Lesson video

Hello, Mr. Robson here.

Rearranging to solve linear equations.

Great choice by you today to go for this lesson.

I'm excited so let's get started.

Our learning outcome is that we'll be able to recognise that equations with unknowns on both sides of the equation can be manipulated so that the unknowns are on one side and then solve the equation.

Some keywords we're gonna need throughout the lesson, equation and equality.

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

Two expressions form an equality if, when substitution takes place, the expressions have the same value.

For example, substituting x equals 5 into 3x minus 1 equals x plus 9 would give us 3 lots of 5 minus 1, which is equal to 5 plus 9.

We have equality.

Two parts to the lesson and we're gonna start by representing equations with unknowns on both sides.

Some pupils are given an equation to solve.

Aisha says, "I've got 3x equals 9.

That's easy." Lucas says, "I've got 3x plus 1 equals 10.

That's okay." Whereas Sam says, "I've got 5x plus 1 equals 2x plus 10.

I've got the hardest one by far." Sofia says, "Hold on Sam, I think your equation is almost the same as Lucas and Aisha's." Can you see what Sofia has spotted? Pause this video and have a look at the three pupils equations.

Let's look at how the equations are connected.

What is it that Sofia spotted? She says, "It's easier to see the link if you write them like this." Add positive 1 to both sides of Aisha's equation and we get Lucas'.

Oh look, 3x and positive 1, 9 and positive 1 turns Aisha's equation into Lucas' equation.

Add 2x to both sides of Lucas' and we get Sam's.

If we add positive 2x to both sides of the equation, we end up at 5x plus 1 equals 2x plus 10.

You can see this building up on our balance scales.

I'll start again with 3x being equal to 9.

Add 1 to both sides.

That's how the algebra looks.

And we end up at 3x plus 1 on the left-hand side, 10 on the right-hand side.

Add 2x to both sides of the balance scales and that's how the algebra looks.

We've created the equation 5x plus 1 equals 2x plus 10 by adding 1 to both sides and then adding 2x to both sides of Aisha's equation.

We can undo this building up in order to solve equations with unknowns on both sides.

This is Sam's equation.

5x plus 1 equals 2x plus 10.

We'll start by adding negative 2x to both sides.

That looks like that.

And then we remove the zero pairs, a positive 2x and a negative 2x will cancel each other out and disappear from our scales.

That leaves us with 3x plus 1 being equal to 10 and I think you can see the step that we want to take next.

We wanna remove that 1 from the left-hand side.

We do that by adding negative 1 to both sides.

That's how the algebra looks.

And then the positive 1 and the negative 1 make a zero pair, they disappear and that leaves us with 3x is equal to 9, therefore 1x is equal to 3.

That's a solution.

If I remove the visual representation of the balance scales and just show the algebra, we started at a 5x plus 1 equals 2x plus 10.

We started with unknowns on both sides.

The unknown term 5x on the left-hand side, the unknown term, 2x on the right-hand side.

We manipulated the equation by adding negative 2x so that we have the unknowns on only 1 side.

That's an important step.

It gets us much closer to our solution.

We then isolated the unknown by adding negative 1 to both sides, leaving just the unknown term on the left-hand side.

They're important steps when we're solving equations with unknowns on both sides.

We can represent the solving of this very same equation using what we call a bar model.

That is the equation 5x plus 1 being equal to 2x plus 10.

My bar in my top row has five x's and 1, the bar in my second row has two x's and 10.

They're the same length and it's important that they're the same length because that demonstrates the equality between the two expressions.

What then happens, when I draw a line there it enables me to focus on that moment.

Effectively I've added negative 2x or removed 2x from both sides of the equation.

We now have 3x plus 1 is equal to 10.

Just our visual representation looks different to the balance scale.

The next question is what is that length? Well, it's 10 minus 1, it's 9.

Oh look, we now have three x's being equal to 9.

Therefore x must be 3.

Bar models are a really powerful way to solve equations with unknowns on both sides.

Quick check that you've got everything I've said so far.

What equation is represented by this balance model? Is it 5x equals 9? 4x plus x equals 3 plus 6? Or 4x plus 3 equals x plus 6.

Pause and tell the person next to you.

I hope you said option C, 4x plus 3 on the left-hand side, x plus 6 on the right-hand side.

