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Hello, Mr. Robson here.
Well done for joining me for maths today.
That was a smart decision.
Especially so because we're problem solving with linear inequalities today.
If you love problem solving and you love linear inequalities, then you are going to love this lesson.
Let's take a look.
Our learning outcome is we'll be able to use our knowledge of inequalities to solve problems. Inequality is a keyword you're gonna hear throughout the learning today and inequality is used to show that one expression may not be equal to another.
There's two different types of problems we're gonna encounter today.
The first one we're gonna look at is what it means to interpret inequalities on a graph.
In mathematics, there are often multiple ways to solve the same problem.
Let's consider this example.
Carlo's Cabs charge a fixed price of £6 per journey then a variable price of £2 a mile.
Carrie's Cabs charge no fixed price, just a variable price of £4 a mile.
Which firm is better value? I'd like to pause this video and have a think about this problem for a moment.
Welcome back.
We were thinking about this problem between Carlos Cabs and Carrie's Cabs.
I asked you to consider which firm is better value.
I wonder if you discovered it's quite an ambiguous question.
One simple option is to try some values.
Let's try the cost of a two mile journey.
When we substitute that into Carlo's Cabs £6 for the fixed price plus two lots of £2 per mile, that's £10.
For Carrie's Cabs, we've just got two lots of £4, that's £8, so we can conclude Carrie's Cabs are cheaper if you're travelling two miles, but Carrie's Cabs aren't always cheaper.
We tried a different value of five miles.
I found that Carlos Cabs charge £16 and Carrie's Cabs charge £20.
So I now conclude Carlo's Cabs are cheaper if you're travelling five miles.
So which firm is better value? Tricky problem.
This approach substituting in values means we need to calculate the cost for both companies every time we want to make a journey.
Graphing the problem is going to solve this problem for any distance.
For Carlo's Cabs we have a constant of £6 and a varying number of £2.
That's £6 is the fixed price and then £2 per mile.
We could graph this cost using c = 6 + 2m.
The C standing for cost an M standing for miles.
In the case of Carrie's Cabs, we just set a varying number of fours.
We could graph Carrie's Cabs using c = 4m.
Take the number of miles, multiply it by four we've got the cost for riding with Carrie's Cabs.
Graphing the problem makes seeing the solution easier.
There's two lines on this graph.
That line is c = 4m.
You'll note that the horizontal axis represents M.
That's the miles travelled.
The vertical axis represents C.
That's the cost of the journey.
This is the line c = 4m.
Remember that's Carrie's Cabs.
The second line is the line c = 6 + 2m, a linear equation.
A Y intercept at six and a gradient of two.
That's c = 6 + 2m.
That's Carlo's Cabs.
We can see this key moment here.
The intersection of the two lines.
The intersection tells us the moment where they charge the same.
That moment is three miles and at that moment they charge us.
Well done, £12.
We can also see the whole region where Carrie charges less than Carlo.
That region is less than three miles.
You can see that the line representing Carrie's Cabs is below that of Carlo's Cabs.
So Carrie's charging us less in this region.
We can also see the whole region where Carrie charges more than Carlo.
That region is greater three miles.
So if you are riding less than three miles, you wanna go with Carrie.
If you're going more than three miles, you want to go with Carlo.
Quick check you've got that.
Carrie and Carlo have changed their prices.
These are the graphs of their new charges.
I'd like you to complete the statements.
There's five blanks.
I'd like you to tell me what's going in those.
Pause and do that now.
Welcome back.
The statements should have read the point of intersection where they both charge the same price is at two miles.
The region where Carrie charges less is less than two miles.
The region where Carlo charges less is greater two miles.
Well done.
Below is the same problem we saw earlier.
We could have solved this more efficiently by forming and solving an inequality.
Our expression for Carlos charge was 6 + 2m.
Our expression for Carrie's charge was 4m.
We could form these inequalities.
The two inequalities ask us two different questions.
The one on the left asks, when is Carlo charging less than Carrie.
