Lesson video

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

My name is Ms. Davis, 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 where we're solving simple linear inequalities.

So equality is used to show that one expression may not be equal to another.

You will have used them before, but you may not have ever solved an inequality before.

So we're gonna start by looking at solving two-step linear inequalities, and then we're gonna look at what happens when we multiply or divide by a negative, and that's quite an interesting result.

Let's get started.

When we solve a linear inequality, we are looking for the range of values which satisfy the equality.

What values of a satisfy the inequality a plus four is greater than nine? Let's look at this one together.

Lucas says "This is true when a is eight." We can substitute to check.

Eight plus four is greater than nine.

12 is greater than nine.

That is true.

So a could be eight.

This is not true when a is five.

Let's check.

Five plus four is greater than nine.

Nine is greater than nine.

Well, that's not true.

Nine is not greater than nine.

So a cannot be five.

Now, eight is not the only value which satisfies this inequality.

What we want is the range of solutions, and we can get that efficiently using algebraic manipulation.

So if I've got a plus four is greater than nine, and I add negative four to both sides of my inequality, that tells me that a has to be greater than five.

Any value greater than five, add four, will be greater than nine.

We've solved this in exactly the same way as an equation, but what's really important is we haven't got a single solution.

We have an inequality which shows the range of solutions.

If we're gonna represent this on a number line, we could draw an open circle at a, an arrow to the right.

A is greater than five.

Now, we've already shown that five is not a solution.

Any value greater than five should be a solution.

So let's try one.

Six plus four is greater than nine.

10 is greater than nine.

That's true.

5.

2 plus four is greater than nine.

9.

2 is greater to than nine.

That's true.

Any value five or less should not be a solution.

We've shown that five doesn't work.

Let's pick one less than five.

Four plus four is greater than nine.

Eight is greater than nine.

No, it isn't.

0.

5 plus four is greater than nine.

4.

5 is greater than nine.

No, that's not true either.

So by testing we can see that values greater than five work, five doesn't work, and values less than five do not work.

Solving algebraically works for inequalities with any of the four equality symbols.

You just need to make sure you are consistent with the symbol you are working with.

So we've got 5a is less than or equal to 15.

If I divide both by five or multiply both by fifth, you get a is less than or equal to three.

Make sure you're consistent with that inequality symbol.

This solution includes three and any number less than three.

Let's check some.

When a is three, five times three is less than or equal to 15.

15 is less than or equal to 15.

That's fine because we've got that less than or equal to.

So it's okay for 'em to be equal.

Let's try a value less than three as well.

So five times two is less than or equal to 15.

10 is less than or equal to 15.

That is true as well.

Here I've picked a value greater than three to see what happens.

Five times 3.

2 is less than or equal 15.

So 16 is less than or equal to 15.

That's not true.

So we've shown that it works for three.

It works for a value less than three.

Doesn't work for a value greater than three.

So our solution a is less than or equal to three, seems to be right.

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

We're solving the inequality.

b plus five is less than 12.

I'm gonna subtract five from both sides of my inequality.

It gets me b is less than seven.

I'm also gonna represent this on a number line.

So I've got an open circle at seven 'cause seven is not included, and I want b is less than seven.

So I've got an arrow to the left.

Can you solve this inequality? X minus 3.

5 is greater than or equal to five.

And then represent that on the number line.

Off you go.

So we need to add 3.

5 to both sides.

So you've got x is greater than or equal to 8.

5.

So we find 8.

5 on the number line.

We need a closed circle and arrow to the right.

When an inequality has multiple steps, we solve it in exactly the same way.

So you've got five x plus seven is less than 42.

If I add negative seven to both sides or subtract seven from both sides, five x is less than 35.

Divide both sides by five.

X is less than seven.

And again, let's check some values.

When x is six, that's less than seven.

So five times six plus seven is less than 42, which gives me 37 is less than 42.

That's true.

Let's try x is three.

That's less than seven and that gives me 22 is less than 42.