Which useful first step starts to solve this equation? 4x plus 3 equals x plus 6.

What first step are we gonna take? Are we're going to add x to both sides? Are we going to add negative x to both sides? Or will we add negative 4x to both sides? What do you think? Pause, tell the person next to you.

Adding negative x will be a really useful first step.

If we add negative x to both sides, it creates a zero pair, which we can remove leaving us with 3x and 3 on the left-hand side and just 6 on the right-hand side.

That enabled us to turn the equation 4x plus 3 equals x plus 6 into the equation 3x plus 3 equals 6.

What step might we take next? Would it be to add negative 3 to both sides? To add positive 3 to both sides? Or to add negative 6 to both sides? What's the most efficient thing we could do? Pause and tell the person next to you.

It was option A, adding negative 3 to both sides.

That will look like that.

Negative 3 and positive 3 become a zero pair and cancel each other out.

We'll remove them from both sides of the equation.

We've now got 3x is equal to 3 therefore a solution x equals 1.

A bar model now.

Which equation is represented by this one? Is it 10x equals 2x plus 13? 5x plus 8 equals 7x plus 5? Or 8x plus 5 equals 5x plus 7? Which one is it? Pause.

Tell the person next to you.

It's option B.

5x plus 8 equals 7x plus 5.

Same equation.

What does this line show us? Does it shows that 5x equals 5x? That 8 equals 5? Or that 8 equals 2x plus 5? Pause and have a think about those three options.

I hope you said it does show us that 5x equals 5x.

It doesn't show us that 8 equals 5 because the 8 and the 5 are not the same length and 8 and 5 are not equal.

And it also shows us that 8 is equal to 2x plus 5.

To the left-hand side of the line we can see five x's and five x's being equal in length.

This is really useful.

Because they're equal we retain equality when we remove them from both sides of our equation.

A manipulation to have the unknowns on one side only is what we read from the right-hand side of the line.

The length of 8 being equal to the length of 2x plus 5 means we've removed five x's from both sides.

We've now got unknowns on only one side.

That means we're much closer to a solution.

What is this length? What length? This length.

Is it 2, 3 or 5? Pause.

Tell the person next to you.

I hope you said 3 because it's 8 minus 5.

So what is our solution? The last step.

Is it x equals 1.

5? x equals 3? Or x equals 3 over 2? Pause and tell the person next to you.

x equals 3 over 2.

We have 2 x's being equal to 3, therefore x is equal to 3 divided by 2, which we write as 3 over 2.

You've noticed something haven't you? 3 divided by 2 can also be expressed as 1.

5.

But it's not 3, the solution is not 3.

Most often you'll see solutions written as a fraction but sometimes in context we might want to call it x equals 1.

5 instead of x equals 3 over 2.

Aisha and Sam is solving the equation x plus 13 equals 6x plus 3.

Aisha says, "You always have to start with the variable terms so I'm gonna add negative x to both sides." Sam says, "You don't have to start with the variable terms. You can manipulate the constants first.

I'm gonna add negative 3 to both sides." Who's right? Pause and have a discussion with the person next to you.

Both of our pupils' methods work.

There are no rules about what you manipulate first.

In Aisha's case, she wants to add negative x to both sides first.

When we do that we have a zero pair where the positive x and the negative x disappear leaving us with 13 on the left-hand side, 5x plus 3 on the right-hand side.

We can then add negative 3 to both sides to turn our equation into 10 being equal to 5x, giving us a solution, x equals 2.

In Sam's case, Sam wanted to add negative 3 to both sides of the equation, manipulating the constants, not the unknown terms. By adding negative 3 to both sides we get a positive 3 and a negative 3, a zero pair.

They cancel each other out and disappear from our model.

We're left with x plus 10 equals 6x.

The next step, add negative x to both sides, which leaves us with 10 being equal to 5x, a solution x equals 2.

Let's compare the two methods.

That was Aisha's method, adding an unknown to both sides to start.

And there's Sam's method, manipulating the constants first.

Compare the two.

What do you notice? Pause.

Have a good look and a good think.

Did you notice same steps, different order? Aisha went for add negative x and then had to add negative 3.

Whereas Sam started with add negative 3 and then later had to add negative x.

Exactly the same steps, just done in a different order.

You also notice the exact same result.

