Lesson video

Hello, Mr. Robson here.

Welcome to Maths.

Lovely to see you again.

Today we're problem-solving with quadratic and linear simultaneous equations.

I know you love problem-solving.

I do, too.

So let's get stuck in.

Our learning outcome is that we'll be able to use our knowledge of simultaneous equations to solve problems in context and interpret the solutions.

Some keywords we're going to encounter, substitution and elimination.

Substitute means to put in place of another.

In algebra, substitution can be used to replace variables with values, terms, or expressions.

Elimination is a technique to help solve equations simultaneously where one of the variables in a problem is removed.

Two parts to today's learning, I'm going to start by solving linear problems. It may not be immediately obvious that solving simultaneous equations will be helpful.

For example, this problem.

"An electrician charges a fixed call out charge then an hourly rate for every hour worked on a job.

They charge £190 for a four hour job and £330 for an eight hour job.

Find their hourly rate and call out charge." There's lots of ways we might tackle this problem.

The most efficient way is to write a pair of simultaneous equations.

Equation One is the call out charge and four hours costing £190.

Equation two, the call out charge in eight hours costing £330.

When we subtract Equation One from Equation Two, we find the difference is four hours and £140, therefore H = 35.

Just like any other simultaneous equation, substitute the H back in and we can find the other variable.

We can find that C = 50.

We just need to put that algebra back into context and we can declare their hourly rate is £35 and their call out charge is £50.

Problem solved and very efficiently, too.

Quick check that you've got that.

"Which pair of simultaneous equations would you form to solve this problem?" I'd like you to pause this video, read that problem, and then decide which pair of simultaneous equations you'd use.

Do that now.

(silence) Welcome back.

I do hope you said Option B.

We'd want that pair of simultaneous equations.

Once we've selected the right pair of simultaneous equations, I'd like you to solve them to find the fixed fee per rental and the daily rate.

Pause and do that now.

(silence) Welcome back.

Let's see how we did.

Hopefully you found that the difference between the two equations, or when you subtract Equation One from Equation Two, you're left with 5D = 200, so D must be 40.

Substitute back into one of the original equations and we find that F = 80.

That means nothing unless we put it back into the context of the question.

The daily rate is £40 and the fixed fee £80.

Some contextual problems may require more work.

For example, "Swizzie Sweets come in small packets and large packets.

Two small packets and five large packets contain 370 sweets.

Three small packets and four large packets contain 345 sweets.

Calculate how many sweets in each pack size." What a mouthful that question was, but we can solve it really quickly and efficiently by writing a pair of simultaneous equations.

We'll write Equation One.

That's two small packets and five large packets contain three 70 sweets.

Equation Two.

Three small packets, four large packets, 345 sweets.

You'll notice we've got no matching coefficients at the moment, so we need to scale these equations in some way.

I'm going to treble all the terms in Equation One, double all the terms in Equation Two.

'Cause what we get now is 6S and 6S present in both equations, matching coefficients.

I can then subtract one equation from the other and 7L = 420.

L must be 60.

Little bit of substitution, we find that S must be 35.

Does that mean anything? Not to a reader unless we put it back into the context of the question.

"Small packets contain 35 sweets.

Large packets contain 60 sweets." Problem solved.

Sometimes simultaneous equations could be used but they're not necessary.

This problem, for example.

"Two numbers have a sum of 20 and a difference of 10.

What are they?" Absolutely could write a pair of equations that will enable me to solve this, but those numbers are pretty straightforward.

We can trial and error this.

Nineteen and one.

Nope.

Eighteen and two.

Nope.

Seventeen and three.

Nope.

Nope.

Yes.

Fifteen and five.

They sum to 20.

They've got a difference of 10.

That problem wasn't that difficult.

A more difficult version of this problem might deem simultaneous equations absolutely necessary.

For example, "Two numbers have a sum of 19 twelfths and a difference of one 12th.

What are they?" You might be able to spot the solution to this, but just in case you couldn't, we could write a pair of simultaneous equations.