That's correct.

Let's try seven itself.

Seven gives me 42 is less than 42.

No, 42 is equal to 42.

It is not less than 42.

This shows us that when x is seven, five x plus seven equals 42.

That's why we've got 42 on both sides of the inequality.

But any value less than seven should make the expression less than 42.

It doesn't matter which side of the inequality side the variable is on, as long as you are consistent and you're careful to keep the values on the correct side of the inequality sign.

So let's try this one.

If I expand my brackets, I've got 32 is less than or equal to 6x minus six.

So 38 is less than or equal to 6x.

I've added six to both sides, then divided both sides by six.

19 over three is less than or equal to x.

Now, what we could do, we could write the inequality with x on the left hand side if we prefer.

Just be really careful to check you've written an equivalent inequality.

So if I move the x to the left hand side, I should have x is greater than or equal to 19 over three.

Again, reading those out loud could be helpful.

Now, leaving our answer as a fraction is absolutely fine.

Converting into a mixed number or a decimal could sometimes help you see which values are included in the solution.

So I'm gonna convert it to a mixed number.

It's six and a third.

So I've got x is greater than or equal to six and a third, and then I can test some values.

Well seven is greater than six and a third.

And that gives me 36 is greater than or equal to 32.

That's true.

I'm gonna try value less than six and a third.

So six is less than six and a third.

This shouldn't work, and that gives me 32 is less than or equal to 30.

That doesn't work.

Right, I'm gonna have a go at one on the left, and then you are gonna try one.

If I subtract five from both sides, I get 2a is greater than or equal to negative two.

Dividing by two, a is greater than or equal to negative one.

I'm gonna check some values.

When a is negative one exactly, I get three is greater than or equal to three.

That's fine.

Let's try a value greater than negative one.

So when a is zero that gives me five is greater than or equal to three.

That's fine as well.

I'm just gonna check value less than negative one as well.

So I've tried negative 1.

2, and that gives me 2.

6 is greater than or equal to three.

That's not true.

So I can see that it doesn't work for negative 1.

2, but that's because my solution is values greater than or equal to negative one.

Your turn.

Can you solve the inequality 3b plus two is less than negative four? Let's have a look.

We subtract two from both sides.

We get 3b is less than negative six.

Divide by three.

B is less than negative two.

Pick a couple of values to check Right, be careful with your negatives here.

I've picked negative 2.

1 'cause that's slightly less than negative two.

And that shows me that negative 4.

3 is less than negative four.

That's true.

I've gone for another value less than negative two.

So negative three.

Negative seven is less than negative four.

That is true.

I'm just gonna check negative two itself.

See what happens.

You get negative four is less than negative four.

That's not true.

And I'll try another one.

Let's try b is zero.

And also not true.

Negative four is not greater than two.

Time to put that all into practise.

I'd like you to solve the inequality, then pick a value in your solution and show it satisfies the inequality.

Then, pick a value not in your solution and shows that it does not satisfy the inequality.

Then do the same for question two.

Off you go.

For question three, I'd like you to solve the inequality and then represent your solution on a number line.

Then, do the same for question four.

Give those ones a go.

For question five, here's an attempt to solve this inequality.

I want you to substitute x equals eight into the inequality and explain why this shows the solution is incorrect.

Then, explain the mistake.

There, you've got the same for question six.

Allow you to substitute x equals 15, and explain why that shows the solution is incorrect, and see if you can spot the mistake that's been made.

Give those ones a go.

And question seven, you've got a whole range of inequalities to solve.

Think carefully about your algebra manipulation.

Off you go.

Let's have a look at our answers.

So we've got 3a is greater than or equal to 12.

So a is greater than or equal to four.

There's all sorts of values that you could have picked.

You could try four itself, and that shows you that 10 is greater than or equal to 10, which is true.

You might have also wanted to check a value greater than four to make sure you've got the correct inequality sign.

