Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Great choice by you to be here.

We're solving linear equations, where brackets are involved today.

That is going to be wonderful.

Let's get started.

Our learning outcome is that we'll be able to recognise that there's more than one way to remove a bracket when solving an equation.

"Equation" being a keyword that you're gonna hear a lot during this lesson.

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

Two parts to today's lesson, and we're gonna begin by using reciprocals to solve equations with brackets.

When we see equations with brackets, we can expand them to solve.

Expanding seven lots of (y+1).

We need seven lots of every term inside that bracket.

We need seven lots of y, and we need seven lots of one.

So the bracket expands to give us the expression 7y+7.

We put that expansion into the equation.

We're now solving 7y+7=35.

This format looks very familiar.

From here we can solve by manipulating.

Let's add -7 to both sides, and then divide through by seven.

We have a solution, y=4.

We can check that by substituting y=4 back into the original equation.

Seven lots of 4+1 is seven lots of five is 35, which is what we were expecting.

We know that solution is correct.

Brackets both sides is no different.

We can expand out to solve.

In this case, I've got a bracketed expression, 2x+5 on the left hand side, with a coefficient of three, and the expression x+4 in brackets on the right hand side with a coefficient of five.

Expand out both pairs of brackets, on the left hand side we'll get 6x+15.

On the right hand side, 5x+20.

From there, we've got a very familiar-looking format, which we can manipulate to solve.

We add -5x to both sides, we'll end up with x+15=20, and from there you can see, x=5.

Check the left hand side by substituting five back in.

Two lots of 5+5 multiplied by three makes 45.

Check the right hand side.

When we substitute five back in, we get 45.

That means x=5 is an accurate solution.

I'm gonna solve a couple of problems like that, and then I'll ask you to do the same.

I'm gonna solve three lots of (6x+2)=60.

Multiply out those brackets, to get the expression 18x + 6 on the left hand side.

Add -6 to both sides.

18x=54, so x must equal three.

Brackets on both sides.

Not a problem.

I'll just expand them to get the equation, 5x+10=10x-20.

Manipulating from there, I would find that 5x=30, so x must equal six.

Your turn now.

I'd like you to solve these two equations.

Pause, and give those a go.

For the first one, I hope you expanded that bracket to get the expression 12x-30 on the right hand side.

Add a positive 30 to both sides.

12x=48.

Divide three by 12.

X=4 should be the solution you arrived at.

For the second equation, multiplying out those pairs of brackets on both sides.

4x+12 on the left hand side.

9x-3 on the right hand side.

Add positive three to both sides.

Add -4x to both sides, and you wiLl find 5x=15.

Divide three by five, and x=3.

Sam spotted something.

Here's an equation we saw solved earlier.

I showed you how to expand that bracket, and then manipulate it to get to the solution y=4.

I wonder what Sam spotted.

I can see a quicker way to solve that one.

Just divide by seven.

Can you see what Sam has spotted? Is there a quicker way to do that problem? What does Sam mean by just divide by seven? Pause and have a think about that.

Seven is a common factor on both sides.

That's what Sam has noticed.

The expression in brackets, y+1, is being multiplied by seven.

Seven we know is a factor of 35.

If you can spot common factors in your equations, there's a quicker way to solve them.

Well done, Sam, for spotting that one.

Where we see common factors, dividing by the coefficient of the bracket could be the quickest solution.

So instead of expanding, we notice seven is a factor on the left hand side.

Seven is also a factor on the right hand side.

The common factor of seven.

Divide both sides by seven.

We'd write that like that, with a fraction bar.

When we divide the left hand side by seven, we did have seven lots of y+1, but by dividing by seven, we're just left with one lot of y+1.

35÷7 on the right hand side is just five.

And from there we've got a very simple equation to solve.

Y must equal four.

The step "divide by seven" is the exact same as multiplying both sides by one seventh.

Hence, you might see this called, "using the reciprocal to solve", because one seventh is the reciprocal of seven.

Quick check you've got that.