A and B sum to 19 over 12, whereas A subtract B is one twelfth.

This is a lovely pair of simultaneous equations because when I sum them together, the positive B and the negative B are eliminated.

I get two A on the left-hand side, 20 over 12 on the right-hand side, so A must be ten twelfths, which, of course, simplifies to five sixths, and B must be nine twelfths, which, of course, simplifies the three quarters.

Oh look, A was ten twelfths B was nine twelfths.

You might have known that a while back, but it's nice to see how that solution was reached using simultaneous equations because we're going to see a similar problem, which is a little bit more difficult shortly.

Quick check.

You've got this.

"Which method would be most efficient for this problem?" The problem being "Two numbers sum to negative 3.

1 and have a difference of 0.

08.

What are they?" Would you go for Option A "Forming and solving a pair of simultaneous equations"? Option B, "Trial and error"? or Option C, "Forming and solving a linear equation"? Pause.

Make your choice now.

(silence) Welcome back.

I do hope you said Option A "Forming and solving a pair of simultaneous equations".

Just out of interest, which equations did you form? Did they look like that? A and B sum to negative 3.

1, whereas A subtract B sums to 0.

08.

From there we can find those values.

Practise time now.

Question One.

"A locksmith charges a fixed call out charge then an hourly rate for every hour worked on a job.

They charge £245 for a two hour job and £425 for a five hour job.

Find their hourly rate and call out charge." Pause and do that now.

(silence) Question Two.

"Fab Frooties sell their sweets in small packets and large packets.

Five small packets and two large packets contains 220 sweets.

Three small packets and five large packets contains 284 sweets.

Calculate how many sweets in each pack size." Pause and do that now.

(silence) Question Three.

"Two numbers sum to 3.

43 and have a difference of 3.

99.

What are they?" I hope you're looking at that problem going, "I'm not going to try and guess.

I'm going to form and solve a pair of simultaneous equations." You can pause this video and have a go now.

(silence) Question Four, get ready for this one.

"Funky Fruits sell a wide variety of bags of fruit.

Two bags of apples, a bag of bananas, and three bags of clementines contain 57 pieces of fruit.

One bag of apples and a bag of bananas contain 13 pieces of fruit.

Five bags of apples and a bag of clementines contain 52 pieces of fruit.

Calculate the number of pieces of fruit in each bag." That was a mouthful.

You're going to have to form several equations and this problem's slightly different to other ones we've seen so far, but I'm sure you'll work it out.

Enjoy.

Pause and give it a go now.

(silence) Feedback time.

See how we did.

Hopefully for Question One, you formed a pair of simultaneous equations that looked just like that and then subtracted one equation from the other.

3H = 180.

H = 60.

Some substitution to find that C must be equal to 125.

And I hope you didn't stop there.

I hope you wrote a sentence explaining "The hourly rate is £60 and the fixed call out charge is £125." If you didn't write that sentence, you wanna copy it down now.

Question Two.

Fab Frooties and their small packets and large packets.

Hopefully you formed a pair of simultaneous equations that look like so and recognised we can't eliminate from here.

We need to scale up those equations and match up, we could have matched up the S terms. I matched up the L terms and I subtract the second equation from the first.

I'm left with 19S = 532.

S is equal to 28.

Substitute that back into one of our original equations and I found L = 40.

Am I finished there? Not yet.

I need to just pop it back into context for the reader.

"Small packets of Fab Frooties contain 28 sweets.

Large packets contain 40 sweets." Problem solved.

Question Three.

"Two numbers sum to 3.

43 and have a difference of 3.

99.

What are they?" I'm definitely not going to try and trial and error this one.

I need to form a pair of simultaneous equations that look like so.

Joyously when I add those equations together, 2A = 7.

42, so A must be 3.

71.

Substitute that back in and I find B is negative zero.

28.

I'm glad I had my ability with simultaneous equations to help me with that one.

That would've taken a while to trial and error those numbers.