For a value not in our solution, so we want a value less than four, I've gone with three.

You get seven is greater than or equal to 10, which is obviously not true.

Again, there's infinite options you could have gone with.

For two, doesn't matter that I've got the variables on the right hand side.

I can still subtract 15 and divide by four.

So you've got negative three is greater than a, but you can write that as a is less than negative three.

So values less than negative three.

You could go with negative 3.

5, negative four, all sorts of options.

I've gone with negative four, and I get three is greater than negative one.

Well, that is true.

You could pick a value not in your solution.

You could go with negative three itself, and that gets you three is greater than three, which is obviously not true.

You might have preferred to pick a value greater than negative three.

So maybe negative two, negative 2.

50, zero, to show that that also does not satisfy your inequality.

Let's a look at this one.

You've got choices.

You can divide both sides by three or you can expand your brackets.

I divided by three, to get a is less than four.

On a number line, I've got open circle at four and an arrow to the left.

For four, I think the best way is to subtract seven first and then multiply both sides by five.

A is less than or equal to negative 10.

On a number line, I need a filled in circle at negative 10 and an arrow to the left.

So if we substitute x equals eight, that is in my solution.

The solution says x is less than nine, so x equals eight should work.

Let's have a look.

Now, that gives me 24 is less than 10.

That doesn't work.

So x is less than nine, so should be a solution.

However, when substituted it does not satisfy the inequality.

The mistake was in the first step.

Eight should be subtracted from both sides, not added.

You should have 2x is less than two, and x is less than one.

Let's see what happens when you substitute x equals 15.

15 over three plus four gives me nine is greater than or equal to eight.

That's true.

Nine is greater than or equal to eight, but 15 was not in our solution.

Our solution said x is greater than or equal to 20.

So 15 shouldn't work.

However, it did, and that shows that inequality is incorrect.

The mistake was in the first step.

If multiplying by three, every term should be multiplied by three.

So it should be x plus 12 is greater than or equal to 24.

Or a more efficient way might be to subtract four first, then multiply by three.

Finally, you had a load of inequalities to solve.

I'd like you to pause video and check your answers.

Just like when you're solving equations, there are often options for the order in which you've done things.

I've tried to be as efficient as possible.

When you're happy with your answers, we'll look at the next part of the lesson.

We're gonna look now at multiplying or dividing by a negative.

Lucas and Izzy are trying to find the solutions to this inequality.

Negative 2x plus four is less than 10.

Lucas does this, subtracts four, divides by two, multiplies by negative one.

Izzy does it this way.

She adds two x both sites, subtracts 10, then divides by two.

Are their answers the same? No, they are completely different.

Lucas says solutions are less than negative three, and Izzy says they are greater than negative three.

They're completely the opposite solutions.

So whose answer is correct? Pause the video.

Can you work it out? Let's try some values to check.

Let's try x is negative four.

That gives me 12 is less than 10.

That is wrong.

Negative four should not be in our solution.

Let's try negative two.

Eight is less than 10.

That is fine.

So negative two should be in our solution.

So we want values greater than negative three it looks like 'cause negative two worked, but negative four didn't.

So it's probably Izzy that is correct.

Let's look at what's happening.

The problem with Lucas's method is that last step where he multiplies by negative one.

So subtracting four, dividing by two, absolutely fine.

It's that last bit where he goes from negative x is less than three to x is less than negative three.

Now, that works for equations.

You've probably done that loads of times with equations.

Because both sides of the equality symbol are the same, that's not a problem.

But here we've got an inequality.

So both sides of the inequality symbol are not the same.

One side is less than the other, and that's a problem when we're multiplying by a negative.

So we've gotta be really careful.

Let's look at why, and then we'll come back and see what Lucas can do to correct his answer.

Let's consider this valid inequality.

Negative three is less than six.

That is true.

If I multiply both sides by negative one, I get three is less than negative six.

That is not true.

Let's try another one.

Five is greater than three.

That's true.