They're both accurate methods so we both led to the same solution.

It's fine to do it either way.

Quick true or false.

When solving the equations you have to manipulate the variable terms first? Is that true? Is it false? Could you justify your answer with one of these two statements? We need the unknowns on one side in order to solve or you can start by manipulating either the variable terms or the constant terms. Pause and have a think about this question.

I hope you said false, because you can start by manipulating either the variable terms or the constant terms. Practise time now.

Question 1, build an equation with unknowns on both sides.

Start with this balance, which shows 2x being equal to 8.

We have unknowns on only one side.

And show what you need to add to both sides to make the equation 7x plus 2 equals 5x plus 10.

You should draw some extra things on your balance scale model and write down the algebraic notation to reflect what you're adding to the balance model.

Pause and do that now.

For question 2, I'd like you to undo this balance model to solve this equation.

Write your algebraic notation at the same time alongside your model.

The equation is 6x plus 4 equals 3x plus 10.

What are you gonna do to your balance model to solve? Make sure you write the algebraic notation that matches whatever manipulations you make to the model.

Question 3, use this bar model to solve this equation.

That's 2x plus 161 being equal to 8x plus 29.

Whilst using the bar model, be sure to write your algebraic notation alongside.

Pause and do that now.

Question 4, draw a balance scale and a bar model for this equation.

Two diagrams please.

And then I'd like you to use both of those diagrams to solve of course, writing your algebraic notation as you go.

Pause and do that now.

Let's see how we did.

We are building an equation with unknowns on both sides.

Transforming 2x equals 8 into 7x plus 2 equals 5x plus 10.

What did we need to do? A good start would be to add 5x to both sides of your balance scales.

The algebraic notation of that would look like so.

We now have 7x on the left-hand side, 5x plus 8 on the right-hand side.

What else do we need? We need to add 2 to both sides.

Adding 2 to both sides algebraically looks like that.

We created the equation 7x plus 2 equals 5x plus 10.

For question two, undo this balance model to solve the equation.

First useful step to undoing, there are multiple steps.

I think a good one's to add negative 3x to both sides.

The zero pairs disappear, leaving you with 3x plus 4 on the left-hand side and 10 on the right-hand side.

I'd like to see that 4 disappear.

I'm gonna do that by adding negative 4 to both sides of my balance model.

When that zero pairs disappear, we're left with 3x is equal to 6, therefore x is equal to 2.

If you manipulated the constants first and still arrived at the same result, that's absolutely fine.

For question 3, using this bar model to solve the equation, we'll start with a line there because we're gonna remove 2x from both sides of this equation.

We're then focusing on that moment, the length of 161 being equal to 6x plus 29.

That's 161 being equal to 6x plus 29.

There's your algebraic notation.

We're then interested in that length.

Well that length's 161 minus 29.

It's 132.

What does it tell us? It tells us that 132 is equivalent to 6x, therefore we have the solution x equals 22.

Question 4, I asked you to draw a balance scale and a bar model for the equation x plus 18 equals 5x plus 2.

Your balance scale should look like so.

Your bar model like so.

I then asked you to use them to solve.

My first step is to add negative x to both sides of my balance scale and to draw that line there on the bar model because that's gonna reflect the removal of x from both sides of the equation.

We've turned our equation into 18 on the left-hand side, 4x plus 2 on the right-hand side, or 18 in the top row of the bar model, 4x plus 2 in the bottom row of the model.

We're then interested in that length.

We find that length by doing 18 minus 2 or adding negative 2 to both sides of the balance model.

We get 16.

If 4x equals 16, then x must be equal to 4.

Second half of the lesson now.

We're gonna solve equations with unknowns on both sides really efficiently.

Andeep and Laura are looking at a problem.

They're trying to solve x plus 3 equals 2x minus 4, and Andeep makes a plea for help.

"I'm not sure how to deal with negatives," he says.

Luckily Laura is here to help.

"Here, let me show you." Laura draws a balance model, x plus 3 on the left-hand side, and 2x and negative 4 on the right-hand side.

Now you add positives to undo the negatives.

Ah, we've added 4 positive 1s to both sides.

It's the same as before.

We've got positive 4, negative 4, a zero pair.

That can disappear from our model.

That means we're left with x plus 7 being equal to 2x.

And look, all the terms are positive now.

