Lesson video

Hello, Mr. Robson here.

Welcome to Maths, superb choice to be here solving complex linear equations involving brackets.

I love it when the lesson title includes the word complex.

It means today's maths is going to be very exciting, so let's not wait.

Let's get on with it.

Our learning outcome is that we'll be able to solve equations involving brackets where simplification is necessary first.

Keyword for today, equation.

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

Two parts to today's lesson, and we'll begin by solving equations with multiple sets of brackets.

Sometimes we'll need to work with equations with multiple sets of brackets.

For example, if the approach of this rectangle is 33 centimetres, can we find x? We can.

First, we need to set up an equation.

You know that the rectangle has two pairs of equal lengths, so we must have two of the bracket 2x + 3 lengths, and two of the bracket 9 - x lengths.

Hence that equation being equal to 33, because the perimeter is 33.

After expanding the brackets, we'll have an expression that has like terms that can be simplified, multiply out two lots of bracket 2x + 3.

I'll get 4x and 6, two lots of bracket 9 - x.

I'll get 18 and negative 2x.

On the left-hand side now, we've got some light terms which we can collect together to simplify this expression.

4x - 2x is 2x.

6 + 18 is 24.

We simplified left-hand side.

Now, we've got an equation in the form, axe + b = c, which can be solved.

We'll add negative 24 to both sides.

Divide three by two and our solution, x = 4.

5.

And we can manipulate an equation into that form, axe + b = c, we can easily solve.

This equation can be solved in a very similar way.

First step, expand the brackets.

That'll give us 3x + 12 + 6x + 14 on the left-hand side.

Next step, we've got like terms that can be collected to simplify our left-hand side expression, 3x and 6x becomes 9x, 12 and 14 is 26.

That is the form, axe + b = c.

A very familiar form from which we can solve.

Subtract 26 from both sides.

Divide through by 9, x = 7.

Visual representations often help us to understand what's going on in mathematics.

Sometimes just writing the algebraic notation can be a little abstract.

This is the same equation.

I'm gonna show you what it looks like in the form of a bar model.

Three lots of brackets, x + 4, that's three lots of x + 4, two lots of bracket 3x + 7.

That's 3x + 7 twice.

Equal to 89.

I'll make a bar of the same length and put 89 in it.

From there, the next thing I wrote is what we got when we expanded those brackets, 3x + 12 + 6x + 14.

Where we got the 3x + 12 from is we gather together the three xs, and we gather together the three 4s.

6x + 14, gather together those six xs, gather together the two 7s, again, equal to 89.

The same bar model, from there, we now gather the like terms. We turn 3x + 12 + 6x + 14 on the left-hand side into 9x + 26.

Because I can take the 3x and the 6x, and gather them together, I can take the 12 and the 14, and gather those together to get the 26, again equal to 89.

From there, how do we solve? Well, we need to know that length.

That length will be 89 minus 26.

A 63, same, 9x is 63, x must be equal to 7.

Sometimes it's really useful to draw out that visual representation.

Even with brackets both sides, we can manipulate the equation into the form, axe + b = c to solve.

I've got brackets, two sets of brackets to be expanded on the left-hand side, and a set of brackets on the right-hand side, doesn't matter.

Step one, expand the brackets, like so, collect the like terms. I can gather the like terms on the left-hand side, positive 5x, positive 14x, that's 19x, negative 5, positive 28, that's positive 23.

I've simplified that left-hand side.

What we do next is we want to gather the x terms on one side.

We'll do that by adding positive 27x to both sides of the equation.

Now, we're in the form, axe + b = c, and we know we can solve from there.

Subtract 23 from both sides, divide through by 46.

We get solution x = 31 over 46, so even with brackets on both sides of the equations, we still followed those same steps.

There may be circumstances in which expanding the brackets might not be your first step.

There's something different about this equation.

Did you spot it? The multipliers 9, 27 and 36 share a common factor of nine, so we want to multiply through by a ninth.

You might think of that as dividing all the terms by nine.

You might think of it as multiplying through by a ninth.

Either is fine.

When we do that, we turn a very complicated-looking equation into something a little more simple.

We end up with one lot of 8x + 7, just four lots of 3 - x and just three lots of 3x - 9.

This is more simple.

I will show you how to complete that from there.

Multiplying three by ninth, we then expand the brackets till four lots of 3 is 12, four lots of negative x is negative 4x, three lots of 3x on the right-hand side is 9x, three lots of negative 9 is negative 27.

Then we wanna gather together, simplify the light terms on the left-hand side, 8x - 4x is 4x, 7 and 12 make 19.

