equal to six.

Onto the second part of the lesson now where we're going to solve complex linear equations.

"But Mr. Robson, we've already solved some complex linear equations.

How much more complex do they get?" This much more complex.

Sometimes we have a fraction on both sides of the equation.

2x plus 1 over 5 on the left hand side, x plus 8 over 10 on the right hand side, and it looks really, really complex, but you'll have to trust me on this one.

It's as easy as multiplying by one and you know what happens when we multiply by one? Nothing really.

Eight times one, still eight, a hundred times one, still a hundred.

1.

342 times one, still 1.

342.

We're just gonna multiply by one.

We don't wanna change the value of the left hand side.

We just wanna change something about its appearance, but I'm not gonna multiply by one.

I'm gonna express one as two over two.

2/2 being one whole, but why two over two? Can you see why? Well done.

You noticed when we multiply those two fractions together on the left hand side, we end up with two lots of 2x plus 1 over 5 times 2, which is 10.

That is really useful, because over 10 on the left hand side, over 10 on the right hand side, we've got equal denominators, which therefore means we've got equal numerators.

We've now got a far more simple equation to solve.

We're just solving two lots of 2x plus 1 equals x plus 8.

Lots of ways we could solve this.

If I expand out that bracket better manipulation, we'll end up with a solution x equals two.

The goal, and we started with that complex looking equation, was to just turn it into a far simpler form.

Let's check you've got that.

Which equivalent fraction to one should be used in solving this equation? The equation 4x plus 1 over 3 being equal to 7x minus 12 over 9.

I'm gonna multiply the left hand side by one, but I won't use one.

I'll use either three over three, two over two, or nine over nine.

Which is the best one to use? Pause and have a think about that.

I hope you said option A, multiplying by three over three.

This is the only one that gives us equal denominators.

If I change that one for three over three and then multiply those two fractions on the left hand side, we'll end up with three lots of 4x plus 1 over 9 being equal to 7x minus 12 over 9, and, crucially, I said over nine for both sides of that equation.

If we've got equal denominators, we've got equal numerators, three lots of 4x plus 1 must be equal to 7x minus 12, we can solve that one from there.

Sometimes we have to manipulate both fractions.

X plus 5 over 4 equals 4x minus 5 over 6.

Looking at the denominators, four is not a factor of six so we're gonna have to do something different this time.

Jacob says, "We need equal denominators.

Four times six is 24, so I'm going to start like this." He's multiplied the left hand side by one but use the fraction six over six instead of the number one.

He's multiplied the right hand side by one but use the fraction four over four.

Aisha says, "The LCM.

." What did she mean by LCM? Ah, lowest common multiple, well remembered.

Aisha says, "The LCM of four and six is 12 so I'm going to start like this." She's multiplied both sides by one but used three over three on the left hand side and two over two on the right hand side.

Who's right? Pause this video and have a little think.

So who is right? Well, neither's wrong.

Both methods are going to work.

There's just a little more efficiency in what Aisha has done by identifying the lowest common multiple of four and six and working towards that.

You'll see how hers is a little more efficient.

Jacob's work is just fine.

Let's multiply those fractions together, we'll end up with that equation, both denominators 24, therefore the numerators must be equal and we'd like to solve six lots of x plus five being equal to four lots of 4x minus 5.

That's fine.

We can do that.

Aisha, however, she ends up with that, with the denominators of 12, equal denominators, equal numerators.

So we're solving three lots of x plus five being equal to two lots of 4x minus 5.

Both methods work.

Aisha is slightly more efficient because she used the lowest common multiple.

From there, we can expand those brackets.

Bit of manipulation, and we'll solve to find x is equal to five.

Quick check now.

True or false? To solve this equation, we need only manipulate the fraction on the left hand side.

Look at the equation and tell me, is that true or is it false? Pause now and have a think.

I hope you said false.

Why is it false? I'd like you to justify that.

Will you justify it with, we need equal denominators and six is not a factor of nine.

6 lots of 9 is 54, so we should make the denominators 54, or will you justify it with, we need equal denominators and six is not a factor of nine.

The LCM of six and nine is 18 so we should use that.

Which justification is correct? The truth is they both work but we like efficiency, so I hope you went for B, we need equal denominators.

Six is not a factor of nine.

The LCM of six and nine is 18 so we should use that.

Practise time now.

Question one, where is the error in this attempted solution? Find the error and write a sentence explaining why it is wrong.

Pause and do that now.

For question two, Aisha successfully solved two equations.

Fill in the missing blanks in her working.

Two very complex-looking equations.