The one the right, when is Carrie charging less than Carlo? We can solve these inequalities to answer these two questions.
So let's start with the one on the left.
When is Carlo charging less than Carrie, I.
e when is 6 + 2m < 4m? There's all sorts of ways that we could solve this.
I'm gonna start by adding negative 2m to both sides of the inequality and I've got 6 < 2m.
I'm gonna divide both sides of the inequality by 2.
So m > 3.
We can conclude Carlo charges less than Carrie when the number of miles is greater three.
On the right hand side I need to solve 4m < 6 + 2m.
I'm gonna add negative 2m to both sides again, divide three by two miles is less than three.
On this occasion I can conclude Carrie charges less than Carlo when the number of miles is less than three.
We can compare these solutions to the graph we saw earlier.
That's the graph we saw earlier of Carrie and Carlo's charges.
Carlo charges less than Carrie when the number of miles is greater than three, that's that region up there and then Carrie charges less than Carlo when the number of miles is less than three.
That answer is verified by our graph.
We can see that fact on the graph.
Quick check, you've got this.
Donnies Drives charge a fixed price of £12 per journey and a variable price of £1 a mile.
Deb's Drives charge no fixed price, just a variable price of £3 a mile.
Which inequality will tell us the region where Deb's Drives charge less than Donnie's Drives.
Four to choose from, pause, see if you can work out which is right.
Welcome back.
Hopefully you said option B.
M + 12 is > 3m.
In option A and C, they contained incorrect expressions.
1 + 12m does not express Donnie's Drives.
3 + m does not express Deb's Drives.
We need to make sure the expressions are correct on each side of our inequality.
That was true for B and D.
The expressions are correct.
However, in the case of D, they're the correct expressions, but this inequality will inform us when Donnie charges less than Debs.
We were asked for where Debs charges less than Donnie.
Hence it was option C.
Next, I'd like you to solve the inequality that we've just set up in order to find the region where Deb's Drives charge less than Donnie's Drives.
Pause, solve that inequality now.
Welcome back.
I'm gonna start by subtracting M from both sides of the inequality and dividing both sides by two.
I get 6 > m, or m < 6.
We can conclude Deb's Drives charge less in the region, miles are less than six.
If you are travelling less than six miles, you wanna go with Deb's Drives.
Another quick check.
Here's a graph to represent those two taxi firms. Whereabouts on the graph can we see this solution to our inequality? Do we see it in A, B or C? Which region is our solution? Pause.
Have a think about this problem now.
Welcome back.
It was A, that's where we see this solution to our inequality.
That's Deb's Drives and Donnie's Drives represented by those lines and it's in this region where we can see that Deb's charges less.
The cost on every single occasion is less than Donnie in this region.
Practise time now.
Question one.
Eesha's e-drives, Ernie's e-drives and Ev's e-drives are all taxi firms. Their charges are graphed.
Write an inequality for the region described in each question.
Distance is given is miles and charges are given in £ like an inequality for part A to describe the distance where Eesha charges more than Ernie.
For B, I'd like an inequality to describe the distance where Ernie charges less than Ev.
And for C, I'd like an inequality to describe the distance where Ev is the most expensive.
Pause and write those now.
<v ->Question two.
</v> Izzy and Jacob are saving for a school trip.
Izzy starts with £80 and saves £5 a week.
Jacob starts with £0 but saves £10 week.
For part A, I'd like you to solve the inequality.
80 + 5w is > 10w.
For part B, I'd like you to write a sentence to explain what your solution to part A tells us.
Pause and try this now.
Question 2 part C.
I'd like you to graph the lines S = 80 + 5w and S =10w.
And write a sentence describing the significance of the point 16,160 on the graph.
Pause and try that now.
Question three.
Frank's Fast Cabs charge a fixed price of £6.
50 per journey then a variable price of £0.
50 PE a mile.
Freda's Fast Cabs charge a fixed price of £4 per journey than a variable price of £1 a mile.
For part A, I'd like you to form and solve inequality to find the region where Frank's Fast Cabs are the cheaper firm to ride with.