Then we want the x terms on one side, so I'm gonna add negative 4x to both sides.

I'll also add positive 27 to both sides to manipulate that into 5x equals 46, so x must be 46 over 5.

Let's check you've got everything I've said so far.

What is the first step to solving this equation? What equation? This equation.

3 lots of bracket x + 4, plus 5 lots of bracket x + 2 = 46.

What's the first step? Are we going to expand the brackets, collect the like terms, or gather x terms on one side? What do you think is the correct first step, pause, tell the person next to you.

I hoped you said option a, expand the brackets.

If we expand those brackets, we get 3x + 12 + 5x + 10 = 46.

We can't collect the like terms until the brackets are expanded.

Hence it wasn't option B.

What will the next step to solving this equation look like? We've expanded the brackets.

What do we do next? That, that, or that? Hm.

Option a, option b, option c, pause and have a think.

I hope you said option c.

Option a, absolute no go.

Unlike terms cannot be gathered.

We haven't got 15x + 15x, 3x + 12 doesn't become 15x.

We can only gather like terms. Option b, useful to rearrange a left-hand side to 3x + 5x + 12 + 10, but it's not yet simplified.

We've still got four terms on the left-hand side.

There's option c, is the like terms gathered together and simplified.

Once we've done that, 8x + 22 = 46, add negative 22 to both sides, divide through by eight, and we're at the solution, x = 3.

Practise time now.

Question one, fill in the blanks to complete these two equations.

In both methods, we're gonna expand the brackets, and then simplify, gather together the like terms and solve, but what have I missed out in the working? Your job to fill that in, pause and do that now.

Question two.

Each solution contains an error.

Identify it, highlight it, then write a sentence explaining why it is wrong.

You should have a sentence for the error in part a, and a sentence for the error in part b.

Pause and do that now.

Question three, solve four equations there for you, varying levels of difficulty.

Give them a go, pause and do that now.

Feedback.

I asked you to fill in the blanks to complete these two equations.

For a, we should have noticed when we're expanding those brackets, two lots of 5x becomes 10x, and in the second bracket, five lots of negative seven becomes negative 35.

When we simplify that left-hand side, positive two and negative 35, that's negative 33.

The next step will be to add positive 33 to both sides, giving you 60 on that right-hand side, and dividing through by 15, the solution x = 4.

For part B, when we expand those brackets, four lots of 2x is 8x.

In the second bracket, seven lots of negative x is negative 7x.

On the right-hand side, two lots of 2x is 4x.

When we simplify, on the left-hand side, positive 8x, negative 7x leaves us with positive x.

Nothing simplified yet on the right-hand side, 4x remained 4x.

The next step's the important one, because we subtracted x from both sides, or added a negative x to both sides, so 4x - 14 became 3x - 14.

Then we add positive 14 to both sides getting 66 equals 3x, therefore x = 22.

Question two, I said each solution contains an error.

I said highlight the error, and write a sentence explaining why it's wrong.

Did you spot it? It was there.

Now, why is that wrong? It looks right, two lots of 3x is 6x, but you might have written the terms 7x and 4 are inside the bracket so they're correctly multiplied by two, but the term, 3x sits outside of the bracket on the left-hand side, it's on it's own, so it should not have been multiplied.

For part b, did you spot this one? There it is.

What's wrong there? You might have written the second bracket was incorrectly multiplied out and this is a common error when we see negative coefficients of brackets.

It should multiply out negative five lots of bracket y + 1.

We'd need negative five lots of both terms. Negative five ys and negative five 1s, meaning we should get negative 5y - 5.

That should be a negative five, not a positive five.

If you spotted that, well done.

For question three, solving.

Good first step for part a, expand those brackets, and on the left-hand side, we get 14x - 6 + 4x + 36.

We've got like terms, we can simplify from there.

That'll simplify to 18x + 30 on the left-hand side.

From there, add negative 30 to both sides.

Divide through by 18, x equals negative 1.

For part b, three brackets that need expanding, they would look like that.

6x + 18 + 5x + 5 = 9x + 27.

Simplify the left-hand side first.

That'll become 11x + 23 on the left-hand side, lots of options we can take from here.

I'm gonna add negative 23 to both sides.

Then I'm going to add negative 9x to both sides.

We end up at 2x = 4, therefore x = 2.

For part c, slightly trickier this one.

I've got a term which is not in a bracket, that might throw a few people.

We can still expand the brackets that we have though.

Four lots of bracket 2a + 7, that became 8a + 28 on the left-hand side.