For Question Four, there was an awful lot going on here with Funky Fruits.

The first bit of information we're given about the apples, bananas, and clementines enables us to write that equation.

2A + B + 3C = 57.

The second bit of information about the apples and bananas, we can say A + B = 13.

Then the third bit of information we get, we can say 5A + C = 52.

It's not immediately clear how we might eliminate from here so we don't eliminate, we substitute.

I can rearrange Equation Two to leave B in terms of A.

And I can rearrange Equation Three to leave C in terms of A.

I can then substitute those into Equation One.

I'll replace B with 13 minus A and I'll replace C with 52 minus 5A and it'll look like that.

You know from here we can expand those brackets and simplify.

Therefore, A = 8.

Once I've discovered that A = 8, I can substitute back into the other equations and I'll find that B = 5 and C = 12.

There were lots of ways you might have solved this problem, but you should have arrived at those same numbers.

From there, we just need to put that information into a sentence.

"A bag of apples contains 8 pieces of fruit.

A bag of bananas contains 5 pieces of fruit and a bag of clementines, 12 pieces of fruit." What a lovely problem.

Onto the second half of the lesson now.

Linear and quadratic problems. I wonder how they're gonna look different? For some problems we need a pair of equations where one is linear and one is quadratic.

For example, "Two numbers sum to 25 and their squares sum to 313.

What are they?" The crucial difference here, "their squares sum to 313" so it sounds and reads a lot like the problems we saw earlier, but it's crucially different.

The sum of the numbers A and B, they equal 25.

The sum of the squares, when we put that into algebra, A squared + B squared = 313.

This is a different kind of problem that we need to solve now.

Solving by substitution is going to be really efficient.

I'm gonna make B the subject of Equation One and then I'll substitute that in to Equation Two.

That's a pair of brackets that need to be multiplied now.

25 minus A squared.

Once I do that, I get that quadratic which I can simplify and rearrange.

Now I can solve this.

2A squared minus 50A + 312 = zero.

I can start to solve by dividing through by 2, the common factor of every term, and rather joyously it factorises.

So I find that A is equal to 12 or 13.

What happens now is if A = 12, I substitute it back in.

I find that B = 13.

But if I said A = 13, I find that B = 12, and I just conclude the two numbers are 12 and 13.

I enjoyed that problem.

Quick check that you can do a problem of a very similar nature.

"Two numbers sum to 1 and their squares sum to 41.

What are they?" Pause and try this now.

(silence) Welcome back.

Let's see how we did.

Hopefully you wrote those two equations, the two numbers summing to one and their squares summing to 41.

And then we can rearrange.

You could make A the subject and substitute that in.

I'm gonna make B the subject and substitute it in.

We'll arrive at the same answer.

When I substitute one minus A into the B position, I get some brackets that need to be multiplied out.

They multiply out like so.

From here I can simplify and rearrange.

Divide three by two.

I've got a quadratic.

A squared minus A minus 20 = zero to solve.

Factorises.

Wonderful.

So we find either A = 5 or A = negative 4 and then if A = 5, B = negative 4.

But if we called A negative 4, then B would equal 5.

Therefore the two numbers were 5 and negative 4.

Next up, the ability to solve pairs of simultaneous equations where one is linear and one is quadratic can be really useful when solving graphical problems. "Show using simultaneous equations that the line X + Y = 8 intersects the curve Y = X squared + X + 5 twice." Well let's start that pair of equations and let's rearrange Equation One to make Y the subject, and we can substitute it into Equation Two.

When we rearrange from there, hurrah a quadratic that we can solve.

Again, this one factorises.

I'll have two solutions, X = 3 and X = 1.

In the context of this problem, we're actually finished at this point.

We can say, "There's two solutions, therefore there's two intersections." We've solved this problem.

If the problem were worded differently, we might be asked to substitute it back in and find the coordinates of the points of intersections.

We would just put those X values back in to one of the original equations and we'd find the corresponding Y values and we'd get these two coordinates.

