Lesson video

Hello, Mr. Robson here.

Welcome to maths, what a lovely place to be.

Today, we're problem solving with simultaneous equations.

You know me, I love problem solving, so what are we waiting for? Our learning outcome is I will be able to use our knowledge of simultaneous equations to solve problems. Some keywords we're gonna hear throughout the lesson, substitute, substitution.

Substitute means to put in place of another.

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

Another word you're gonna hear a lot is elimination.

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

Two parts to today's lesson and the first part is absolutely delightful.

We're going to solve some puzzles.

In this puzzle, the number at the end of each row or column is the sum of the value of the symbols in the boxes.

Can you give this one a go and figure out what the value of each symbol is? By all means, give it a go and I'll be back in a moment with the solution.

Pause now.

How did you get on? I hope you enjoyed giving it a go.

Well done if you tried.

A useful please to start is that middle row there.

If we know that those three symbols add up to 18 and each symbol has to have the same value, because it's the same symbol, it must be worth six.

Wherever we see that symbol now, we can substitute six into the grid.

From there, we can find out what a star is worth.

A star, add six, add six equals 17, that's the right-hand column.

Those stars must worth five.

Substitute a five in wherever we saw the star and we can find out what an arrow is worth.

The arrow must be worth two.

When we substitute that two into our grid, we can check our work.

It's always important to check your work in maths and frequently we can.

When I check this, every row and every column adds up correctly.

So I've worked out the value of each symbol correctly.

That's reassuring.

Quick check that you can do that accurately for yourself.

In this puzzle, the number at the end of each row or column is the sum of the value of the symbols in each box.

Can you find the value of the L, of the M and of the V? Pause and try that now.

Welcome back, let's see how we did.

The top row was a useful place to start.

If L plus L, plus L equals 21, then each L must be worth seven.

In the middle row there, if we know that an L is worth seven, two L plus V equals 15, the two Ls are worth 14.

That means V must be worth one.

From there, we can turn our attention to that bottom row.

M plus V, plus V equals six.

Well, M plus two V equals six is a way we could describe that bottom row and if V equals one, then M must equal four.

We can check our work again.

If we substitute in that value for L, the value for V, the value for M, every row, every column correctly sums. We know that we're right.

In order to solve this puzzle, we were forming and solving equations.

That's a nice application of our algebraic skills.

In some puzzles of this nature, the starting point is not so obvious.

In this case, there's no row or column containing only one symbol, so where do we start? What do you think? Can you spot a starting point? Can you figure out the value of any of these symbols? By all means, pause and try for yourself or stick around and I'll help you out.

A useful starting point might be that middle row.

We have the addition symbol, two of them and the arrow symbol equaling 23.

Why am I interested in that row? It's because of the left column where we're got an arrow symbol, an addition symbol and another arrow symbol being equal to 25.

What we've got here is two equations involving the same two variables, but we can't eliminate yet with no matching coefficient.

What can we do then? Well, we can double that second equation.

Rather than having two arrow signs and one addition sign, we could have four arrow signs and two addition signs.

And if we're doubling everything on the left-hand side of the equation, we have to double the right-hand side, two lots of 25 is 50.

I'm going to rearrange that equation as that.

Why do I want it in that order? It's for this reason.

When I line equation one up with the equation that we've created, we can ask a key question.

What's the difference between our two equations? Can you see it? Absolutely you can.

The difference is the three arrow symbols on the left-hand side and 27 on the right-hand side.

We've created an equation.

Three arrow symbols must be worth 27.

In that case, each individual arrow is worth nine.

When I substitute that in, the whole thing becomes unlocked.

In the left column, I can see that the addition symbol must be worth seven.

Substitute that in and I can finish the problem by finding out that two blue stars in that top row are worth 20, each blue star must be worth 10.

When I substitute those values back into those symbols' positions, I can tell every row, every column adds up correctly, we know that we're right.

When we solve a similar puzzle, using letters instead of symbols, you can see that we're actually forming and solving a pair of simultaneous equations.

This problem's similar.

There's no row or column with the same letter.

So let's look at that top row, e plus two g equals 29.

And that left column, two e plus g equals 40.

And let's call it equation one and equation two.

From here, we can double equation one to turn it into two e plus four g equals 58.

Why do we want do that? Because when we substitute equation two in underneath, elimination is about to occur.

I'll subtract equation two and we'll have eliminated the e variables.

Two e minus two e leaves us with no e's.

They've been eliminated.

Four g minus g is three g, 58 minus 40 is 18.

If three g equals 18, g equals six.

From there, we substitute that g value back into one of the equations and we can find that e is worth 17.

Once we know e is equal to 17, all that's left is to find the value of f.

Look at that bottom row there, two e's and an f equal 55.

Those two e's will be worth 34, so f is equal to 21.

In order to check here, if we look at our middle row, g plus e plus f equals 44, we can substitute our values for g and e and f in and when they equal 44, that's when we know we're correct.