If I multiply both sides by negative one, negative five is greater than negative three.

That is not true anymore.

So looks like multiplying by a negative does not keep the inequality valid.

Let's look at this on a number line.

So at the moment the triangle has a larger value than the square.

If I multiply them by negative one, that's like reflecting them over the line at zero because of the symmetry of directed numbers around zero, when we multiply by negative, the respective positions flip.

See how five is greater than three? But when we multiply by negative one, negative five is less than negative three.

So there's nothing incorrect with Lucas's choice.

He's just got to do something when he multiplies by negative one.

So here when he multiplies by negative one to turn the negative x into x and the three into negative three, the inequality symbol is going to change.

The respective sizes has changed.

Now, if you solve this the same way as Izzy by adding the x term to both sides, you don't need to worry about your inequality symbol.

So a recap, when you're multiplying by a negative value, your inequality symbol is going to change.

We can prove this is always the case using algebra.

A is greater than b if and only if a subtract b is greater than zero.

And that's true.

If a subtract b is positive, then a is greater than b.

Let's assume c is negative.

So c is less than zero.

When we multiply c by a minus b, that's also going to be less than zero.

That's because a negative multiplied by a positive is negative, and c is negative, and a minus b is positive.

Right, if I expand those brackets, ca minus cb is less than zero.

If I add cb to each side, then ca must be less than cb.

And that also comes from our first statement as well.

If ca minus cb is less than zero, that means ca minus cb is negative.

And ca must be smaller than cb.

Now, we started with a is greater than b, so we started with a is greater than b and we ended up with ca is less than cb.

Therefore, when multiplying both sides of an inequality by a negative number, the respective size of the two numbers swap.

Quick check, then.

Which of these is equivalent to negative x is less than five? It's that bottom one.

If we've chosen to multiply both by negative one to get x and negative five, then the inequality sign is going to change.

X is now greater than negative five.

Let's have a look at doing this the other way.

If we add x to both sides, we get zero is less than five plus x.

Then, subtract five.

Negative five is less than x.

Now, there are often choices of methods when solving inequalities.

It is personal preference which you find most efficient.

So with this equality, what you could do is subtract five from both sides, then we can divide by three, and then we can divide by negative one.

Or you might have done that all in one step and divided by negative three.

Now, when we divide by negative one, 'cause we're dividing by a negative, the negative two becomes two, the negative x becomes x, but our inequality sign is going to change.

Let's try it another way.

So if you're not sure about remembering what happens with the inequality symbol, we can avoid having to divide by a negative by starting by adding 3x.

So add 3x to both sides.

Then, add one.

Then, divide by three.

So your x is less than or equal to two.

Both of those answers are the same.

Have a think.

Which method do you think you are going to prefer? Right, now we can deal with negative unknowns, we can solve inequalities with unknowns on both sides.

Let's have a look at this one.

Now, to avoid a negative x wherever possible, we can start by subtracting 2x.

So from both sides, then I can add nine.

Then, I can divide by three.

Exactly like solving an equation, except we've got a range of solutions.

So x is greater than or equal to four.

Let's check.

When x is four, we can substitute it into both sides of the inequality.

So two lots of four plus three is less than five lots of four minus nine.

So 11 is less than or equal to 11.

And that is true.

Let's also check a value greater than four.

So when x is five.

And that's true as well.

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

My preferred method is to start by adding 9x to avoid negative xs.

I've got 9x plus 14 is less than or equal to eight.

9x is less than or equal to negative 6.

X is less than or equal to negative two thirds.

Other methods are absolutely acceptable.

That's my preferred method.

Have a go for this inequality.

If you do it the same way as me, you can start by adding 5x, subtracting four, and dividing by five.

So we've got x is less than three over five.

You can write either way.

Check your inequality symbol is the correct way round.

If you started by subtracting seven and then divided by negative five, you've got to remember that inequality sign is going to change.

Let's try this one.

So I'm gonna subtract 4x from both sides, add nine, and divide by six.

