Lesson video

Welcome and well done for making a decision to learn using this video today.

My name is Ms. Davies, and I'm gonna be helping you as you work your way through this lesson.

There's lots of opportunities for you to really think today and test your understanding.

Make sure you are pausing the video to give yourself that chance to think, and then I'll help you using any hints and tips as we go through.

If you've got everything you need, then let's get started.

Welcome to this lesson on solving more complicated linear inequalities.

If you haven't already looked at solving simple linear inequalities, you may wish to do that first as we're building on those skills today, if you need to remind us to what an equality is, pause the video and have a read through that now.

So, we're gonna start then by solving simple multiple inequalities.

We can solve inequalities using the rules of algebraic manipulation.

So, we have this equality.

We can add negative nine to both sides and we get negative five is less than x or x is greater than negative five.

So, the solutions are when x is greater than negative five.

We can solve combined inequalities in the same way.

So, if you have an equality like this where we've got two constraints, this means that when you add nine to x, the value is greater than four, but less than or equal to 12.

What do you think we could do to find the possible values of x? Right, well just like before, we can add negative nine to each expression.

We already know that x is greater than negative five because we previously solved x plus nine is greater than four.

But let's look at the other constraint as well.

So, if we add negative nine to each expression, there's our negative five on the left hand side that we saw previously, so we've got x is greater than negative five, but less than or equal to three, 12, add negative nine.

Don't forget, we can represent our solution on a number line.

We've got an open circle at negative five, so it's not included in our solution, but a closed circle at three because it is included in our solution and we can check some values within our solution to see if they are valid.

So an x is three, for example, we've got three plus nine is greater than four and less than or equal to 12, which means 12 is greater than four and less than or equal to 12.

That's fine.

Let's try another value.

Let's try two.

We can see on the line that's in our solution, two plus nine is 11 and 11 is greater than four and less than or equal to 12.

Let's try negative three.

That gives us six and that is greater than four and less than or equal to 12.

And we can try other values as well.

There's negative 4.

5, and all these values were in our solution and they satisfy the original inequality.

Now, we can check some values not in our solution and make sure they are not also valid.

So, let's use negative five.

And you can see we then get the statement four is greater than four and that is not true.

Let's try negative six.

<v ->So that gives us three is greater than four,</v> that's not true.

It says that three is less than or equal to 12.

That's fine, but it's that lower constraint that doesn't work for that one.

And if we try four, we get this statement 13 is greater than four but less than or equal to 12.

And again that top constraint is now the problem.

So we tried the values which were the constraints, so the negative five and the three and we tried values between these and we tried values outside these.

We can be fairly confident we have the correct range of solutions.

I'm gonna have go at one on the left hand side and then you are gonna have a go.

So we need to find the range of solutions for this inequality.

Now, I can start by adding three to all expressions.

So, I've got 2x is greater than or equal to six, but less than 10.

Then I can divide all expressions by two.

So I've got x is greater than or equal to three but less than five and I can represent the solution on a number line.

So I've got a closed circle at three, an open circle at five, and I want all the values between them, so I'm gonna connect them with a line.

By your end, can you find the range of solutions for this inequality? When you're done, can you represent it on a number line? Let's see how you got on.

So, if we subtract seven from every expression, we've now got 3x is greater than negative three but less than nine.

And if we divide each expression by three, we've got x is greater than negative one but less than three.

We need an open circle at negative one and open circle at three and a line connecting them.

So we can use this method to solve trickier inequalities with two constraints as long as we apply the same operation to all expressions.

Let's have a look at this one.

So, if we start by multiplying each expression by five, so now we've got two lots of x minus four is greater than 10, but less than 30, all three expressions have been multiplied by five.

Now, I'm gonna divide each expression by two.

You could also expand the brackets in the middle if you wanted to.

I think divided by two is quicker, so I've got x minus four is greater than five, but less than 15.

And now I can add four, x is greater than nine, but less than 19.

To check, you can substitute nine and 19, and see what happens and then pick a number between them, which is in our range of solutions to make sure we've got the correct constraints and the correct inequality symbols.

Andeep is trying to solve this inequality.

"I'm not sure what to do as there is x in every expression." Jacob says, "See if you can apply the same operation to every expression." What could Andeep do then? Well done if you said subtract 3x from each expression.

Let's see what happens.

We subtract 3x from the left-hand side, we get eight.

We subtract 3x from the middle, we get x.

