Lesson video

Thank you for joining us in today's lesson.

My name is Ms. Davies and I'm going to be helping you every step of the way today.

We've got lots of exciting algebra to look at, so let's get started.

Welcome to this lesson where we're going to be checking and securing your understanding of solving linear equations.

Our focus today is going to be solving equations involving brackets.

In order to solve equations, we're going to be using the idea of additive inverses and reciprocals.

If you're not sure about those, pause the video and have a read through.

We're going to start then with solving equations involving a bracketed expression.

When we are solving an equation, we are looking to find a value for the unknown which makes the equation balanced.

We call that value the solution.

To do that, we need to isolate the unknown so that we can see what its value is on its own.

Let's have a look at an example.

I'm looking for a value for x, which when substituted, both sides of the equation will be equal.

So what we can do is we can add the additive inverse of positive 7, and that'll tell us the value of 5x.

So the left hand side of our equation becomes 5x + 7 + -7.

And to maintain equality, we need to do the same to the other side of the equation.

So that becomes 42 + -7.

Now 5x + 7 + -7 gives you 5x, because 7 and negative 7 are a zero pair.

That's why we chose negative 7 in the first place.

Now that we know that 5x equals 35, this is a one step equation.

We want to find the value of 1x.

We currently have the value of 5x.

So what we can do is we can multiply by the reciprocal of 5 to find the value of 1x.

The reciprocal of 5 is a fifth.

Remember that multiplying by a fifth is the same as dividing by 5.

So you may have looked at the previous step and said, "I would divide both sides of the equation by 5." That is fine.

That is the same thing as multiplying by a fifth.

So 5x over 5 equals x, that's why we've chosen that operation.

And 35 divided by 5 is 7.

There is a solution then when x is 7.

We can check this solution by substituting 7 into the original equation.

So I've got 5 lots of 7 plus 7, and I want to see if that's equal to 42.

Or that's 35 + 7 and that is equal to 42.

The left hand side and the right hand side are the same.

So that was solving a two step equation.

We can do the same when we've got equations involving brackets.

When we see an equation involving brackets, what we can do is expand the brackets and then solve.

So 6 lots of x minus 1 is 6 lots of x plus 6 lots of negative one.

That gives us 6x minus 6.

The right hand side of our equation has not changed.

We've not done any operations at the moment.

We've just changed the format of the left hand side.

Now we can solve doing exactly what we did on the previous slide.

We can add the additive inverse of negative 6, which is positive 6, and that gives us 6x equals 38.

From there, we can divide both sides by 6, or multiply both sides by 6, which gives us x equals 38/6.

The answer in this case is not an integer.

It's easier to leave it as a fraction than it is to try and turn it into a decimal.

However, 38/6 could be simplified to 19/3.

In general, it's good to write fractions in their simplest form.

Let's have a go then.

I'm going to have a go at one on the left hand side.

I'd like you to follow my working out, then you're going to try the same on the right hand side.

We're going to solve this equation by expanding the brackets.

So I've got 3 lots of 2x plus 8 equals 6.

Expanding the brackets gives me 6x plus 24.

Then I can add the additive inverse of 24, which is negative 24, and 6 add negative 24 is negative 18.

Now I want to multiply by the reciprocal of 6, which is 1/6.

So x is going to be negative 18 multiplied by a sixth, or negative 18 divided by 6.

Negative 18 divided by 6 is negative 3.

I'm going to check my solution by substitution.

So I'm going to put the value that I'm substituting in in brackets.

So I've got 3 lots of 2 times negative 3 plus 8.

2 times negative 3 is negative 6.

Negative 6 plus 8 is 2.

So I end up with 3 times 2.

And I want to check that that's the same as my right hand side and it is.

3 times 2 is 6.

I will leave that one there.

I'd like you to have a go at this one on the right, and I'd like you to check your solution by substituting.

Give it a go.

Well done.

So check how you've expanded your bracket.

Our additive inverse is negative 80.

You should end up with a negative answer.

That's absolutely fine.

Solutions can be negative.

And dividing by 12 gives you x equals negative 5.

Pay particular attention to how we've laid out our working down the page.

It's really easy to see every step of our solution.