Pause and try this now.
Question 3 part B.
I'd like you to graph both firm's charges and write a sentence to describe how your graph relates to the solution to your inequality in part A.
The bottom left of the screen there, I've reminded you of both Frank and Freda's charges.
Pause, get graphing.
Feedback time.
Let's see how we get on for problem one.
For part A, we were writing an inequality to describe the distance where Eesha charges more than Ernie.
That would be less than seven miles or rather the inequality m < 7.
For part B we should have written m > 2.
5.
That's the distance where Ernie charges less than Ev.
For part C, the distance where Ev is the most expensive, that's anything over four miles, so you should have written the inequality m > 4.
For question two, we are looking at Izzy and Jacob's savings.
For part A I asked you to solve the inequality 80 + 5w > 10w.
We could have added negative 5w to both sides, divided 3 by 5 to find that 16 is > w or w < 16.
For part B, I asked you to write a sentence to explain what that solution means.
You might have written 80 + 5w is an expression for Izzy's savings and 10w is an expression for Jacob's savings, so the solution to 80 + 5w is > 10w is the region where Izzy has greater savings.
This region is w < 16, I.
e up until 16 weeks Izzy has the most savings.
Question 2 part C.
I asked you to graph two lines.
That's the line s = 80 + 5w.
That's the line s = 10w.
That's the point 16, 160.
I asked you to write a sentence describing the significance of this point.
You might have said the point 16, 160 is the intersection of the two graphs.
It is the moment 16 weeks when they both have the same savings.
Before this moment weeks is less than 16, Izzy has more savings.
After this moment when weeks is > 16, Jacob has more savings.
Question three, we're looking at Frank's Fast Cabs and Freda's Fast Cabs.
For part eight I asked you to form and solvent inequality to find the region where Franks are the cheaper firm to ride with.
Your inequality should have looked like so.
And you could have set about solving it like this.
And you would've found that 5 < m, or m > 5 I.
e Frank's Fast Cabs are cheaper for a journey longer than five miles.
For question 3 part B, I asked to graph both firms charges and write a sentence to describe how your graph relates to the solution to your inequality in part A.
Your graph should have looked like so and then you might have written, we can see in this region that Frank's Fast Cabs are cheaper for a journey of greater five miles.
Well done.
Onto the second half of our lesson now where we're going to look at something called project constraints.
Using graphs of inequalities can help us solve complex looking problems far more simply.
Let's have a look at a complex looking problem.
Oak Academy is making two versions of a game.
The standard version takes 20 minutes to produce, the deluxe version takes 30 minutes to produce.
It takes five minutes to package the deluxe version and 10 minutes to package the standard version.
There are 20 hours available to make the games and five hours available to package them.
Whats possible combinations of games could be made.
Now I don't expect you to be able to answer that question just yet because like I said, this is a complex problem, but we use mathematics to try and produce complexity to make complex problems simple.
Let's have a look at how we might do that in this case.
That's the exact same problem.
Can we make 10 deluxe games and 10 standard ones? This is simple substitution to answer this question On production we know it takes 20 minutes to produce the standard version and the deluxe version takes 30 minutes to produce, so we need 10 lots of 20 minutes and 10 lots of 30 minutes.
That's 500 minutes.
We're later told that there are 20 hours available to make the games.
20 hours is 1200 minutes.
500 minutes is less than 1200 minutes, so we can produce 10 deluxe games and 10 standard ones.
Can we package them? It takes five minutes to package the deluxe version and 10 minutes to package the standard version.
So we need 10 lots of five minutes, 10 lots at 10 minutes.
That's 150 minutes.
We're later told there's five hours available to package them.
Five hours is 300 minutes.
150 minutes is less than 300 minutes.
We can package them.
Yes we can, but what we saw in that maths is there'd be a lot of spare time left unused.
We have 20 hours available to make the games, five hours available to package them and we didn't go to the limits of either.
It would be nice to know what greater volumes we could produce without having to do all of that maths every time.