And if we subtract 3x from the right-hand side, we get 15.

So, we've got x is greater than eight, less than 15, and that gave us our solution straight away.

Again, we can check some values.

I'm gonna pick 10, so I need to do three lots of 10 plus eight, four lots of 10 and three lots of 10 plus 15 and that gives me this inequality.

40 is greater than 38 but less than 45 and that is true.

I'm gonna try 14 as well and that works as well.

Time for you to have a practise.

I'd like you to explain the mistake in each of the attempts to solve the inequality.

Off you go.

For question two, I'd like you to solve each inequality and then represent your solution on a number line and then you've got three more to do here.

Off you go.

Well done.

So you've got some slightly trickier expressions here, but again, remember all you need to do is apply the same operation to all your expressions.

Give those ones a go.

Let's have a look at our answers.

So, this first one, the lower constraint is correct, but the same operations need to be applied to the upper constraint.

So that first step, we've subtracted four from the seven, we've subtracted four from the expression in the middle.

We also need to subtract four from the 10 and then we need to divide all the expressions by three.

If you did that correctly, your answer should be x is greater than one, but less than two.

For the second one, all the steps of working were good apart from the inequality signs.

The upper constraint should be a less than or equal to, so the second row should say 3x plus five is greater than two and less than or equal to 10.

Then we need to use that same inequality symbol throughout the question.

Final answer should be x is greater than negative one, less than or equal to five over three.

Let's see how you got on.

So, the first one you should have x is greater than one and less than three.

The second one, x is greater than or equal to one, less than or equal to four.

The third one, x is greater than three and less than or equal to five.

Pause the video and just check you've got the correct diagrams on your number line.

For D, if you subtract five and then divide by two, you get x is greater than or equal to zero and less than two.

For E, if we add three and then divide by four, you get x is greater than negative one but less than one.

And for F, you can do this in a number of different ways.

I have divided by three, then subtracted eight, then divided by two, and you get x is greater than negative two, less than or equal to four.

Again, just pause the video and make sure you've got the correct diagrams on your number line.

For question three, if we add four and then divide by three, we get x is greater than or equal to seven and less than eight.

For B, we need to multiply by five, then subtract four, then divide by seven.

You get x is greater than negative two, less than 36 over seven.

If you want to change that into a mix number, you get x is greater than negative two, less than five and a seventh.

The C, if you start by subtracting x, you get x minus three is greater than five, but less than 10.

Then you can add 3, x is greater than eight, less than 13.

Fantastic, so you've got the basics of how to solve simple multiple inequalities.

We're now gonna look at when that method doesn't work and this idea of separating inequalities in order to solve them.

Izzy is trying to find solutions to this inequality.

Negative x is greater than two, but less than five.

Jun says, "You can think of examples of values which work to help you." Can you find a value that satisfies this inequality? Remember, negative x means negative one times x.

Right, there's lots of values you could have picked.

I went for x is negative four.

So if x was negative four, that means negative x would be four and that satisfies that inequality four is greater than two, but less than five.

But that doesn't find us all solutions.

We've just found one example.

Lucas says, "To find all solutions, you can multiply by negative one, remembering to correct the inequality symbols." So, if I multiply each expression by negative one, we get negative 2, x negative five.

Now, I wonder if you remembered when you multiply by a negative the respective size of your expressions swap.

So if negative x is greater than two, x isn't greater than negative two, x is less than negative two.

So we've multiplied them all by negative one.

So the inner quality signs are going to change.

Now this is not a nice way to write an inequality.

We would prefer to write the smaller value on the left and that would make it consistent with a number line.

So, I'm just gonna rewrite this.

So, if I put the negative five on the left, I've got x is greater than negative five, but less than negative two.

Izzy says, "When there is negative x in an inequality, I like to add x to both expressions to avoid dividing by a negative." When we solved a simple linear inequality, we said you've got that option, haven't you, of adding the negative x term rather than having to divide by negative one.

However, when she tries to do it here, it doesn't seem to work.

We've now got x in two expressions and zero in the middle and that doesn't really give us a solution.

So, sometimes it can be difficult to solve an inequality with two constraints by applying the same operation to all expressions.

What we can do instead is separate the inequality into two separate inequalities and solve them individually.

So, let's consider the two restrictions separately.

So, if negative x is greater than two, we can write that as a statement.

And then negative x is less than five, that's another statement.

Now, let's use Izzy's preferred method to see what happens.