Let's have a go at checking the solution then.

So we've got 3 lots of negative 5, which is negative 15, plus 20.

Negative 15 plus 20 is 5.

So we have 4 lots of 5 is 20, and that was the right hand side of our equation as well.

Well done.

We can solve an equation now involving brackets.

So Jun reckons he saw a quicker way with the last question.

Can you see what it is that Jun has spotted? There was nothing wrong with the way we solved the previous question, but having more than one way of solving things is really helpful for us to check our answers.

So what Jun has spotted is that 4 is a common factor in the expressions on both sides of the equation.

The left hand expression is 4 lots of 3x plus 20, whereas the right hand expression is 4 lots of 5.

So where we can see a common factor, we can actually divide by the coefficient of the bracket, rather than expand the bracket, and sometimes that is quicker.

Instead of expanding our bracket, we can divide both sides of the equation by 4, which leave us with 3x plus 20 equals 5.

Now we can add our additive inverse, which is negative 20, to get that 3x is negative 15, and therefore x is negative 5, and that's what we got by expanding our brackets as well.

So Jun is going to solve more equations with brackets.

He's going to try it by expanding first, and by multiplying by the reciprocal or dividing by the multiplier of the bracket.

We're going to look at both methods, and he's going to discover which one's easier in each case.

So expanding our bracket first this time.

We've got 3x - 6 = 15.

Adding our additive inverse, we've got 3x is 21.

Dividing by 3, we've got x equals 7.

So we solved that essentially in three steps.

Seemed fairly straightforward.

Let's have a look doing the other way.

So we can multiply both sides by 1/3, that's why we say multiply by the reciprocal.

Remember that's the same as dividing by 3.

That gives us x minus 2 equals 5.

Then we can add 2 to both sides, which gets x equals 7.

So that example, we only had two steps.

In that case because there was a common factor, dividing by 3 was a slightly quicker way.

Jun reckons his found an example where expanding the brackets will be easier.

Let's have a look.

So let's try expanding first.

We've got 8x - 24 = 20.

So 8x is 44.

So x is 11/2 or 5.

5.

Let's look at multiplying by the reciprocal.

So we've got to multiply by an 8, so that gives us x - 3 = 20/8.

You might want to simplify that at this point and get 5/2.

Okay, so now what we need to do is we need to add 3 to 5/2.

Because we're dealing with fractions here, I've written 3 as 6/2, just to make it a little bit easier.

So we've got x - 3 + 6/2 = 5/2 + 6/2.

That gives us x = 11/2, or if you prefer decimals, 5.

5.

In this case, that second one seemed a little bit trickier.

Because we've got a fraction early on, because 20 divided by 8 is not an in integer, it meant more calculations with fractions.

It's entirely up to you which of those you found easier.

It is totally personal preference which method you find easiest for each question.

Generally though, when the coefficient of the bracket is a fraction, expanding the bracket can be time consuming and lead to some tricky calculations.

Let's have a look at this example.

3/5 of x + 7 = 30.

If we were going to expand the bracket, we'd have to do 3/5 times x and 3/5 times 7.

So 3/5x + 21/5 and that's equal to 30.

Now we want to add our additive inverse to 21/5, which is negative 21/5.

So 3/5 of x equals 30 subtract 21/5.

So we may need to write 30 as a fraction of a denominator of 5.

So that's 150/5.

So 150/5 subtract 21/5 gives us 129/5.

So 3/5x equals 129/5.

Then we need to divide by 3/5x or multiply by the reciprocal, which is 5/3.

So 129/5 multiplied by 5/3 gives us 43.

Take a second to check your fraction skills, make sure you're happy with where the 43 came from.

There is nothing wrong with that method, we just had quite a lot of fraction calculations to do.

Let's have a go by dividing both sides of the equation by 3/5.

Remembering that that's the same as multiplying by 5/3.

So x plus 7 must equal 30 times 5/3.

So you could do 30 divided by 3 then times by 5, or 30 times 5 then divide by 3, whichever you prefer.

So, x + 7 = 50.

Subtracting 7 from both sides gives me x equals 43.

I think in this case that was a lot easier than expanding the brackets.