Quick check that you can do this now.

Which pair of equations would you use to begin solving this puzzle? There's equation A, equation B, equation C.

Pick two of them.

Which two do you need to begin to solve this? Pause and have a think about this now.

Welcome back.

I hope you said A and C.

We have two equations with the same unknowns.

The variable a and the variable c are in those two equations.

Once we've selected those two equations, we can start to model a simultaneous equation being solved.

I'd like you to complete the working to find the values of a, b and c.

I've started you off by calling one of them equation one, one of them equation two and you'll notice I'm asking you to double equation one as your next step.

I'd like you to continue this work and solve this puzzle.

Pause now, give it a go.

Welcome back.

Let's see how we did.

In doubling equation one, we should've got four a plus two c equals negative 20.

And then when we substitute equation two underneath and then subtract, two c minus two c leaves us with no c's, they've been eliminated.

Four a minus a is three a.

Now our right-hand side.

Nothing serious here.

Negative 20 subtract negative five, well, that's going to be negative 15.

Be really careful with negative values.

If three a is equal to negative 15, a is equal to negative five, we can substitute that back into that equation to find the value of c, which is zero.

From there, we can use the equation two c plus b equals 13 to find that b is equal to 13.

We can check by adding up that right-hand column, a plus c plus b equals eight.

And our values for a, c and b mean we've got a balanced equation.

We know that we're correct.

Practise time now.

I'd like you to find the value of each symbol or letter in these two puzzles.

Pause, figure them out now.

Question two, I'd like you to find the value of each letter in these two puzzles.

These two are a little trickier, but I'm sure you'll find them thoroughly enjoyable.

Good luck.

Feedback time.

Question one, puzzle A.

Did we find two crescent moons and a heart equals 37, two hearts and a crescent moon equal 29? If we double up that second equation and then consider the difference between the two, we find three hearts are worth 21.

If that's the case, each heart is worth seven.

That's where I started.

You might've started somewhere differently.

One I knew that each was seven, I could unpick the rest of the puzzle to find that the pentagon was three and the crescent moon was 15.

Whichever method you chose, you should've got the same values for the symbols.

For part B, I started by setting up a pair of equations, two f plus h equals 3.

55 and f plus two h equals 2.

9.

From there, I doubled the first equation, then aligned the second equation and eliminated by taking the second equation away from double the first equation, two h minus two h left me with no h's.

They were eliminated.

If three f equals 4.

2, f equals 1.

4.

Substitute that back in and I was able to find that h equals 0.

75, g equals 2.

5.

You might not have done this problem in the exact same way as me, but if you've done it accurately, you should have the same values.

For question two, part A, I started by writing the equation a plus three b equals 28 and then the equation two a plus two b equal 24.

If I double equation one and then subtract equation two from there, I eliminate the a's and find four b equals 32, therefor b equals eight.

From there, I can substitute back into my first equation and find that a equals four and then I can substitute into another equation, two a plus c plus b equals 13, that's the left column there.

And I find that c is equal to negative three.

One more little step to go.

I need to find out what d is worth.

I can substitute that into two b plus c plus d equals 22, that's the third column.

Once I've substituted, I find that d is equal to nine.

Question two, part B.

This one was a real brainteaser, because I can write these three equations and they're the only equations I can write, but where do we go from here? It's not quite so clear how to eliminate from this position, so can we eliminate or do we do something different? In this case, you could rearrange the equations.

If I rearrange my equation three to read two a equals 18 minus c, look at equation one.

It contains the term two a.

So I can substitute 18 minus c into it.

A little rearrangement from there and I get c as the subject or c in terms of b at an integer value.

I can substitute that into equation two to get two b plus five plus b, which is what we know c is, equals 14.

From there, that simplifies and rearranges to three b equals nine, b must equal three.

From there, we can substitute to find that a equals five and then c equals eight.

Trickier puzzle that one, but I'm sure you'll agree a beautiful one.

On to the second part of our lesson now, problems in context.

It may not be immediately obvious that solving simultaneous equations will be helpful.

For example, an electrician charges a fixed call out charge then an hourly rate for every hour worked on a job.

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

Find their hourly rate and call out charge.

There's lots of ways we could figure this problem out.

Simultaneous equations is a very efficient one.

Equation one is c plus four h equals 190.

That is the fixed call out charge and four hours of work, costing 190 pounds.

Can you guess what I'm going to write next? Well done.

The eight hour job.

That's the call out charge and the eight hours of work equaling 330 pounds.

From here, we can eliminate.

If I do equation two subtract equation one, I'm left with four hours of work being worth 140 pounds.

Every hour must therefor be 35 pounds, substitute that back in and we find our call our charge is 50 pounds.

To answer in context.

Their hourly rate is 35 pound and their call out charge is 50 pounds.

Quick check that you can set up a pair of simultaneous equations like that.

