Lesson video

Hello, thank you for joining me today.

My name is Ms. Davis and I'm gonna help you as we work through this lesson.

I'm really looking forward to working alongside you.

There's lots of exciting things that we're going to have a look at.

I really hope there's bits that you enjoy and bits that you find a little bit interesting.

Maybe there'll be some things you haven't seen before as well.

Let's get started then.

Welcome to our lesson on solving simple linear equations with an additive step.

By the end of the lesson, you'll be able to solve a linear equation requiring a single additive step.

A few keywords that are gonna be really important to us today.

Firstly, a solution.

A solution to an equation with one variable is the value of that variable which maintains the equality between the expressions.

We're also gonna talk lots about additive inverses.

The additive inverse of a number is a number that when added to the original number gives the sum of zero.

So a number and its additive inverse make a zero pair.

Some examples, (-7) and 7 are additive inverses because (-7) and 7 is zero.

They are a zero pair.

x and (-x) is zero.

0.

3 and (-0.

3) is zero.

And (-2/5) and 2/5 are a zero pair.

We're gonna start with looking at additive steps using representations.

An equation with an additive relationship can be represented with a bar model.

Looking at this bar model, what equations can you write from this? You may have found a + 6 = 20.

6 = 20 - a, or a = 20 - 6.

When there's only one variable, it is possible to use a bar model to find the solution.

The solution will be a value for a, which makes both sides of the equation equal.

One of these equations, that we've already written, makes it clear what the solution is.

Have a look at them.

Which one do you think it is? Well done.

That bottom one tells us exactly what we need to do to calculate the value of a.

So 20 - 6 gives us the solution.

The solution to all these equations is when a is 14.

We can substitute a is 14 just to check.

14 = 20 - 6.

14 = 14.

That's balanced.

14 + 6 = 20.

20 = 20.

That one's balanced.

6 = 20 - 14.

6 = 6.

So that last one is balanced as well.

What operation have we done to both bars to get the value of 14? So we looked at rearranging our bar model, but what operation is it that we've done to the bars to find 14? Right.

We've subtracted 6 or added -6 and that's shown us that a has the value of 14.

Let's have a go with another bar model.

What equations can you write this time? Well done if you've got all three of those.

Which one tells us what the solution is? That last one tells us that the solution can be found by doing 20 - 12.

The solution then is when b is 8.

It does not matter that I've got 8 on the left-hand side of my equation and b on the right.

That is still a solution.

What operation have we done this time to get the value of 8? Right, this time we've subtracted 12 or added -12 and that gave us 8.

This bar model looks slightly different.

What equations can you write this time? Well done if you found all three of those.

Which equation makes it clear what the solution is? That first one this time p = 20 + 3, we can see clearly then p has a solution when it's 23.

We can check this works for the other equations.

23 - 3 = 20.

20 = 20.

23 - 20 = 3.

3 = 3.

This solution was possibly easier to find than the previous two and that's because the bar representing the p was on its own.

We did not need to rearrange the equation or apply an operation.

We literally could see from the diagram that p was equivalent to 20 + 3.

The process of finding the solution for an equation is called solving.

When solving an equation, we're looking for the value of the unknown.

To do this, we need to isolate the unknown to see its value on its own.

Quick check.

Which of these equations are represented by this bar model? When you're happy with your answer, what is the solution to those equations? Can you put into words how you know? There was five to find, I wonder if you managed to find all of them? The one that gives us the clearest idea of what the solution is is probably that top one that says r = 31 - 12.

Then I can do 31 - 12, and I know that r must be 19.

This balance represents the equation, x + 3 = 5.

What have I done to the left scale pan? Right, I've added one.

What would we need to do to the right-hand side to balance the scales? That's correct.

We have to add one to the right-hand side as well.

This time I've added -1.

If you've got algebra tiles with you, you might want to do this alongside me.

What will happen to the left-hand side if I add -1? Well, 1 and -1 are zero pairs.

They add to 0, so that means the -1 and 1 of the 1 tiles will cancel out.

Adding -1 one is the same as subtracting 1.

In order to maintain equality, I need to do the same thing to the right-hand side.

So add a -1, 1 and -1 are zero pairs, so that gives me 4 on the right-hand side.

Our new equation is x + 2 = 4.