We end up with a solution x equals 7.

Quick check you've got that.

Which useful first step starts to solve this equation, x plus 3 equals 4x minus 6? Is it, add positive 3, add negative 6 or add positive 6? Pause.

Tell the person next to you.

I hope you said option C, add positive 6.

When we add positive 6 to both sides of the equation, we'll transform it into x plus 9 equals 4x.

From there, we can add a negative x to both sides and we'll get 3x being equal to 9 therefore x equals 3.

Andeep says, "Can you do negatives with bar models?" Laura's a bit of an expert.

"Absolutely, it's more difficult to set them up, but if you master it, they can be easier to solve." That's x plus 3.

That's nice and straightforward.

2x minus 4.

Hmm.

There's 2 x's or 2 positive x's.

2x's moving in the positive direction towards the right.

The negative goes in the negative direction.

Can you see that is a move of 4 backwards or in the negative direction? That's 2x minus 4.

Those bottom bars.

From there x equals 7 as a solution is really rather obvious.

It's just a little trickier to draw this bar model.

Let's check you've got that.

Which equation does this bar model represent? Hmm? Is it x plus 13 equals 3x minus 5? x plus 18 equals 3x? Or 2x equals 18? Which equation is represented by that model? Pause and have a think about those three.

I hope you said option A, x plus 13 equals 3x minus 5.

Why? Because that's x plus 13.

And that's 3x is moving in the positive direction and 5 moving in the opposite direction or the negative direction.

Hence that's 3x minus 5.

Did you also say I can see x plus 18 equals 3x? I hope so because you can.

There is our 3x minus 5.

When we apply the x plus 13 there can you see x plus 18 on the top bar and three x's on the bottom bar? Did you also say I can see 2x is equal to 18? Can you see it? You can when you remove x from both bars.

The answers were the steps to solving this equation.

If I write those three answers there, the step from A to B was to add positive 5 to both sides of the equation.

The step from B to C was to add negative x to both sides of the equation.

Our solution, x equals 9.

Andeep and Laura are solving another equation.

"Whoops, I've started to solve 8x plus 4 equals 3x plus 19, but I've made a mistake.

I'll have to start again." What mistake has Laura made? Oh, she's added negative 8x to both sides.

4 equals negative 5x plus 19.

She thinks she's made a mistake and she wants to start again.

Andeep says, "No.

Keep going.

Add negative 19 to both sides." Do you agree with Andeep's advice or do you think Laura should start again? Pause and have a think.

It is okay to say yes.

It's also okay to say no.

We need to be really fluent when we're solving equations.

Laura could start again and maybe add negative 3x to both sides of the equation, but there's no need to start again.

Andeep's right.

If we add negative 19 to both sides, we'll get there.

Add negative 19 to both sides.

Negative 15 equals negative 5x.

Divide through by negative 5 and we get the solution x equals 3.

You don't have to manipulate the smallest x first.

When we look at that equation, 8x on the left-hand side, 3x on the right-hand side, the temptation is to add negative 3x as a first step.

That's fine.

It's efficient.

We don't have to.

Adding negative 8x works perfectly as valid.

Quick check you've got that.

Which of these are valid starts to solving this equation? 8x plus 7 equals 29 minus 3x.

Is it, add 3x, add negative 7, add negative 3x, add negative 8x? There may be more than 1 correct answer.

Pause and have a think.

Option A was a good one.

Add 3x to both sides.

Option B was a good one as well.

Add negative 7 to both sides.

Option C, not good.

Option D, absolutely.

Add negative 8x to both sides.

Good start.

So adding 3x to both sides will transform the equation into 11x plus 7 equals 29.

We're closer to a solution.

Adding negative 7 to both sides of the equation will transform it into 8x equals 22 minus 3x.

We're closer to a solution.

Adding negative 3x to both sides, no, no good.

We'll be left with 5x plus 7 equals 29 minus 6x.

We've still got unknowns on both sides of the equation.

We're no closer to our solution.

Adding negative 8x gives us 7 being equal to 29 minus 11x.

Another valid start.

Sometimes a model is inappropriate.

In this case we just use a fluency in manipulation.

If I said draw a balance model or a bar model for one fifth of x minus 7 equals 13 minus 4x, it's a really tough diagram.