Aisha done them wonderfully, but I've just blanked out some of her work.

Can you fill in the missing gaps? Pause and try that now.

Some feedback now.

Question one, where is the error in this attempted solution? We've reached a solution, x equals six, but if you were to substitute that back into the original equation, you'll find it's not right.

So where did it go wrong? I hope you noticed that moment there and why is that wrong? I asked you to write a sentence to explain.

You might have written something along the lines of multiplying the left hand side by a fifth and not doing the same to the right hand side means our equation has lost equality.

Remember every example we've seen so far, we were just multiplying by one to manipulate the fractions.

Question two, filling in the blanks in Aisha's superb solving of equations here.

You should have got in the missing spaces these values.

Crucially, on the left hand side, multiplying by five over five to give us that equal denominator of 10.

Once we've got these equal denominators of 10, we're just solving five lots of 3x minus 5 equals 2x plus 1.

On the right hand side, multiplying through by three over three on the right hand side, giving us equal denominators of 15.

Therefore we just need to solve those equal numerators, five lots of five minus x equals three lots of three minus x.

You might just wanna pause this video and just check that you got all those little gaps filled with the right values.

Do that now.

Onto the third part of the lesson then, solving simple rational equations.

I wonder what a rational equation is.

How exciting! Jun and Sofia are discussing this equation, x over 10 equals 2.

Jun says, "I see this as x divided by 10 equals 2, so I know x equals 20, but how do I write that using notation?" Sofia says, "I start by writing the multiplicative inverse of the denominator." The multiplicative inverse of the denominator, that denominator over 10 means we're dividing by 10.

What's the multiplicative inverse? To multiply by 10.

So that's how Sofia has written it, to multiply both sides of the equation by 10, turning it into x equals 20, which is the same solution Jun got.

Jun says, "I see.

You had x divided by 10, and you multiplied it by 10 to leave you with just x." That's a lovely observation.

Jun and Sofia now try this equation, and can you see it slightly different? 10 over x equals 2.

"I'm very stuck on this one.

It looks strange.

I've never seen an unknown as a denominator before.

If you feel the same as Sofia, that's absolutely fine, but we can work our way out of this.

Jun says, "Isn't the multiplicative inverse of divide by x going to be multiply by x?" What do you think? Would Jun's idea work? It's going to work beautifully if we multiply both sides by x.

It'll look like that when we write the algebra.

Crucially, on the left hand side, that is 10 divided by x inside the brackets and when we multiply it by x, we're just left with 10.

So the left hand side of our equation just reads 10, on the right hand side of our equation, multiply that by x, two lots of x is 2x.

Well, that is a far more simple equation to solve.

We took some thing very complex looking and did one manipulation to make it a whole lot more simple.

From there, x is equal to five.

Sofia says, "Yes, Jun! That works! 10 divided by 5 equals 2." Sofia's done that really sensible thing.

If we think the solution to this equation is x equals five, let's substitute it back into the starting point, 10 over 5 equals 2, we know that we are correct.

That's good practise, Sofia.

Well done.

I'm gonna solve a couple of equations like this now, and then ask you to have a go at a couple yourself.

So if I solved 7 equals 35 over y, I'll start by multiplying both sides by y.

That's the multiplicative inverse of divide by y to multiply by y.

On the left hand side, nice and easy.

That's gonna be 7y.

On the right hand side, I've got 35 divided by y, multiplied by y, which will just leave me with 35.

So I've turned a very complex-looking equation into a far more simple one.

7y equals 35, therefore y must be equal to five.

My next equation is a little more tricky because I've got 3x as my denominator.

Does that change things? Not really.

What's the opposite of divide by 3x? Multiply by 3x, I'll multiply both sides by 3x.

On the left hand side, 4 divided by 3x times by 3x just leaves me with four.

On the right hand side, 3x multiplied by 11 is 33x.

That resolves as x being equal to four divided by 33 or four over 33.

Your turn now.

I'd like you to have a go at these two equations.

Your working out should look very similar to mine.

Pause and try that now.

How did we do? Did we start by multiplying both sides of the first equation by d? If you did, well done.

What did that look like after we'd done that? 48 divided by d, multiplied by d, just leaves you with 48.

Eight multiplied by d is 8d.

From there, the solution is d equals six.

Substitute that back in, 48 divided by 6 is 8.

You know you are right.

What about 5 over 9e? The multiplicative inverse of divide by 9e is multiply by 9e.

That will give us the equation 5 is equal to 18e.

Divide e by 18, e equals five over 18.

Like many parts of maths, there are multiple ways to solve rational equations.