We're gonna have a look at a couple of other examples.

Think about what I've done to the equation, x + 3 = 5.

So what I've done here is I've added -3, that is the same as subtracting 3.

So I end up with x = 2.

This time I've added + 3, so I now have the equation x + 6 = 8.

Let's look at all four examples.

Which was the most useful operation for finding the solution? Well done if you said it was that top right.

We've maintained equality with all four of those operations, but by adding -3, we have seen the value of x.

What it has done is isolated one + x so we can see what it is equal to.

We can see clearly now that there's a solution when x is 2.

Let's look at this balance.

It represents the equation x + 1 = 5.

What operation will isolate x and why? Pause the video and answer this one.

Well done if you said adding -1, adding -1 is the additive inverse to + 1.

1 and -1 are zero pairs.

So by adding -1, the left-hand side will just give us x.

If you said subtract one, it's the same thing.

The solution now is clearly when x is 4.

This time I've represented the equation, 15 = x + 12, does not matter which side of the equation my unknown is on, the process is the same.

I'd like you to take your time to think about what operation will isolate x and why? And then we'll look at it together.

We need to add -12 as it is the additive inverse to + 12, 12 and - 12 are zero pairs.

Adding -12 will isolate x.

Any of those statements would've been a good reason.

So let's show add -12 to both sides.

They're zero pairs, so on the right-hand side I'm left with x, on the left-hand side 15 + -12 is 3.

I can clearly see that my equation is 3 = x.

So there's a solution when x is 3.

Some pupils are trying to solve this equation.

Sam says, "I'm going to add 3." Jacob, "I'm going to add -x." and Sophia, "I'm going to add -7." Will any of these options isolate x? What do you think? Let's look at them.

Sam adds 3.

They've maintained equality by adding the same value to both sides.

We are left with 2x + 10 = 19.

Sam's choice does not isolate x however.

Let's look at Jacob's, adding -x.

The left-hand side becomes x + 7 and the right-hand side (-x) + 16, there was no x term there to collect the right terms with.

So I'm left with (-x) + 16.

Sophia's.

Sophia adds -7 to both sides, is left with 2x = 9.

So Sophia's choice also does not isolate one x.

We have maintained a quality with each of those, but none of the options has shown us what x has a value of.

Sophia says, "What about if I combine my idea with Jacob's and now add -x?" Let's have a look.

If she adds -x then that does leave us with an x on the left-hand side.

However, we've now got 9 - x on the right-hand side, so it looks like we've isolated x but there's still an x term on the other side.

It doesn't tell us what x is.

The reason these three students struggled is this is actually an example of an equation where we cannot isolate x with just an additive step.

Time for you to have a go.

Which operation will isolate x in each of these? You've got some choices on the right-hand side.

First one you should have had, + (-1), then + (-5).

That's the top right.

+ (-4).

So the bottom left.

And + (-9).

That's the bottom right.

Well done.

You're ready for the next bit.

Time for a practise, I'd like you to find the value of the unknown in each bar model.

The important part of this is to explain how you know.

It may be that you can just look at some of these and tell me the answer straight away.

The important part with additive relationships is getting used to writing your working and explaining why.

The reason being, you might well combine this skill with other skills later on where you can't just spot the value straight away.

Give those a go and we'll look at the next bit.

Good work on your bar models.

We're now gonna look at some balances.

These balances work down the page.

So your original equation is on the top balance and then I would like you to think about how to show the additive step that would give us the solution.

Try it for those three and then we'll look at the answers.

So the first one, this represents the equation, 15 - 9 = a.

So 6 = a and there's a solution when a is 6.

For b, we could write the equation, b = 41 - 33, so b is 8 and there's a solution then where b is 8.

C, we could write this equation as c = 27 + 31.

That means there's a solution when c is 58.

And finally, this represents the equation 8.

4 - 5.

7 = d, 2.

7 = d.

So there's a solution when d is 2.

7.

For our balances we need to + (-2).

Then we've got x = 5, solution when x is 5.

B, you need to choose to + (-11) this time.

That leaves us with an x on the left-hand side and 19 on the right-hand side, solution when x is 19.

And finally you need to + (-8).

So on the left-hand side, 19 + (-8) is 11, on the right-hand side 8 + (-8) is 0.

So we just get left with x.

