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 "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 bracket, 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.

7 lots of 4 + 1 one is 7, lots of 5 is 35.

Which is what we are 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.

2 lots of 5 + 5 X 3 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 I'll ask you to do the same.

I'm gonna solve 3 lots of bracket 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 = 3.

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 = 6.

Your turn now.

I'd like 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 +30 to both sides, 12x = 48.

Divide 3 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 +3 to both sides.

Add -4x to both sides and you'll find 5x = 15.

Divide 3 by 5 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 7 lots of y + 1.

But by dividing by seven, we're just left with 1 lot of y + 1, 35 ÷ 7 on the right hand side is just 5.

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

Why must equal four? "This 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, and 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 3 lots of bracket 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 1 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 6 is 3.

And we've just got 1 lot of 2x - 5 on the right hand side, which means that 2x must be = 8, so x must be = 4.

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 3 lots of 6x + 2 to just having 1 lot of 6x + 2, which must mean that the right hand side was divided by three also.

60 ÷ 3 being 20.

That's the most efficient way to write this.

Practise time now.

Question one.

"I'd like 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 the 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 pupils 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 = 30.

So 2x must be = 16, x must be = 8.

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 with our 40x + 49 on the right hand side, add -49 to both sides.

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

Therefore, x must be = -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 X.

The equation 3 being = 2x + 7.

So 2x must be equal to -4x, x = -2.

"Whilst there were as many steps either way with this equation, the method where we multiplied by the reciprocal gave us simpler equation to solve upon reaching the form axe + 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 = 50x - 55.

And a better manipulation from there, we'll end up with 12x being = 52.

So x = 52 ÷ 12, which cancels down to 13 over 3.

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 equation.

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 5 lots of bracket x - 10 over 5 = 45 over 5.

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

Laura however was incorrect.

"She simply erased the brackets rather than multiplied them out." It's a common error and you have to make sure that you multiply both terms inside the brackets.

5 lots of brackets x - 10 should multiply out to 5 lots of x, and 5 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 with x + 4 = 10, so x must = 6.

Izzy says, "It seems to me that the reciprocal method is always the quickest way!" Let's see a different equation this time.

3 lots of bracket 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 a solution by expanding the bracket first.

3 lots of bracket x + 4 became 3x + 12.

From there, 3x must = 8, and x = 8 over 3.

Watch what happens when we multiply by the reciprocal first.

The bracket 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 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 3.

We'll add that to both sides to get to the solution.

X = 8 over 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.

4 lots of bracket 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 axe + 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 over 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 4, 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 15ths of bracket 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 equation.

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 1 lot of x + 4.

The right hand side's not quite so easy.

We've got one third of brackets 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." Same.

I should start again.

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

We end up with 3 lots of bracket x + 4 on the left hand side and just 1 lot of x + 8 on the right hand side from there.

Well that's wonderful, Izzy.

We'll expand that bracket.

Better 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.

3 lots of bracket 2x - 1, being = 9 lots of bracket 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 bracket 2x - 1 = 9 halves of bracket x - 2.

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

For option C, 3 over 9 cancels to a third, so on the left hand side we end up with a third of bracket 2x - 1.

The right hand side, super simple, x - 2.

But one third of bracket 2x - 1.

Not impossible, just not much more simple.

By contrast, if we multiply it by a third on both sides, we end up on the left hand side with 2x - 1 = 3 lots of bracket 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 bracket x - 4 = 20.

Izzy's gonna expand, that gives us two thirds of x.

Two thirds of -4 is -8 over 3, and we've got 20 on the right hand side.

We can add eight thirds to both sides.

We'll end up with 2x over 3 = 68 over 3.

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 3 over 2.

What's the reciprocal of 2 over 3? 3 over 2.

She multiplies both sides by 3 over 2.

You should notice, 3 over 2 by 2 over 3, is 1.

On the left hand side we've just got 1 lot of x - 4.

On the right hand side we've got 30.