What we could do is use X to represent standard games and Y to represent deluxe games and write some inequalities.
It takes 20 minutes to produce the standard version, 30 minutes to produce the deluxe version.
20x + 30y therefore is an expression for how long production takes.
Take the number of standard games, X, multiply that by 20, take the number of deluxe games, Y, multiply that by 30 and add them together.
That's how many minutes production's going to take.
We know that production is limited to 20 hours, that's 20 hours available to make the games.
20 hours is 1200 minutes, so our expression 20x + 30y has to be less than or equal to 1200 minutes.
How does that look as an inequality? Like so.
20x + 30y is less than or equal to 1200.
We can graph this.
That's the graph of 20x + 30y equals 1200.
We can now graph the region whereby 20x + 30y is less than or equal to 1200.
That's this region here.
So can we produce 10 standard and 10 deluxe games? Yes, we can because this point 10, 10 is within this region.
Could we produce 20 standard and 20 deluxe games? Yes, because that point too is within the region.
We answered that second question in a matter of seconds.
This maths has made this problem far more simple.
For every point in the region, the coordinate values satisfy the inequality.
Let's have a look at what that means.
For the point 10, 10, that's X = 10, Y = 10.
If we substitute those values into our inequality, we find 500 is less than or equal to 1200.
That's absolutely true.
The values X = 10, Y = 10 have satisfied the inequality.
In the case of 20, 20, we'll substitute X = 20, Y = 20 into our inequality and again we're going to find that that inequality is satisfied.
Could we produce 40 standard games and 20 deluxe games? No.
Why not? Because that point is not in the region.
If we substitute those values into our inequality, X = 40 and Y = 20 we get 1,400 is less than or equal to 1200.
No, it is not.
That pair of values does not satisfy our inequality.
Quick check, you've got this.
Which of these combinations could we produce? Pause and take your pick.
Welcome back.
Hopefully you said A, yes we can.
There it is on the graph within the region.
Hopefully you said B, we can.
That point is also within the region.
For C, we can, just.
It's on the line but that's allowed in this case.
For D, it was a no.
That point is outside of our region.
We cannot produce 10 standard and 45 deluxe games.
Production time was not our only project constraint.
If you read the problem again, we were also constrained on packaging.
It takes five minutes to package the deluxe version and 10 minutes to package the standard version and there's five hours available to package them.
Again, we'll use X to represent standard games and Y for deluxe ones.
10x + 5y is an expression for how long production takes and it has to be less than or equal to our five hours.
That's 300 minutes.
We can use the inequality 10 x + 5y is less than or equal to 300.
There's our graph of production.
We already have the region, which is our constraint on production.
We can now map the region 10 x + 5y is less than or equal to 300, which is our packaging constraint.
We'll map it on the same graph and it'll look like so.
Can we package five standard games and 40 deluxe games? Absolutely we can.
We can package them.
Have you spotted the problem? We can package them but we can't produce them.
They're within the region of constraint for packaging but not within the one we need to satisfy production constraints.
In order to meet both our production constraint and our packaging constraint, the coordinate pair needs to be in the region which satisfies both inequalities.
Example this point, 10 standard games, 30 deluxe games we could produce and package 10 standard games and 30 deluxe ones.
But look at this point, 30 standard games, 10 deluxe ones.
We could produce 30 standard games and 10 deluxe ones, but we would not be able to package them.
Quick check, you've got that.
Which of these combinations could we both produce and package? Four to consider.
Pause and do so now.
Welcome back.
Hopefully you said A, absolutely we can produce and package that amount of games.
For B, we cannot.
We can produce it but we can package it.
For C, we can neither produce nor package that many games and for D, we can produce 20 standard and 20 deluxe games.
Testing these points gives us some interesting results.
There's three coordinate pairs there.
I'm gonna substitute them into the inequalities.
Let's start with the coordinate pair 20, 10.
When we substitute them into our production constraint inequality, we find 700 is less than equal to 1200.
Absolutely it is.
The inequality is satisfied.