Solution then when x is 11.

Well done.

Brilliant.

All those skills that we've just looked at using our bar models and our balances, we're now gonna apply to questions without representations.

So let's solve the equation x + 17 = 53.

Remember what I said? It's not about just spotting the answer.

We want to think about what calculations we're doing.

When we write mathematics, we want people to be able to follow our work as if it was like an argument about why our answer is right.

So it needs to be nice and clear even if we know in our head that we're definitely right.

So we've got x + 17 = 53.

The additive inverse of + 17 is + (-17).

So I'm gonna write this step in and it's important, at this stage, that you are writing this step in.

Maybe when you get onto longer equations you might want to get rid of this step, but it's important at this moment in time.

So x + 17 + (-17) = 53 + (-17).

This leaves us with x on the left-hand side and 36 on the right-hand side.

When we solve an equation, we write our working out down the page.

Each equivalent equation goes on a new line and we have shown clearly that we've maintained equality by adding the same value to both sides.

Your challenge today is to try and present your working in an equally clear and obvious way.

Let's try a few more.

How can we solve the equation y + 0.

7 = 6.

5? Well, we need the additive inverse of 0.

7 which is -0.

7.

So y + 0.

7 + (-0.

7) = 6.

5 + (-0.

7).

The left-hand side is left with y.

That's why we've chose to + (-0.

7).

The right-hand side, 6.

5 + (-0.

7) is 5.

8.

If you feel like you need to do more working out, so for example a column edition or subtraction method at any point, that's absolutely fine.

Please do write that down on your page but don't include it as an equivalent equation.

Let's try this one.

How could we solve x + 2/3 = 5/6.

Again, we want the additive inverse.

The additive inverse of 2/3 is -2/3.

Now this time, I want to convert my fractions to have the same denominator to make it easier for me to do that addition.

So I can write the left-hand side as x, but I've just rewritten the right-hand side as 5/6 - 4/6.

I've got an equivalent fraction 2/3 and 4/6 are the same.

That's fine to write it as part of my working out 'cause it's still an equivalent equation.

Underneath, I can write that as x = 1/6.

This also works when the constant is negative.

We're gonna solve the equation d - 17 = 53.

What do you think the additive inverse will be this time? Well done if you said that the additive inverse of -17 is + 17.

So what we're gonna do is we're gonna add 17 to both sides of the equation.

The left-hand side says d - 17 + (17).

Well, if I have something and I - 17 and then I add 17, I get back to where I started.

So the left-hand side is just gonna be d and the right-hand side 53 + 17 is 70.

See if you can have a go at this one.

Lovely.

Let's look at our working out.

We want to add 5.

3 to both sides, which gives us a = 16.

1.

Right, I'm gonna show you an example on the left-hand side.

I would like you to watch carefully and then you are going to have a go.

We're gonna solve the equation x + 18 = 12.

First, I'm gonna identify the additive inverse of + 18, which is -18.

So I need to add on -18.

If I do that to the left-hand side of my equation, I need to do the same to the right.

That gives me x on the left-hand side and then 12 + (-18) is -6.

So the solution this time is when x = -6.

I'd like you to have a go at the one on the right-hand side.

Pay attention to your working out so that somebody else has followed clearly what you've done.

Well done.

Bonus marks for perfect Working out.

Let's have a look, x + 20 = 17.

Now we are showing that we found the additive inverse and that we've done the same thing to both sides.

So x + 20 + (-20) = 17 + (-20).

Now we can write our solution as a solution when x = -3.

Let's have a go at this one, x - 18 = -10.

Our additive inverse is + (18).

So I'm gonna add it to both sides of my equation.

That leaves you with x on the left-hand side, <v ->10 + (18) is 8.

</v> I'm gonna check it this time by substituting.

So 8 - 18 = -10.

8 - 18 is -10.

So our solution must be when x is 8.

Have a go at solving this equation and then show how you would check your answer.

Lovely.

Let's go through it together.

We're adding 21 to both sides, <v ->6 + (21) is 15.

</v> To check our answer, we're gonna do (15) - 21 = -6 and (15) - 21 is -6.

The solution is when x is 15.

Make it clear what your solution is.

Otherwise people might think that your solution is -6.

Remember, the solution is the value for x that makes both sides balance.