Adding x to both sides, we get x plus two is less than zero, so x is less than negative two.

Let's look at the other inequality.

Adding x to both sides gives us zero is less than five plus x, negative five is less than x.

Now, we just need to combine our solutions and this is where you've got to be careful.

What might help here is reading the inequalities in words or drawing on a number line.

So, if you read them in words, this says x is less than negative two, x is greater than negative five.

If we draw them on a number line, there's x is less than negative two, there's x is greater than negative five.

To satisfy both, we need that section in the middle.

So, we can write this single inequality.

Quick check, which is the correct way to combine the solutions, x is less than four and x is greater than zero? Well done if you picked B.

Which is the correct way to separate this inequality into two inequalities? Give this one again.

This one is really important.

So, what we've got here is we've got the statement that five minus 2x is greater than one and five minus 2x is less than three.

So, those should be our two separate inequalities.

Remember to keep five minus 2x as a complete expression because it's that whole thing which is greater than one and that whole thing that's less than three, but I'd like you to fill in the missing steps to Izzy's working to find this solution.

Off you go.

Let's have a look.

Just like before, she's preferred to add the negative x term, so she's added 2x to both sides to get 2x plus one is less than five, then she's subtracted one, 2x is less than four, then she's divided by 2, x is less than two.

The inequality on the right hand side, we should have five minus 2x is less than three.

Then again, if we add 2x to both sides, we've got five is less than three plus 2x.

Then if we subtract three, 2x is greater than two and x is greater than one.

We can either draw them on a number line or read them out loud.

So, we've got x is less than two and x is greater than one.

So, putting that in the correct order, we've got x is greater than one but less than two.

This method is also helpful when our inequality has multiple x terms with different coefficients.

Have a read of the inequality below.

Why will subtracting 2x from all expressions not work? Can you see this one? Right, well there'll still be x terms in two expressions.

So, even though the left hand expression will only now be negative one, we'll still have 3x in the middle expression and x in the right hand expression.

What we're looking for is to have x in one expression and numbers in the other.

Okay, so subtracting 2x didn't work.

How about subtracting 3x? Why will that not work? Same reason, there'll still be x terms in two expressions.

So, just subtracting x terms from all three expressions did not seem to work this time.

So, what we can do is separate into two inequalities and solve them separately.

So, we could say that 5x minus 10 is greater than 2x minus one and 5x minus 10 is less than or equal to 3x plus 22.

And now we can subtract 2x from both expressions and add 10.

So, we get 3x is greater than nine and x is greater than three.

If we look at the other inequality, subtract 3x from both sides, we get 2x minus 10 is less than or equal to 22, 2x is less than or equal to 32, x is less than or equal to 16.

And then, we can combine our solutions at the end, x is greater than three, less than or equal to 16.

And remember, you can always pick a value to check.

I've gone with five and then our final inequality is true.

If you wanted to be sure we were correct, we would need to test both constraints.

So, see what happens when x was three, when x was 16, and also test a value in the middle and that would make sure we've got the correct constraints and the correct inequality symbols.

Right, let's have a look at another example.

What inequalities could we separate these into? What would our first step be? Right, we've got 2x plus three is greater than or equal to 3x minus two and 2x plus three is less than or equal to 5x minus six.

Just like before, we can now solve these separately.

Let's look at solving that left-handed inequality.

So, subtracting 2x from both sides of the inequality.

Adding two, we've got x is less than or equal to five.

Let's look at the other inequality.

Subtracting 2x, adding six, and dividing by three.

We've got x is greater than or equal to three.

How could we combine these solutions? Be careful this time Right, I've drawn on a number line to help me, and I can see that x has gotta be greater than or equal to three and less than or equal to five.

So, notice that the lower constraint was determined by the second inequality this time and the higher constraint was determined by the first inequality.

This sort of swapped over, but that's not a problem.

We can just combine our solutions at the end.

Time for a check.

I'm gonna show you one on the left and then you are gonna have a go.

So, looking at this inequality, I'm gonna want to split it into two separate inequalities.

So, we've got 2x plus five is greater than or equal to 3x minus four and I can solve that one.

So, subtracting 2x and adding four.

Then my other inequality, 2x plus five is less than or equal to 4x plus one, subtracting 2x, subtracting one, and dividing by two.

Then, if I read those out loud, it'll help me combine my solutions.

I've got x is less than or equal to nine and x is greater than or equal to two.

So, I can write my final inequality that way round.