True or false? When solving equations with brackets, we have to expand the brackets first.

Is that true, or is it false? And can you justify your answer with one of these two statements? "You always expand the brackets first", or, "We don't have to.

It might be more efficient to multiply by the reciprocal if we have common factors on both sides." Pause, tell the person next to you, true or false? And which statement will you use to justify? I hope you said false, and justified that with, "We don't have to expand the brackets.

It might be more efficient to multiply by the reciprocal if we have common factors on both sides." I'm gonna practise one now, and then ask you to do the same.

I'm gonna use the reciprocal to solve three lots of (6x+2)=60.

The reciprocal of three is a third.

I've multiplied both sides by a third.

You might think of that as dividing both sides by three.

From there I'm just left with one lot of 6x+2 on the left hand side, and one third of 60, 20, on the right hand side.

If 6x+2=20, 6x=18, x=3.

Your turn now.

Use a reciprocal to solve that equation.

Your method should look a lot like mine.

Pause, and do that now.

I hope for a starting point, you divided both sides by six, or multiplied both sides by one sixth.

From there, 18 over six is three, and we've just got one lot of 2x-5 on the right hand side, which means that 2x must be equal to eight, so x must be equal to four.

For efficiency, you will often see it written without the fraction bar being explicitly written.

Remember, from my equation, I divided both sides by three, and demonstrated that with a fraction bar.

You'll probably just see it written like this, straight from three lots of 6x+2, then just having one lot of 6x+2, which must mean that the right hand side was divided by three, also.

60 divided by three being 20.

That's the most efficient way to write this.

Practise time now.

Question one.

I'd like you to solve these equations in two different ways.

Firstly, I'd like you to expand the brackets as a first step.

Secondly, I'd like you to multiply by the reciprocal to solve.

It's really powerful to be able to solve the same problem in more than one way.

Pause and give these questions a go now.

Question two, solve by expanding first.

Two equations there.

Both of them have bracketed expressions on both sides.

Pause and have a go at those two equations.

Question three.

Laura and Izzy have solved an equation by removing the brackets.

Laura's work on the left hand side, reaching a solution of x=11.

Izzy's work on the right hand side, reaching a solution of x=19.

Both pupils started with exact same linear equation, yet somehow they've reached different solutions.

Only one pupil is correct.

Write a sentence explaining why the right method worked, and write a sentence explaining what went wrong in the other pupil's work.

Pause and do that now.

Feedback time.

I asked you to solve these equations in two different ways, firstly by expanding the brackets, secondly, by multiplying by the reciprocal.

Let's have a look at expanding that bracket.

That would give us two lots of x and two lots of seven.

That's 2x+14, being equal to 30.

So 2x must be equal to 16, x must be equal to eight.

Multiplying by the reciprocal means we're gonna take half of both sides of that equation.

Half of the expression on the left, half of the value on the right, leaving us with x+7=15.

Therefore x=8.

If I don't write the fraction bar, you can see that using the reciprocal, there's only two steps to solving it.

That makes it the more efficient method on this occasion.

For part B, expanding the bracket first, we'll still have 21 on the left hand side, but we'll have 14x+49 on the right hand side.

Add -49 to both sides.

14x is equal to -28, or negative 28.

Therefore x must be equal to -2.

What about if we multiply by the reciprocal first? I'm gonna multiply both sides by one seventh.

That will leave us with the equation three being equal to 2x+7, so 2x must be equal to -4.

X is equal to -2.

Whilst there was many steps either way, with this equation, the method where we multiplied by the reciprocal gave us a simpler equation to solve upon reaching the form ax+b=c.

For question two, solving by expanding.

For part A, when we expand those brackets, we get the equation 5x-5=9x-33.

With a little bit of manipulation, we'll reach the solution x=7 from there.

For part B, expanding those brackets, we'll get the equation 3x-3=15x-55, and a bit manipulation from there, we'll end up with 12x being equal to 52.

So x is equal to 52÷12, which cancels down to 13 over three.

Question three, Laura and Izzy were solving an equation, and only one of them was correct.