Some pupils are trying to solve the equation, x - 2 = 7.

I would like you to look at their three methods and identify what mistake they have made.

Alex's one was a little bit difficult to spot.

He seemed to have tried to add 2 and suddenly his x has become 3x.

He's actually added 2x as well as adding 2.

Jacob has picked the right operation but he forgot to add it to the right-hand side of his equation as well.

So he hasn't maintained equality.

And Aisha has added -2 instead of + 2, she needs to add the additive inverse, which is + 2.

It is okay to make mistakes, that's how we learn.

As long as we've laid out our working in a way that makes it easy for us to spot our mistakes and then having a method to check our answers is really useful.

So what have these pupils done well and what advice could we give them to help them in the future? So things they've done well, they've all set out their working out going down the page, making it clear.

That's how we could see what part of their working had gone wrong.

What you might want to suggest to them is they actually write the operation they are doing on both sides of their equation.

They've kind of missed out that middle step.

They should have written x - 2 + 2 = 7 + 2, and then it's gonna be really clear to spot where they've gone wrong.

Have they made the wrong choice of what to add or have they just done their calculation wrong? Time for a check.

What is the additive inverse for each of these? Right, we've got + (-3.

4), + 0.

5, + 3/5, and + (-1).

Which of these is a solution to x + 13 = 8? And how do you know? Well done if you spotted it's -5.

There's two ways we could work it out from the equation.

So we add the additive inverse, so we add -13 to both sides.

8 + (-13) is -5.

Equally we could check by substitution.

(-5) + 14 does give us 8.

Which of these is a solution to x + 6 = -10? Again, how do you know? Well done if you spotted it's -16.

Just like before, we could work it out by adding the negative inverse, -10 + (-6) is -16.

Equally we might want to check by substitution.

So is (-16) + 6, -10? Yes, it is.

So both sides of our equation are balanced.

Time to put all that into practise.

I would like you to solve the following equations and I want you to show each step of your working.

Come back when you're ready for the next bit.

Well done, this time I would like you to spot the mistakes in the following methods.

Off you go.

Come back when you're ready for the answers.

Well done.

Hopefully you're feeling really confident now with solving equations with an additive step, particularly paying attention to our working.

So the first one we want to + (-1).

We get a solution when x = 11.

Give yourself a big tick if you're working looks neat like mine.

B, we want to + (-3.

6).

So x is 14.

2.

C, we want to + 18 so x is 60.

D, we want to + (-13) so x is -2.

E, we want to + 10, <v ->8.

4 + 10 is 1.

6.

</v> The final one, I wanted a bit more space for my working so I can make it really clear.

So I want to + (-3/5).

Right, then when I've tried to + (-3/5), I thought it'd be helpful if they had a common denominator.

So I found an equivalent fraction as 6/10.

So I've rewritten my equation as x = -9/10 - 6/10.

<v ->9 - 6 is -15.

</v> So -9/10 - 6/10 is -15/10.

I've got a solution of x = -15/10, but it's always good to write our answer in its simplest form.

So I've written it as -3/2.

If you wanted to change it into a decimal and write it as -1.

5, that's absolutely fine.

Generally, fractions are a nicer form to leave your answers in, but if it doesn't specify, then leave it in the form that suits you best.

Let's see if we can spot these mistakes.

So in a, we should have added -7, not 7.

In b, whoever did this forgot to add the -7.

1 to the right-hand side.

For c, they haven't isolated x yet.

They've added 10 to get an equivalent equation of x + 5 = 20, but we haven't got a solution.

What we could now do from this stage is we could now add -5 or you might have said that they needed to just add 5 in the first place, not add 10.

For d, they have subtracted 23 instead of added 23 to the right-hand side.

So the left-hand side looks good, but we've added 23 to the left-hand side, we need to do the same to the right-hand side.

If you found the correct solutions, they're written on the screen to check.

Outstanding effort today.

Let's look at what we've learned.

Equations with a single additive step can be solved with a bar model.

Equations with a single additive step can be represented with a balance and solved by maintaining a quality.

We then applied that to equations written algebraically.

We know to isolate x and solve, the additive inverse of the constant must be added to both sides of the equation.

And we're now experts at showing our working out and we know how important that is so people can follow our work and we can spot our own mistakes if we've made them.

I really look forward to seeing you again.