X = negative 3, Y = 11, X = 1, Y = 7.

You don't necessarily need to draw this.

I'm just showing you this because you'd be interested.

Those two intersections will be there.

That's the parabola X square + X + 5 and the straight line X + Y = 8, and you can see those two points of intersection.

Quick check now.

Let's see if you can do a similar problem to that one.

I'd like you to show using simultaneous equations that the line Y minus 5X minus one = zero intersects the curve Y = X squared + X + 5 just once and therefore is a tangent.

What an interesting problem.

You can pause and try it now.

(silence) Let's see how we got on.

Hopefully you started by writing out the two equations rearranging Equation One to make Y the subject, substituting it into Equation Two, and rearranging, so we had a quadratic equal to zero.

From there it factorises.

Wonderful.

It factorises to X minus 2 squared.

From here we can see there's only one solution, X = 2.

If there's any one solution, there's any one intersection, therefore it is tangent.

Again, you didn't need to go this far because we weren't asked to in the question, but that intersection we could find the Y coordinate to be 11 and there it is.

The curve and the line and the intersection.

I'm just showing you that because I know you are interested in this problem.

Practise time now.

Question One.

"The line X + Y = 6 intersects the curve Y = x square + 2X + 6 twice.

Find the distance between these two intersections." For Question Two, I'd like you to "Find the area of the triangle whose vertices are the point (9,0) and the intersections of the line 3Y = X and the curve Y squared = X.

There's a lot going on in that second question, but it's lovely.

You'll want to find those intersections and then maybe a visual representation will help you put it all into context.

Pause and give those problems a go now.

(silence) Welcome back.

Let's see how we did.

For Question One, start by writing down the equations and then there's all sorts of ways we can solve this.

Can you see how I'm doing it? We could have rearranged and made X the subject, we could have rearranged and made Y the subject, but X + Y = 6, so I can replace the 6 in Equation Two with X + Y and then I can remove Y from both sides, or add negative Y to both sides, and then simplify and we're left with X square + 3X = zero.

That factorises rather delightfully to X bracket X + 3.

So either X = negative 3 or X = zero.

I hope you found those same values and I hope you know we're certainly not finished here.

We need to find the corresponding Y values, so substitute that back in.

When X is negative 3, Y is 9.

When X is zero, Y is 6.

We get these two coordinates.

We were asked to find the distance between these two intersections, the distance between those two coordinates.

Lots of ways we could work it out.

I'm going to use the visual, myself, just to get a picture of what's going on here.

That's those two coordinates.

That's how I'm going to find the distance between.

I'm gonna use Pythagoras.

3 squared at 3 squared is 18.

The square root at that, my calculator tells me, is 3, lots of root to 2, which is 4.

24 and so on.

For Question Two, we were asked to find the area of a triangle, but we don't know all of its vertices yet.

We're told two of the vertices of the intersections of the line 3Y = X and the curve Y squared = X, so find those intersections.

We can substitute Equation Two into Equation One, rearrange, we got quadratic equal to zero.

We can solve this.

Factorise and we find Y = zero and Y = 3.

We need to find the X coordinates.

Substitute back in, and we find the X coordinates of zero and 9.

We found the other two vertices.

Zero, zero and 9, 3.

A visual representation is gonna help me solve this problem.

You wouldn't necessarily need to draw that line and that curve.

I'm just showing you these because I know you'd be interested.

But you would know these coordinates, you know, zero zero, you know, 9, 3 and you were told in the question the other vertice is the point 9,0.

So it's that triangle that we're interested in.

This visual gives you the dimensions of that triangle.

I just plug that into the area of a triangle formula.

The area is gonna be half multiplied by 9 multiplied by 3.

That gives us an answer of 27 over 2.

You might see that then declared as 13.

5 square units.

That's the end of our lesson now, sadly, but we learned that we can use our knowledge of simultaneous equations to solve a wide variety of problems in context.

I've enjoyed it.

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

Goodbye for now.