We'll put those values into our packaging constraint and again, the inequality is satisfied.
Well the inequality satisfied, but there's a lot of wasted time.
Potentially we've got 1200 minutes of production time.
We're way short with just 700 minutes.
What about the point 20,20? When we substitute in the production inequality, we find 1000 is less than or equal to 1200.
The inequality is satisfied.
When we substitute into the packaging inequality, something interesting happens.
300 is equal to 300, so it satisfies the inequality.
These values satisfy both inequalities.
There is spare capacity on production.
We've still got 200 minutes to play with but the packaging is optimised.
I wonder if you can see from that point why packaging is optimised but production is not.
What about this last point, 15,30.
When we substitute those values into production, we find 1200 is equal to 1200.
Satisfies the inequality, just.
When we substitute the packaging, again, 300 is equal to 300.
The inequality is satisfied, just.
The values satisfied both inequalities with no spare capacity.
Both production and packaging are optimised.
This value is not at the limit of either inequality.
You can see that it's not.
It's well within that region.
This value, however, is at the limit of our packaging inequality but not at the limit of our production one.
The intersection of the two lines is the only point which is at the limit of both inequalities.
Therefore, this was the moment when we had no wastage on either packaging or production.
We were optimised at this point.
Quick check, you've got this.
Complete this statement.
The intersection is, statement A, statement B, or statement C.
Pause and take your pick.
Welcome back.
Hopefully you said A.
The intersection is the only point, which is the limit of both inequalities.
B was not right.
It's a point which only satisfies one of the inequalities.
C was another correct option.
The intersection is also the only point with no wastage.
Production and packaging time are at their limits.
No minutes are wasted.
Practise time now.
Here's another problem similar to the ones we read earlier.
I'd like you to do is read that and then use X to represent standard games and Y for deluxe ones and decide which two inequalities should be graphed.
Pause and choose two of those inequalities now.
Question two.
The lines 15x + 10y equals 540 and 5x + 10y equals 240 are graphed.
The shaded region is the region where both inequalities are satisfied.
I'd like you to show by substitution that we can make 20 standard games and 10 deluxe ones.
Pause and do that now.
Question three.
Which of these combinations can we make? For each combination I'd like you to write a sentence to justify your answer.
Question four.
I'd like you to write at least one sentence to explain the significance of the point 30,9.
You should include some mathematics to justify your answer.
Pause and do this now.
Feedback time.
Question one.
I asked you to decide which two inequalities we should graph.
That was 15x + 10y is less than or equal to 540 to represent the production constraint and 5x + 10y is less than or equal to 240 to represent the packaging constraint.
For question two, I ask you to show by substitution that we can make 20 standard games and 10 deluxe ones.
When we substitute those values into the production inequality, we find 400 is indeed less than or equal to 540.
The inequality is satisfied.
Substituting those values into packaging again, we find that the inequality is satisfied.
The coordinate pair 20,10 satisfies both inequalities.
Question three, I asked you which of these combinations we can make.
For A, you should have said yes, this point satisfies both inequalities.
For B, you should have said no.
This point satisfies neither inequality.
For C, you should have said no, we can produce this many but we can't package that many.
For D, you should have said no.
We can package that many but not produce.
For question four, I asked you to write at least one sentence to explain the significance of the point 30,9.
You might have written, the point 30,9 is the intersection of the two lines.
It is the only point which is at the limit of both inequalities.
Therefore, we have no time wasted.
I did say you need some mathematics to justify your statement.
That might have looked like this.
When we substitute in x = 30, y = 9 to both inequalities we find that we are optimised.
We're at the end of the lesson now.
What we've learned is that we can use our knowledge of inequalities to solve problems. We can see the link between the algebraic solution to an equality such as 80 + 5w is > 10w and the graphical representation.
We can see how graphic inequalities can be used to more efficiently solve problems in context such as project constraints.
I hope you found this lesson interesting and I hope you can see how you are gonna use maths like this in the future.
I hope to see you again soon for more mathematics.
Goodbye for now.