On the right-hand side, we still started with the term 12a, and then we had negative six lots of a, and negative six lots of eight.

I can see something to be simplified there.

12a - 6a will leave us with 6a.

From there, lots of options.

If we add negative 6a to both sides, we'll turn it into the equation 2a + 28 equals negative 48.

We can solve from here.

Let's add negative 28 to both sides, then divide through by 2a equals negative 38.

For part d, expanding those brackets.

Be careful with any negative terms here.

We should have got 21x - 28 + 16 - 40x being equal to 9x + 54.

Some simplifying we can do here.

The left-hand side becomes negative 19x.

That's 21x - 40x, negative 19x - 12 being equal to 9x + 54.

Lots of what manipulations we can make from there.

Add positive 19x to both sides, and we'll get 28x + 54 on the right-hand side, and leave just negative 12 on the left-hand side.

Add negative 54 to both sides.

Divide three by 28.

We'll get a solution, x equals negative 66 over 28, which simplifies to negative 33 over 14.

Onto the second half of the lesson.

Simplifying before solving.

We mathematicians love simplifying.

We want life to be as easy as possible.

Jacob and Jun are solving this equation.

Jacob says, "Expand first, then collect like terms." Jacob expands, collects the like terms, simplifies that left-hand side, gets all the way down to the solution, x = 3.

That's wonderful mathematics, well done, Jacob.

Jun says, "I think I see a quicker way." He writes the equation again, and then writes five lots of bracket x + 5 on the left-hand side.

Do you agree with Jun's idea? Pause and have a conversation with the person next to you.

It was an awesome idea.

That's a really nice bit of simplifying we've done before moving on to the next step, expanding.

If we turn two lots of x + 5 and three lots of x + 5 into five lots of x + 5, we've just got one bracket to expand, 5x + 25 = 40.

Now, look, we've got the exact same solution at the end of our equation.

Jun's method got to the same equation, the same solution more efficiently because he simplified before expanding.

He was only able to simplify before expanding, because we had identical expressions inside the brackets.

We're talking about how many lots of x + 5 have you got.

The expressions in the brackets are identical, so it can be simplified.

Two lots of x + 5, I could represent in a bar model, that's x + 5 trice.

x + 5 three times will look like that.

It's equal to 40.

That's five lots of x + 5.

Can you see the one, two, three, four, five lots of x + 5? That's a visual representation of this simplification.

We're adding that identical expression, the x + 5.

By collecting identical expressions, Jun has simplified our equation which will make our solution more efficient.

Jacob and Jun are solving another equation.

Five lots of bracket x - 3 plus three lots of bracket x - 3 = 40 minus two lots of bracket x - 3.

Jacob says, "I see it, we can simplify like this before expanding." I wonder what he's seen.

He's seen on the left-hand side identical expressions, x - 3, five lots of x - 3 and three lots of x - 3 is eight lots of x - 3.

"You can simplify even further than that, Jacob." Says, Jun.

I wonder what Jun means.

Can you see what Jun means? Can you spot how we can simplify this any further? Pause and tell the person next to you.

The x - 3 on the right-hand side is also identical.

If we add two lots of x - 3 to both sides of the equation, we end up with ten lots of x - 3 on the left-hand side and just the numerical value 40 on the right-hand side.

This is a lovely form from maths to solve now.

We can expand the bracket.

Add 30 to both sides, divide 3 by 10, x = 7.

A really complicated-looking equation made far more simple by gathering together the identical expressions, but we're not done yet.

Jacob says, "We could do it even more efficiently." I wonder what he means.

If we go back to this step, Jacob says, "10 is a common factor of 10 and 40." What does that enable us to do? Well, we can divide both sides by 10, or multiply through by a 10th.

That turns 10 lots of bracket x - 3 = 40 into one lot of x - 3 being equal to 4.

Therefore x = 7.

We're always seeking efficiencies in the mathematics that we do.

Jacob noticing that 10 is a common factor, and dividing three by 10 got us to that solution even more efficiently.

Quick, check you've got that.

Which of these equations could be simplified before solving? Equation a, equation b, equation c, or equation d? Pause and have a think about those four.

I hope you said option a could be simplified.

Option b, nope.

Option c, yes.

Option d, yes.

Why? In equation a, we've got identical expressions of x - 2.

We've got five lots of x - 2, and we're taking away two lots of x - 2, leaving us with three lots of x - 2.

For equation b, we haven't got identical expressions, x - 2, x + 2, and not identical expressions, so we can't really simplify before we solve that one, we'd have to expand first.