Jun agrees with me.

Let's have a look then.

True or false, the equation 5 lots of x plus 4 equals 50 must be solved by expanding the brackets first, what do you reckon? Yeah, we've just seen a couple of examples where we've done it a different way, haven't we? Which of these then is the correct justification for why that is false? Well done if you said that 5 and 50 have a common factor, so dividing by the coefficient might be easier.

It's not the case that all equations are easier to solve if you divide by the coefficient of the bracket first.

However, we have seen some where that has been easier.

Generally, it's where there's been a common factor, or where we've had to divide by a fraction.

I'd like you think about this one now.

So which first step would make this equation easier to solve? What do you think? Okay, this one is personal preference, however, 7 and 32 do not have a common factor, so you may have found that expanding the brackets was easier.

Let's have a look at the top where we've expanded the brackets.

You can see we have 7x + 21 = 32.

We're going to be able to solve that in two steps.

Adding 3 to both sides does not help us solve this equation.

Multiplying by 7th or dividing by 7 are the same thing.

They would help us solve this equation.

The next step would give us x + 3 = 32/7.

It's personal preference then whether you are happy to do 32/7 add negative 3, or if you prefer to expand the brackets.

Time for you to have a go.

In each of these three questions, someone has had a go at solving these equations.

You'll see that their working out are presented really neatly down the page.

What I'd like you to do is spot the mistake in each method.

Give that a go, come back for the next bit.

Okay, three more for you to have a look at.

Pay particular attention to those two examples with a fraction.

Can you spot those mistakes? Lovely.

Your turn to have a go by yourself.

So I'd like you to solve these equations, and I'd like you to use both methods we've looked at today.

So first, expand the brackets and solve the equation.

And second, can you do it by multiplying by the reciprocal or dividing by the coefficient of the bracket? I'd like you to finish by writing a sentence to compare the efficiency of the methods.

Which one did you think was more efficient? Make sure you're explaining your reasoning.

Off you go.

Well done.

You've got another equation to solve.

Again, can you use both methods and write a sentence telling me which one was most efficient? You've got an equation this time where the coefficient of your bracket is a fraction.

Again, I would like you to try both methods and tell me which one you thought was the best.

If you present your working down the page, like we saw in the previous examples, it'll be easier for you to get each of your terms correct.

Give that one a go.

Well done.

Let's see if we've got these mistakes.

First one, the brackets were not expanded correctly.

The second line should say 5x + 50 = 55.

For B, this person has tried to divide the equation, but when dividing the left hand side by 2, the terms in the bracket should not be divided by 2 as well.

Essentially what's happened there is you divided by 2 and then divided by 2 again.

Dividing 2 lots of 4x plus 10 by 2 gives you 4x plus 10.

C, it should be the additive inverse of negative 3 which is added, which is positive 3.

So we should have x = 6 + 3, or x = 9.

For D, the brackets have not been expanded properly again.

The 5 should be multiplied by 3 to get 15x + 10.

E, the reciprocal of 1/3 is 3.

So we need to be multiplying both sides of that equation by 3 to get 2x - 5 = 45.

What this person's done is they've multiplied the left hand side by 3, but divided the right hand side by 3, so we have not maintained equality.

For F, it would definitely have been easier to multiply by the reciprocal first.

6 times 4/3 gives us 8, therefore making the equation easier to solve.

However, we can still follow this person's working out and see where it's gone wrong.

It's actually gone wrong at the very end with our negative values.

Negative 24/4x equals negative 3/2 is correct, but we need to multiply by negative 4/24 if we want to get to x.

I suggest doing this in two steps.

So the penultimate line should read negative x equals negative 3/2 times 4/24.

So that negative x equals negative 1/4.

Positive x equals positive 1/4.

Take a second to check back through that person's work if you need to.

Well done.

For this one, it's really important that you've got the working out correct.

So pause the video and check your working out.

Your sentence might have been something like, "The second method requires a lot more fraction skills, so the first method is likely to be more efficient." Can you now do the same for B? Pause the video and check your steps of working.

Your sentence this time does depend on personal preference.

Both expressions have a common factor of 15, so dividing by 15 first makes the value smaller.