3 over 2 multiplied by 20, well that'll become 60 over 2 which is 30.

Wonderful.

Look how much more simple this is now.

x - 4 = 30, x must be = 34.

"Aisha's method is most efficient this time.

When we have a non-integer coefficient multiplied 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 bracket 7x + 1 = 12.

Would it be to multiply both sides by 3? To multiply both sides by 4 over 3? To multiply both sides by 3 over 4? "Which first step makes this equation the most simple?" Pause, and have a think about that.

I would said option C.

Multiplying both sides by 3 over 4.

4 over 3 is "Not the reciprocal." If you multiply both sides by 4 over 3, 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 X 4 over 3.

That's just one on the left hand side we'll end up with 1 lot of 7x + 1, and on the right hand side, three quarters of 12 is 9.

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 4 lots of bracket 7x + 1 on the left hand side, and multiplying by three on the right hand side, 36.

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

Practise time now.

Question one.

"Solve this equation using both methods and right 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, looks remarkably similar.

"Solve this equation using both methods and write a sentence comparing the efficiency of those methods." On the left hand side, I'd like you to expand the brackets first.

On the right hand side, I'd like you to multiply by the reciprocal.

I wonder if your sentence is going to change.

A sentence comparing the efficiency.

Will it be different this time? We'll find out.

Pause, and give it a go.

Question three.

"Solve both of these equations twice using two different methods." For part A, we've got the equation 2 lots of bracket x + 6 = 10 lots of bracket 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.

Question four.

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 bracket to get 5x + 65 = 40.

5x must therefore = -25, x = -5.

"Multiplying by the reciprocal first." And reciprocal of 5 is 1 one fifth.

Take one fifth, the left hand side were left with x + 13, one fifth of 40 is 8x = -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.

It looked like a very similar problem.

Solve by expanding, solve by using the reciprocal.

And then compare the efficiencies.

So why was this one different? When we expand those brackets, we get 3x + 39 = 40.

So 3x must be = 1, x must be = one third.

When we come to multiplying by the reciprocal, taking one third of both sides gives us x + 13 = 40 over 3.

We have to do 40 over 3 - 13, which we.

From there we will get x = one third.

It's just not as clean and easy.

"You might have written, 'For this equation expanding first was the most efficient method because there was no common factor between 3 and 40." Question three.

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 by both sides by a half.

Common factor of 2 and 10 is 2, divide 3 by 2 on both sides, or multiply by half.

You might say, "We'll leave you with the equation x + 6 = 5, lots of bracket 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 5 have a common factor of 5.

Let's multiply both sides by a fifth.

Or divide 3 by 5.

When we do that, we end up with 4 lots of bracket x + 2 on the left hand side and just 1 lot of 7x - 1 on the right hand side.

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

x = 3.

Hopefully, you're able to do both of 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.

For four.

Solve.

3 over 5 lots of bracket 7x - 4 = 21.

I think I'll use the reciprocal.

The reciprocal of 3 over 5 is 5 over 3.

Multiply both sides by 5 over 3.

That will lead me with 1 lot of 7x - 4 on the left hand side.

5 over 3 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 over 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, 3 over 2 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 -15, multiply those terms by fifth, we get x - 3.

From there, the reciprocal of 3 over 2 is 2 over 3.

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 3 to both sides.

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

So x must = -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 5 lots of bracket 5x - 15 on the left hand side, and 30 over 2 lots of bracket x + 4 on the right hand side.

That's useful.

10 over 5, that's just 2.

30 over 2, that's just 15.

So, we can rewrite the equation here as 2 lots of bracket 5x - 15 = 15 lots of bracket 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 lesson now.

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

For example, the equation 17 = 2 lots of bracket y - 8, could be solved by first expanding the bracket out, or alternatively, by multiplying both sides by the reciprocal of two." Hope you've enjoyed this lesson as much as I have.

And I hope I'll see you again soon for more mathematics.

Good bye for now.