Now, your turn.

Have a go at solving this inequality.

Let's have a look.

So, we can do x plus three is greater than or equal to 3x minus seven.

First subtract x, add seven, and divided by two.

You get x is less than or equal to five.

Let's look at the other part of the inequality.

x plus three has to be less than or equal to 2x minus one, subtracting x and adding one, you get x is greater than or equal to four.

So you can write your final inequality like that.

Fantastic, well done if you've got that one by yourself.

Not all inequalities have valid solutions.

Laura says, "Aisha made up this equality for me to solve, but I don't think it works." What inequalities could we separate this into? Let's check what's going wrong here.

You have 3x plus five is greater than or equal to 2x plus three.

3x plus five is less than or equal to x minus five.

It's not necessarily clear at this stage why this didn't work for Laura.

So, let's carry on solving.

And then the right hand inequality, we can subtract x, subtract five and divided by two.

What is the issue then with this solution? Right, well, our solution says, "x is greater than or equal to negative two, but less than or equal to negative five." There are no values that satisfy both simultaneously.

How can a value be greater than or equal to negative two but less than or equal to negative five? If we draw that on a number line, we can see that there is no overlap between the two inequalities, so there are no valid solutions to that original inequality.

Laura has written this inequality, but she doesn't think it has any solutions either.

Should we have a look together? So we can split those into two inequalities, solve like normal, we get x is less than one, and if we look at the right hand side, we get x is less than four.

What will these solutions look like on a number line? Are there any values which satisfy both? Do these have solutions? Yeah, these do have solutions.

The solutions are when x is less than one.

So, any value less than one will satisfy both inequalities.

We can try some values to check.

Laura was incorrect this time.

This inequality did have solutions.

They just didn't have two constraints.

We ended up with a solution where x was less than one, but can be any value less than than one.

There's not a lower constraint this time, and that's absolutely fine, that is a solution.

All right, time to put that into practise.

I'd like you to find all the solutions to these inequalities.

Then you've got four more to do and some slightly more complicated expressions at the end.

Off you go.

For question two, I'd like to know which of these have valid solutions and if they do have valid solutions, can you write the range of solutions just like you did previously? Off you go.

For three, Jun is trying to solve this inequality, so have a read through his working.

I'd like you to substitute x equals four into the inequality to show that the solution is incorrect and then I'd like you to explain his mistake and write the correct solutions.

Off you go.

Let's have a look.

Now, the quickest way to do this first one would be to subtract 2x from each expression.

We don't need to split it into two inequalities this time.

So, you get x is greater than or to one, less than or equal to 17.

For B, you could have solved this by subtracting six and then dividing by negative five.

You just have to remember that your equality signs are gonna change or you could split it into two inequalities and use Izzy's method of adding 5x.

Then you can write the combined solution.

For C, get solution of x is greater than or equal to two, less than 21.

And for D, x is greater than or equal to negative one, less than or equal to three.

Pause the video and have a look through my working if you need to.

Okay, for E, F, G, and H, pause the video and have a look through those answers.

For H, I've left mine as improper fractions.

If you converted yours to decimals or mixed numbers, that's fine as well.

For question two, so A, there were no solutions.

We get x is greater than 10 and x is less than five.

There are no values that satisfy both of those.

For B, we do have solutions.

Solutions when x is greater than negative one and less than three.

For C, there are solutions.

There are solutions when x is less than negative seven.

There's no lower constraint this time.

And for D, no solutions.

We can't have x is less than negative two, but also greater than five.

Jun says, "The solution is when x is greater than negative one, but less than five." So, x equals four should work.

Let's try it.

We get 16 is greater than 11, but less than 15.

That doesn't work, 16 is not less than 15.

There's an issue here.

The mistake is actually in the final solution.

If we read those out loud, it says that x is greater than negative one and x is greater than five.

So, his final solution should be any value greater than five.

Remember, not all solutions are gonna have two constraints.

Well done with that final task.

There's lots of ideas that you are bringing together.

You now become experts in solving linear inequalities.

Let's have a look at what we've learned today.

So, we've said that we can solve some inequalities with two constraints by applying the same operations to all expressions, but then we looked at when that didn't work and said that we can separate an inequality and then solve and combine the solutions.

And then we looked at examples that do not have valid solutions and examples that have solutions with only constraint at the end.

Thank you for joining me today.

You've worked incredibly hard and I look forward to you joining us again for another lesson.