45 divided by 15 is 3, it's an easier value to work with.

Otherwise we're working with 60, 105 and 45.

Just maybe a little bit trickier with our mental arithmetic.

However, both methods had a similar number of steps, so it's entirely up to you which one you preferred.

And finally, this is similar to the example we had a look at in the lesson.

Expanding first meant we had to deal with a lot of fractional calculations.

In my opinion, it was a lot easier to do the multiplicative step first, multiply by the reciprocal of 6/5, which is 5/6.

Especially easy because 30 is divisible by 6.

That gets us 3x + 4 = 25, which is a lot easier to solve in my opinion.

I wonder if you agree with me.

Well done.

We're now going to put those skills to the test and we're going to have a look at solving equations where we have multiple brackets.

So let's have a quick reminder of how to solve an equation with unknowns on both sides of the equals sign.

I've got 5x + 1 = 2x + 10.

Often the easiest first step is to manipulate the equation so there are only unknowns on one side of the equation.

In this case, if we add negative 2x, because it's the additive inverse of 2x, that'll mean the next step of our working will only have x's on the left hand side of the equation.

That gives us 3x + 1 = 10.

Now this looks like the equations we solved in the first part of our lesson.

Adding negative one and multiplying by a third.

Especially useful this time to substitute 3 into both sides of the equation and see whether both sides of the equation are the same.

So 5 lots of 3 plus 1, and 2 lots of 3 plus 10.

Well, that gives us 15 plus 1, and 6 plus 10, which is 16 and 16.

Both sides of the equation are balanced when x equals 3.

This allows us to solve equations where we've got brackets on both sides.

So we can start by expanding the brackets.

That gives us 6x + 80 = -4x - 10.

We can add the additive inverse to negative 4x, which is positive 4x.

That gives me 10x + 18 = -10.

Now we've got a two step equation which we can solve.

10x equals negative 28, x equals negative 28/10.

You can leave that as negative 14/5 or negative 2.

8.

Lucas has a go at solving this equation.

I'd like you to explain what he has done at every step.

Have a read.

Hopefully by seeing how well he's presented his answers, we can spot what he has done at every step.

Let's have a look.

He has chosen to expand both brackets.

He has then added negative 6x to both sides.

Negative 4x add negative 6x is negative 10x.

Then he's added negative 6 to both sides.

Then he's divided both sides by 10.

And then because he's got negative x equals 2.

4, he can multiply by negative one or divide by negative one, which gives you x equals negative 2.

4.

I'd like you to pause the video and think carefully about this question.

Is there anything that you would do differently? Right, I wonder if it's on that second step where Lucas decided to add negative 6x, you might have decided to do something different.

You might not.

Alex says, "There's a common factor on both sides of the equation.

What would happen if I start by dividing by 2?" What do you reckon? Is this going to work? Let's have a look then.

So if you divide two lots of 3 minus 2x by 2, you get one lot of 3 minus 2x.

The one in front of the bracket is not necessary.

You could write that as 3 minus 2x.

Dividing 6 lots of x plus 5 by 2 gives you 3 lots of x plus 5.

Then we can expand the brackets.

Pause the video.

What would you do next? There's many options at this stage.

I wonder if you said add 2x, or if you said add negative 3.

We could have added negative 15, or added negative 3x.

All four of those options would've helped us solve the equation.

Alex is going to start by adding 2x.

I wonder if, like Alex, you like the idea of all your terms being positive.

So that gets him 3 equals 5x plus 15.

Added negative 15, it gives you negative 12 equals 5x.

So x equals negative 12/5, or negative 2.

4, which is the way Lucas chose to write it.

Let's see if we can do this then.

I'm going to show you one on the left hand side, and I'd like you to have a go on the right.

We have got no common factors.

I'm going to start by expanding both brackets, then I'm going to add 12x to both sides of my equation, then I'm going to add 20 and then divide by 20.

I'm going to do that in two steps to make it easier for myself.

I'm going to divide by 2 first and then divide by 10.

Make sure you're happy with each step, and then have a go at this one yourself.

Let's have a look.

Expanding your bracket should look like this.

You have options now with what you'd like to do.