In c, we've got identical expressions of 3x + 6.

We could add seven lots of 3x + 6 to both sides.

We'll be left with just four on the right-hand side, and we'll have nine lots of 3x + 6 on the left-hand side, we could expand and solve from there.

The last one was a little trickier, but whilst x + 2, and x - 9 are not identical expressions, we do have two identical expressions in those x + 2s.

We could add three lots of bracket x + 2 to both sides of the equation.

We'll just be left with eight on the right-hand side, and we'll have eight lots of x + 2 on the left-hand side.

That's made it a little more simple.

Practise time now.

I'd like you to solve these equations twice.

Firstly, by expanding as your first step, then secondly by simplifying before expanding.

After you finish those equations, I'd like you to write a sentence to compare the efficiency of those two methods.

Write a sentence for part a and another sentence for part b.

Pause and do that now.

Question two.

I'd like you to solve these equations.

There are a lot of ways that you could solve them.

I wonder if you can spot the most efficient way.

I'll say no more so as not to give too many clues away.

Pause and try those two equations now.

Feedback time.

Question one, solve these equations twice once expanding.

Secondly, simplifying before expanding.

If we expand, we get that on the left-hand side, which simplifies to 11x + 44.

Let's add negative 44 to both sides.

Divide 3 by 11, x = 5.

But what if we simplify first? x + 4, x + 4, identical expressions inside those brackets, so we've got 11 lots of x + 4 in total.

From there, what do we notice? 11, 99, a common factor, 11.

Let's multiply both sides through by one 11th.

You might think of that as dividing by 11 on both sides, leaving you with one lot of x + 4 on the left, and just nine on the right-hand side.

Therefore x = 5.

Do you notice? Two very different routes to the same solution.

In terms of comment on the efficiency, you may have written simplifying identical expressions was more efficient with simpler steps on this occasion.

For part b, if we expand, we get to that equation.

Be careful of any negative terms. Make sure yours matches mine.

When we simplify from there, the left-hand side simplifies to negative eight, positive 5x, the right-hand side, 6x - 18.

From there, add 18 to both sides.

Add negative 5x to both sides, X = 10.

If we simplify before expanding, we'll start by looking at the right-hand side.

We've got eight lots of x - 3, and we take away two lots of x - 3.

We're left with six lots of x - 3.

We've got identical expressions in brackets on both sides.

We can subtract five lots of x - 3 from both sides.

That'll leave us with seven on the left-hand side, and just one lot of x - 3 on the right-hand side.

Therefore, x = 10.

In terms of efficiency, you may have written simplifying identical expressions was again more efficient, and also avoided a lot of manipulating negative terms. That's why we see a lot of errors.

Pupils frequently make mistakes when expanding brackets involving negative coefficients, negative terms. We avoided that by simplifying, adding identical expressions first.

Question two, part a, solve.

Lots of ways you could have done this.

You might have expanded the brackets.

That's fine.

I didn't.

I did that.

Can you see what I've done? 4x + 24 on the left-hand side is four lots of bracket x + 6.

I've actually factorised.

Why have I done that? Because I'm now talking about identical expressions of x + 6.

On the left-hand side, 13 lots of bracket x + 6.

Now, we can subtract 11 lots of x + 6 from both sides, and we get to a solution, x = 6.

Like I said, you might have started by expanding all the brackets.

If you arrived from that method at x = 6, your calculations were probably correct.

Question two, part b, solve this equation, and what a beautiful equation it is.

You might have noticed we've got identical expressions, y + 5, but on the right-hand side, half of y + 5.

Ouch.

What do we do there? Well, half of y + 5, what's the reciprocal of a half? Two.

If we multiply both sides by two, this happens.

We have twice as much on the left-hand side, we've got two lots of 13, and when we multiply that last term, that half of y + 5 by 2, we're left with one lot of y + 5.

From there, we've got identical expressions, y + 5 on both sides.

I'm gonna add a y + 5 to both sides.

I'll get 17 lots of y + 5 on the left-hand side, and I'll just be left with 26 on the right-hand side.

From there, we can expand, add negative 85 to both sides, divide through by 17, and y equals negative 59 over 17.

Sadly, we're at the end of the lesson now, but in summary, we've learned that equations involving brackets don't need to be expanded as a first step if they can be simplified first.

For example, seven lots of bracket y - 11 plus three lots of bracket y - 11 = 18 can be simplified to 10 lots of bracket y - 11 = 18 before solving.

Hope you've enjoyed today's lesson as much as I have, and I hope to see you again soon for more mathematics.

Goodbye for now.