Laura is the one who's wrong on this occasion.

Izzy's solution is right.

Why did Izzy's work? Well, Izzy's correct because she's multiplied both sides by the reciprocal, one fifth.

What would that look like? That's five lots of (x-10) over five equals 45 over five.

Multiplying both sides by one fifth left her the simple equation, x-10=9.

Laura, however, was incorrect.

She simply erased the brackets, rather than multiplied them out.

This is a common error, and you have to make sure that you multiply both terms inside the brackets.

Five lots of (x-10) should multiply out to five lots of x and five lots of -10.

That expression on the left hand side should read 5x-50, not 5x-10.

Onto the second part of the lesson now.

We're gonna look at efficient methods to solve equations.

Izzy is solving more equations with brackets.

She expands, to get 2x+8=20, 2x=12, x=6.

Then she tries multiplying by the reciprocal.

Multiplying both sides by a half leaves her with x+4=10, so x must equal six.

Izzy says, "It seems to me that the reciprocal method is always the quickest way." Let's see.

A different equation this time.

Three lots of (x+4)=20.

Do you think Izzy's statement is correct? Is the reciprocal method always the quickest way? Pause and have a little think.

This is the solution by expanding the bracket first.

Three lots of (x+4) became 3x+12.

From there, 3x must equal eight.

And x=8/3.

Watch what happens when we multiply by the reciprocal first.

The bracketed expression on the left hand side is being multiplied by three, so we're gonna divide both sides by three or multiply both sides of the equation by a third, and it'll look like that.

The left hand side is really neat now.

We've just got x+4.

On the right hand side, we've got a fraction, 20 thirds.

We now have to add -4 to both sides.

How do we do that? Oh yes, we'll use, instead of -4, we'll express it as -12 over three.

We'll add that to both sides to get to the solution x=8/3.

So there was no common factor.

The reciprocal method was less simple this time.

It wasn't impossible.

It was just less clean and easy, because when we look at the equation, we've got a multiplier of three on the left hand side, and three was not a factor of the 20 that we had on the right hand side.

Quick check you've got that.

What might be an efficient first step for solving this equation? What equation? This equation.

Four lots of (x+11)=15.

Option A is those brackets expanded.

Option B is to multiply both sides by a quarter.

Option C is to multiply both sides by one 15th.

What might be an efficient first step? A, B, or C? Pause and tell the person next to you.

I hope you said option A is really efficient this time.

Expanding gives us the form ax+b=c.

A, b, and c are all integers.

That one's gonna be easy to solve from there.

We'd get 4x=-29, and from there, x=-29/4.

Option B, it works, but it gives you non-integer values earlier.

The left hand side is lovely and clean.

We just end up with x+11.

But the right hand side becomes 15 over four, so we're working with a slightly more complicated number.

Option C, it could be done from there, but it would be really inefficient.

We all simplify the 15 over 15 to one.

We can write the bracket on the left hand side as four fifteenths of (x+11).

We could solve it from there, but it's not going to be very easy.

Izzy is joined by Aisha to solve this equation.

"I'll expand first," says Izzy.

"I'll multiply by the reciprocal of the coefficient," says Aisha.

For Izzy, she expands both expressions, gets the equation 6x+24=2x+16, which manipulates to 4x=-8, with a solution of x=-2.

Aisha starts by multiplying by the reciprocal.

She multiplies both sides by a sixth.

The left hand side is lovely, just left with one lot of x+4.

The right hand side, not quite so easy.

We've got one third of (x+8), which is not impossible.

We can expand that bracket.

But Aisha concludes, "I'm not sure this is the best way." Izzy says, "The common factor of six and two is two.

Try using that reciprocal instead." So, Aisha starts again.

Dividing both sides by two instead of dividing by six this time, and watch what happens.

We end up with three lots of (x+4) on the left hand side, and just one lot of x+8 on the right hand side.

From there, well, that's wonderfully easy.

We'll expand that bracket.

Bear the manipulation to find that 2x=-4, so x must be -2.