So if you did it a different way, that's absolutely fine.

You should get to the same solution.

I added 8x, then added -7, which gave me 22x equals 33.

I've solved this in two steps again, firstly by dividing by 11, and then dividing by 2.

You get 3/2 or 1.

5.

So when solving an equation, there are often options for which order to complete steps in.

We've seen that already.

You should choose the way that is easiest for you.

Do not be swayed by other people.

Your method's going to depend on what makes the most sense for you.

It's useful then to use a different method to check and see if you get the same answer.

Let's try this one.

Six lots of x plus 8 minus 3 lots of x plus 8 equals 9.

Lucas is going to start by expanding the brackets, then collecting like terms, and solving his equation in two steps.

Alex is going to start by dividing both sides by 3.

Remember, we need to divide the whole of the left hand expression by 3.

So we've got 2 lots of plus 8 minus 1 lot of x plus 8 equals 3.

Then he's expanded his brackets, making sure he's subtracting both terms in the bracket, and then he's solved a one step equation.

He got the same answer as Lucas.

Izzy reckons that that was a very clever way of Alex doing it, but she thinks she's seen an easier way.

Have you spotted the same thing as Izzy? Let's have a look.

Those brackets are the same, so we can actually unitize them and simplify into a single bracket.

If I've got 6 lots of x plus 8, I'm subtracting 3 lots of x plus 8, I have 3 lots of x plus 8.

We could now divide both sides by 3, and solve our one step equation.

All 3 methods gave us the exact same answer.

I reckon Izzy's was the quickest.

Alex's looked the most complicated, but if that's what made sense to him, that's absolutely fine.

Which method did you prefer? If you've got someone to discuss this with, share your ideas with each other.

If not, share your ideas with me.

Have a look at these four questions.

Which of these could be simplified before solving? What do you reckon? Let's have a look.

So the first one could be written as a single multiple of the bracket x minus 8, 'cause we've got the same bracket.

The second one, there's no common factor in all those expressions.

So what we're going to have to do is expand the brackets before solving.

For C, our brackets are the same, so two lots of 2x minus 1 minus one lot of 2x minus 1.

This bottom one was a little bit tricky.

The brackets aren't the same, but there is a common factor of 3.

So you might have said we could simplify both sides by dividing by 3 before solving.

Time to see if you've been paying attention with how we lay out our working.

I have told you the solution to these equations are at the bottom of the page.

I would like you to fill in the steps to show how to get these answers.

Off you go.

Three more to do.

If you don't get the solution, that's absolutely fine, 'cause you'll be able to look back through your working and see if you can spot where you've gone wrong.

Give it a go.

Well done.

This time I'd like you to solve both of these equations in two different ways.

So you're going to need space to do your working out.

Try expanding your brackets first, and then see if you can simplify before expanding.

Do that for question A.

And then B, I'd like you to write a sentence to compare the efficiency.

Give it a go.

Superb.

Pause the video now and make sure you look through these lines of working.

You may have made different decisions to me at different points.

As long as you can justify the decisions you've made and why they lead you to the correct answer, that's absolutely fine.

And do the same for D, E, and F.

Give yourself time to check your working.

And then for 2a, here are two methods.

Your answer might have read, "Both methods were reasonably efficient.

Simplifying first meant a common factor could be seen and that eliminated a step." Because 4 lots of x plus 5, plus 3 lots of x plus 5 was 7 lots of x plus 5.

Both sides of the equation then had a common factor of 7.

For this second one, simplifying first did seem to be more efficient.

There was quite a lot of brackets to expand otherwise.

Doing it that way meant no expanded brackets was actually necessary, and less manipulating of negative numbers was needed.

Read those two methods.

Which one would you prefer? Well done today.

You have shown that you can solve equations with brackets perfectly, particularly pay attention to your working out.

So you should be really proud of yourselves.

Hopefully what you've seen today is there's not one way of solving an equation.

As long as we do appropriate steps to maintain equality on both sides of our equation, there are plenty of options for how to solve it.

Find ways that work for you, but keep your mind open to other ways 'cause they do help you check your answer and make sure you haven't made any mistakes.

Really enjoyed working with you.

I look forward to seeing you again.