So your x is less than seven over three.

See if you can do the same for this inequality.

Let's have a look.

Subtracting 3x, adding three, divided by five.

Your x is less than two.

You can write it either way round.

Right, Lucas has started to solve this inequality.

What should his final answer be? Off you go.

Right, well if we carry on with the method he's chosen, so divide both sides by two, and then multiply by negative one.

We've gotta remember that that inequality sign is going to change.

You should have x is less than or equal to three.

So, let's put that into practise.

I'd like you to solve each of these inequalities, and then choose a value within your solution to check your answer.

Off you go.

For question three, Alex has tried to solve this inequality.

He's checked his answer, and he thinks it's correct.

I'd like you to substitute a equals six to show that it's incorrect, and then explain where he is gone wrong.

Can you give him a suggestion on how to improve? Off you go.

For four, you've got loads of inequalities to solve here.

Think carefully about your negative values.

Remember, if you're multiplying or dividing by a negative to look at that equality sign.

You might be able to avoid doing that by choosing to add your negative variable.

Have a go with those.

There's some tricky ones at the ends.

See how far you get.

Off you go.

Let's have a look.

So you should have x is greater than three, x is greater than or equal to negative two, and x is less than four.

So all sorts of values you could have chosen to check your answer.

I'll show you some that I've gone for.

For the first one, I've gone with 3.

1.

For the second one, I've gone with negative one.

And for the third one, I've picked three.

For question three, if we substitute a is six into the original inequality, we get negative 12 plus 15 is greater than or equal to five, and three is greater than or equal to five, which it's not.

So what we've done is we've picked a value within Alex's solution, six is greater than five, but it didn't work.

Now, where he is gone wrong is in his second step when he multiplied by negative one.

And we know that when you multiply by negative one, you need to adjust your inequality sign.

This means his final inequality symbol is the wrong way round.

It should be a is less than or equal to five, and that's why substituting five worked for Alex to check, but substituting six did not.

He got the right constraint.

He's just got his inequality symbol the wrong way around.

So some suggestions.

He could substitute a second value to check, not just the smallest possible or largest possible value.

He needs to remember that multiplying by a negative means he has to adjust the inequality sign.

Or, he could avoid having to multiply by a negative by starting by adding 2a.

Let's have a look at question four.

Should have a is less than four.

For B, you should have b is less than negative six.

For C, you're gonna want to expand your brackets first.

Then, if you add 12c to both sides, then you get c is greater than or equal to two.

Right, then we go onto the trickier ones.

I'm gonna start by expanding some brackets.

Then, I can multiply both sides by three.

And then I can subtract 20x, add 75, and divide by 22.

You get x is less than five.

Well done, if you got that one.

For E, again, I'm gonna want to expand the brackets, add y to both sides.

And then, divide by five.

You get y is greater than or equal to one.

And finally for f, you're gonna want to expand your brackets before you start.

So we're doing half of the first bracket and three quarters of the second bracket.

Then, we can collect like terms. We've got five minus 9z is less than or equal to 22.

I'm gonna add 9z to both sides, and then subtract 22, and you get z is greater than or equal to negative 17 over nine.

Pause the video if you want to check my working in more detail.

Make sure you've got your inequality symbols the correct way round.

Fantastic.

We can now solve simple linear inequalities.

We've shown this exactly the same as solving equation.

We follow the rules of algebraic manipulation.

However, we get a range of solutions to finish.

It's really important that we're using the correct inequality symbol.

We know that when we multiply or divide by a negative, it reverses the inequality sign, and that's due to the reflection in the number line at zero.

And then we know that there's options where we can avoid having to divide or multiply by a negative by adding the term with the negative variable.

You've worked incredibly hard today.

Some of those inequalities that you had to solve in that last task were really tricky.

However, you've got all the skills that you need now to go on to solve more complex linear inequalities if you wish.

Thank you for joining us, and I hope you choose to learn with us again.