And we know we're right, because we've got the same solution as Izzy had a few moments ago.

Aisha says, "That was much easier.

Thanks Izzy!" You would've noticed that the manipulation didn't change the fact we still had to expand some brackets to solve, but it did give us an equation that was more simple to solve.

Quick check you've got that.

Which first step makes this equation more simple? Which equation? This equation.

Three lots of (2x-1), being equal to nine lots of (x-2).

Will we multiply both sides by a half? Multiply both sides by a third? Or multiply both sides by a ninth? What do you think? Pause and tell the person next to you.

I hope you said option A, not a particularly efficient start.

Option B, wonderfully efficient first step.

Option C, not an efficient first step.

For option A, you'd end up with three halves of (2x-1) equals nine halves of (x-2).

It's not impossible from there, it's just not more simple.

For option C, three over nine cancels to a third, so on the left hand side we end up with a third of (2x-1).

The right hand side, super simple.

X-2.

But one third of (2x-1), not impossible, just not much more simple.

By contrast, if we multiply by a third on both sides, we end up on the left hand side with 2x-1 equals three lots of (x-2).

That is a really simple form from which to go on and solve.

For this equation, Izzy expands first, and Aisha uses the reciprocal.

Two thirds of (x-4)=20.

Izzy's gonna expand.

That gives us two thirds of x.

Two thirds of -4 is -8 over three.

And we've just got 20 on the right hand side.

We can add eight thirds to both sides.

We'll end up with 2x over three equals 68 over three.

2x=68, x=34.

Well done, Izzy! That one was tricky.

What did Aisha do? Aisha uses the reciprocal.

Hmm.

What reciprocal? Oh! The reciprocal three over two.

What's the reciprocal of two over three? Three over two.

She multiplies both sides by three over two.

You should notice three over two by two over three is one.

On the left hand side, we've just got one lot of x-4.

On the right hand side, we've got 30.

Three over two multiplied by 20, well that will become 60 over two, which is 30.

Wonderful.

Look how much more simple this is now.

X-4=30.

X must be equal to 34.

Aisha's method is most efficient this time.

When we have a non-integer coefficient, multiplying by the reciprocal can be really efficient.

Quick check now.

Which first step makes this equation most simple? What equation? This equation.

Four thirds of (7x+1)=12.

Would it be to multiply both sides by three, to multiply both sides by four over three, to multiply both sides by three over four? Which first step makes this equation the most simple? Pause and have a think about that.

I hope you said option C.

Multiplying both sides by three over four.

Four over three is not the reciprocal.

If you multiply both sides by four over three, you'd end up with that equation.

Very inefficient.

Not impossible, just very inefficient.

Multiplying by the reciprocal, by contrast, is very efficient.

Three quarters of both sides, well three quarters times four over three, that's just one.

On the left hand side, we'll end up with one lot of 7x+1.

And on the right hand side, three quarters of 12 is nine.

I fancy solving that one.

I know I can solve that one.

7x+1=9.

Multiplying by three, well that's a reciprocal of one third, not the reciprocal of four thirds.

So would that have made it a good start or not? It would've been okay.

We would've ended up with four lots of (7x+1) on the left hand side, and multiplying by three on the right hand side, 36.

It's not a truly inefficient start, it's just not as efficient as option C.

Practise time now.

Question one.

Solve this equation using both methods, and write a sentence comparing the efficiency of those methods.

On the left hand side, I'll ask you to solve by expanding first.

On the right hand side, the same equation, but I'll ask you to solve by multiplying by the reciprocal first.

Pause, and give those two methods a go.

Question two.

Solve both of these equations twice using two different methods.

For part A, we've got the equation two lots of (x+6) equals 10 lots of (x-2).

I'd like that equation solved twice, and I'd like you to use a different method each time.

Same for part B, when you get onto that one.

Pause and try those now.

Straightforward solving.

Or maybe it's not straightforward solving.

They look like fun equations.

Good luck! Pause and give them a go now.

Feedback.

Question one.