You've just seen me solve this equation, but was there another way to do it? Absolutely, and it's useful to be able to solve the same problem in multiple ways.

The multiplicative inverse of the denominator, that's method one.

Alternatively, I can just multiply by one, and we've seen how effective that can be.

What do I mean by multiply by one in this context though? What I'm going to do is multiply seven by one, but instead of using one, I'm gonna use y over y.

Y over Y is one because anything divided by itself is one.

So all I'm doing is multiplying the left hand side of the equation by one.

Why would I want to do that? Because I will turn the left hand side into 7y over y.

Can you see why? Pardon upon.

Because we've got equal denominators.

If we've got equal denominators of y, we've got equal numerators, and we're back at that stage again, 7y equals 35, which means we're incredibly close to solution y equals five.

Can you see how those two different methods are really rather similar? They just look slightly different.

You can do this either way and either way is absolutely fine.

I'm gonna solve a problem like that one now using the multiply by one method and ask you to do the same.

So negative 19 equals 2 over y.

I'm gonna multiply by one, one being y over y, something divided by itself, that's one.

The left hand side will now look like negative 19y over y, and I'm in that position where I've got equal denominators, y.

Therefore I must have equal numerators, negative 19y equals 2, so y must be negative 2 over 19.

Your turn.

I'd like you to solve three over x equals negative eight, and I'd like you to use the multiply by one method.

Your working should look pretty similar to mine.

Pause and do that now.

I hope you started with multiplying the right hand side by one and using x over x as your one.

From there, you'll have negative 8x over x on the right hand side, you've got equal denominators.

Your numerators must, therefore, be equal, so your solution should be x equals negative three over eight.

One more to solve and it looks a little trickier again, but is it any trickier? No, it's the same.

We're gonna multiply by one, just the one's gonna look slightly different this time.

I'm gonna multiply the left hand side by one, and I'm using y plus four over y plus four as my one.

Why do you think that is? Because I want to get to a position where I've got equal denominators like so.

Once I'm in that position, I've just got to solve those equal numerators, three lots of y plus 4 equals 15.

I can divide both sides through by three.

Therefore, y must be equal to one.

Tricky one for you to do now, but it's as easy as multiplying by one.

Pause and try that equation.

Multiplying through by one on the left hand side and using y minus three over y minus three as your one is the way to start.

From there, you'll end up with that equation.

Equal numerators, equal denominators, we're solving six lots of y minus 3 equals 24.

You can divide through by six from there, y minus three must be equal to four, so y must be equal to seven.

You might wanna pause, just make a note of any little bits that you missed there.

Practise time now.

Question one, I'd like you to finish each method to solve this equation in two different ways.

The equation 54 over e equals 6.

I've given you the starting point in two slightly different ways.

I'd like you to finish working those two out.

You should arrive at the same solution.

In fact, that's how you'll know you are right.

Pause and try that now.

Question two, some straight up solving, three equations for you to solve.

There's a few different ways you might do this.

You use the method that's comfortable for you.

Pause and solve those three now.

Feedback time.

Finishing solving these equations in two different ways.

So left hand side, we went for multiply both sides of the equation by e, to turn it into the equation, 54 equals 6e, therefore e must be equal to nine.

On the right hand side, I went for the multiply by one method using e over e as my one.

So the right hand side should now look like 6e over e.

Equal denominators, equal numerators.

54 is equal to 6e, e is equal to nine.

Part two, question two, some solving.

In A, lots of ways you could do this.

I'll go for the multiply by one method, 12 multiplied by x over x, meaning I've got 12x over x.

Equal denominators, equal numerators.

Therefore, x must be equal to seven.

You might have used a slightly different method, but if you arrived x equals seven, there's a good chance your method is correct.

For part B, 11 equals 84 over x.

What's the multiplicative inverse of divide by x, multiply by x, multiplying both sides by x.

That turns that complex-looking equation into 11x equals 84, so x must be 84 over 11.

For question C, slightly trickier because there's not an x coefficient of one, there's an x coefficient of three, but it doesn't matter.

I'm just gonna multiply three by one.

I'm gonna use 3x over 3x as my one.

That means on the left hand side, I've got negative 21x over 3x.

Importantly, equal denominators, equal numerators, negative 21x equals 84, divide through by negative 21 and I'll have x equals negative four.

That's the end of the lesson.

In summary, complex linear equations, even those including reciprocals, can be manipulated and solved using our knowledge of manipulating fractions.

I hope you enjoyed the lesson as much as I did, and I hope to see you again soon for more mathematics.

Goodbye for now.