So we just need to be fluent in our algebraic manipulation.

A sensible start here is to multiply both sides of our equation by 5 because multiplying by 5 is the inverse of multiplying by a fifth.

Multiplying both sides by positive 5 means we no longer have a fraction in our equation.

We multiply out those brackets, we get x minus 35 equals 65 minus 20x.

Then we can add 20x to both sides, meaning we now have a positive unknown term on one side.

From there, add 35 to the equation and we'll get x equals 100 over 21.

Quick check you've got that.

Which of these is a useful first step to solving this equation? 7 minus 3x equals 13 minus a quarter x.

Would it be to multiply by positive 4, multiply by positive quarter, or multiply by negative 4? Pause and have a think.

I hope you said option A is a good one, multiplying by positive 4.

I hope you said option B is not a good one, and I hope that option C is a wonderful one.

Not efficient to multiply by positive a quarter.

Why? Because we'll end up with that.

Multiplying by positive 4 was a really efficient start.

Multiply all those terms by 4 and we're left with 28 minus 12x equals 52 minus x.

We can solve that.

Arguably even more efficient was to multiply by negative 4 because you're left with 12x minus 28 equals x minus 52.

Our unknown terms are positive, a little easier to work with.

Practise time now.

For question one, I'd like you to solve these equations in two different ways.

Pause and give this a go now.

Question 1 part B.

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

Pause and give this a go.

For question two, find the perimeter of this rectangle.

Pause and think about this one.

For question three, I'd like to finish solving this equation in both ways.

Pause and do that now.

Question 4, solve one eighth of x minus 7 equals 3 minus 7x.

Pause and give it a go.

Some feedback now.

Solving this equation in two different ways.

The method I've shown you there is to start by manipulating the unknown terms. I've added negative 3x to both sides of the equation first, turn it into 2 equals 4x minus 10 and I'll add positive 10 to both sides to leave us with 4x being equal to 12, therefore x equals 3.

My other different way was to manipulate the constants by adding positive 10 to both sides and then adding negative 3x to both sides, arriving again at 4x equals 12, therefore x equals 3.

There were multiple other methods you might have used, which should have led to the same solution.

For part B, again, I've started by manipulating the unknown terms first.

I think a positive 5x to both sides is quite an efficient way to start.

Turns the equation into 7x plus 2 equals 23.

Add negative 2 to both sides.

7x equals 21, so x equals 3.

What if we manipulated the constant terms first? Adding negative 2 to both sides, add the positive 5x to both sides, therefore x equals 3.

Question 2, finding the perimeter of the rectangle.

If we know that the top length, 17 minus 2x is the same as the bottom length, 3x plus 2, we can make an equation out of that.

Those two lengths are equal.

If we make the equation 17 minus 2x equal 3x plus 2, you can solve like so.

We find that x has a value of 3.

We'll substitute that into the length, 3 lots of x plus 2, 3 lots of 3 plus 2.

That length must be 11.

If the base length is 11 and the height is 7, our perimeter must be 2 lots of brackets, 11 plus 7.

Our perimeter must be 36.

For question 3, I ask you to finish solving this equation in two different ways.

I've added negative 2x to start, which turns the equation into 10x plus 30 equals 70.

From there it's sensible to add negative 30 to both sides creating 10x equals 40, therefore x equals 4.

If we go down the other path and add negative 12x, it works.

We'll end up with 30 being equal to 70 minus 10x.

We'll then add negative 70 to both sides and we'll get negative 40 equals negative 10x.

X equals 4 therefore.

Exact same solution.

We just got there in a different way.

For question 4, solving this beautiful equation, what's the opposite of one eighth? Hmm, multiplying by 8.

So multiplying both sides by 8 will look like that and then when you multiply out those brackets, you'll get x minus 56 equals 24 minus 56x.

From there, you can add 56x to both sides of the equation and then add 56 to both sides of the equation.

57x equals 80.

Hmm, that's a tricky one, or is it? Just divide through by 57.

Our solution x equals 80 divided by 57, which we write as 80 over 57.

What a beautiful equation.

In summary, equations with unknowns on both sides can be manipulated in many ways so that the unknowns are on one side and we can therefore solve.

I hope you've enjoyed today's lesson.

I have and I look forward to seeing you again soon for more mathematics.

Goodbye for now.