Solving the equation using both methods and then comparing the efficiencies.

Expanding first, expand the brackets, get 5x+65=40, 5x must therefore equal -25, x=-5.

Multiplying by the reciprocal first.

The reciprocal of five is one fifth.

Take one fifth for the left hand side, we're left with x+13, one fifth of 40 is eight.

X=-5.

Comparing the efficiencies, you may have written, "For this equation, multiplying by the reciprocal was the most efficient method because in one step it gave us a very simple equation to solve." You might have written something along the lines of, "Five and 40 had a common factor.

So multiplying by the reciprocal was more efficient on this occasion." For question two.

Solving these equations twice, using two different methods.

I will say to you on many occasions, the ability to perform multiple methods fluently is very powerful in mathematics.

So, did you have two ways of solving this equation? Well, option one was to expand the brackets, and turn it into that equation.

Simplifying, manipulating from there, will lead you to a solution of x=4.

That's solving by first expanding the brackets.

A second way would be to multiply both sides by a half.

Common factor of two and 10 is two.

Divide three by two on both sides, or multiply by half, you might say, will leave you with the equation x+6 equals five lots of (x-2).

We'll expand that bracket on the right hand side, manipulate, and we'll get to that same solution.

X=4.

For part B, expanding the brackets was an option.

Manipulating from there will lead you to a solution of x=3.

Was there an alternative? Absolutely.

20 and five have a common factor of five.

Let's multiply both sides by a fifth, or divide through by five.

When we do that, we end up with four lots of (x+2) on the left hand side, and just one lot of 7x-1 on the right hand side.

Expand the bracket, manipulate, we get to that same solution.

X=3.

Hopefully you were able to do both those equations using two different methods.

If so, that makes you a really powerful mathematician.

You might wanna pause now.

Just make any notes of any little things you might have missed.

Solve.

Three over five lots of (7x-4) equals 21.

I think I'll use the reciprocal.

The reciprocal of three over five is five over three.

Multiply both sides by five over three.

That will lead me with one lot of 7x-4 on the left hand side, five over three multiplied by 21 is 35.

From there, 7x=39, so x must be 39 sevenths.

What a beautiful equation.

Other methods were available.

I think this one, multiplying by the reciprocal, is the most efficient.

If you did it another way, and still got to x=39/7, well done.

That's awesome.

For part B, this was tricky.

Fractional coefficients of the bracketed expressions on both sides.

One fifth of the expression on the left hand side, three over two lots of the expression on the right hand side.

Ouch, tricky.

What were the options then? So there's always the option of noticing one fifth of 5x is x.

One fifth of -15 is -3.

Did you spot that? Nothing changed on the right hand side, but on the left hand side, we could have expanded that bracket.

5x and -15, multiply those terms by a fifth, we get x-3.

From there, the reciprocal of three over two is two over three.

Multiply both sides by the reciprocal, and we'll end up with that.

Manipulating from there, I need to add -4 to both sides and add -2x over three to both sides.

I'll get one third of x being equal to -6.

So x must equal -18.

That was a lot of work.

Was there a better way? Absolutely.

If you look at those fractional multipliers, one fifth, three over two, denominators, five and two.

Five and two have got a common multiple.

10.

If I multiply both sides of this equation by 10, I'm going to end up with 10 over five lots of (5x-15) on the left hand side, and 30 over two lots of (x+4) on the right hand side.

That's useful.

10 over five, that's just two.

30 over two, that's just 15.

So we can rewrite the equation here as two lots of (5x-15) equal to 15 lots of (x+4).

You know what to do from here.

Expand the bracket, rearrange, x=-18.

That's a far more efficient way of doing it.

End of the lesson, then.

To summarise, there is more than one way to solve an equation with brackets.

For example, the equation 17 equals two lots of (y-8) could be solved by first expanding the bracket out, or alternatively by multiplying both sides by the reciprocal of two.

I hope you've enjoyed this lesson as much as I have, and I hope I see you again soon for more mathematics.

Goodbye for now.