Which pair of simultaneous equations would you form to solve this problem? The problem being, a car rental company charges a fixed fee per rental then a daily rate.

A two day rental costs 160 pounds and a seven day rental 360 pounds.

Find the fixed fee and daily rate.

You're gonna choose option A, option B or option C.

Pause and have a think now.

Welcome back.

I hope you said option B.

Why? Because if f is the fixed fee and d is the number of days, equation one there is the fixed fee and two days of rental costing 160 pounds.

Option B is the fixed fee and seven days of rental costing 360 pounds.

From here, now that you've selected the correct pair of equations, I'd like you to solve them and find the fixed fee per rental and the daily rate.

Pause and do that now.

Welcome back.

When we lay out the equations like so and subtract equation one from equation two, we see five days cost 200 pounds, each day must be 40 pounds.

Substitute back in and we find the fixed fee is 80 pounds.

We can conclude the daily rate is 40 pounds and the fixed fee is 80 pounds.

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 are in each pack size.

What we need to do is set up a pair of equations.

Scenario one or equation one, my two small packets and my five large packets give me 370 sweets.

Equation two is three small packets and four large packets giving me 345 sweets.

It's not so immediately obvious how I might eliminate or substitute here.

I have to scale up the equations.

If I triple everything in equation one and double everything in equation two, I end up with this.

And the joy of that is I've got matching s coefficients.

From there, I can subtract one from the other, six s minus six s leaves you with no s's, 15 l minus eight l is seven l and 1110 minus 690 left 420.

Seven l equals 420.

What that tells us is each large packet contains 60 sweets.

Substitute back in and we can find that the small packets contain 35 sweets.

To finish, we need to take our lovely algebra and put it back into a couple of sentences.

We'd need to write, small packets contain 35 sweets, large packets contain 60 sweets.

Problem solved.

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

For example, two numbers have a sum of 20 and a difference of 10, what are they? We absolutely could solve this algebraically, but if we just have a play with some numbers, 19 and one, sum of 20, but not a difference of 10.

18 and two, 17 and three, we're not there yet, 16 and four, 15 and five.

Bingo.

A sum of 20 and a difference of 10.

We didn't need to use algebra for that one.

Now, sometimes the nature of the numbers means simultaneous equations are absolutely necessary.

I'll give you a similar problem and you'll spot a difference.

Two numbers have a sum of 19 over 12 and a difference of one twelfth.

What are they? What do you think? Could you trial and error this one? I don't think I can, so I'm going to write a pair of simultaneous equations.

Two numbers, a and b, they sum to 19 over 12 and the difference between the two of them is one twelfth.

This is quite a nice pair of simultaneous equations, because I just add the two together and I get two a equals 20 over 12.

So a must be 10 over 12, which cancels down to five sixth.

Substitute that in and you'll find that b is nine over 12, which simplifies to three quarters.

Quick check you've got this now.

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

1 and have a difference of 0.

08.

What are they? If you were solving that problem, would you form and solve a pair of simultaneous equations, would you use trial and error or would be forming and solving a linear equation? Which one do you think it is? Pause, tell the person next to you or tell me aloud at the screen.

Welcome back.

I hope you said it's option A, we'll be forming and solving a pair of simultaneous equations.

The equations would look like a plus b equals negative 3.

1 and a minus b equals 0.

08.

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 pounds for a two hour job and 425 pounds for a five hour job.

Find their hourly rate and their call out charge.

Pause and work this one out now.

For 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 this now.

Question three, two numbers sum to 3.

43 and have a difference of 3.

99.

What are they? Pause, work this one out.

Question four, 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 in total.

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.

You've got all the information you need now to calculate the number of pieces of fruit in each bag.

Tricky on this, good luck.

Feedback time now.

Question one, our locksmith.

We should've started by setting up a pair of simultaneous equations that read like so.

From here, we can eliminate.

Take equation two and subtract equation one, you're left with three hours of work costing 180 pounds.

An hour of work is 60 pounds.

Substitute back in and we can find that call out charge is 125 pounds.

Again, we're gonna turn our beautiful algebra back into a sentence and write, the hourly rate is 60 pound and the fixed call out charge is 125 pounds.

For question two, Fab Frooties.

You should've set up a pair of simultaneous equations like so and recognised it wasn't so easy to eliminate on this occasion.

We needed to five fold equation one, double equation two and then when we subtract one from the other, we're left with 19 s equals 532, s is equal to 28.

Substitute back in and we find l is equal to 40.

Then we turn that beautiful algebra back into a sentence.

Small packets of Fab Frooties contain 28 sweets.

Large packets contain 40 sweets.

Problem solved.

Question three's a lovely one.

Two numbers sum to 3.

43 and have a difference of 3.

99.

How do we go about solving this? A pair of simultaneous equations obviously.

We can add the two equations together to get two a equals 7.

42, a must therefor be 3.

71.

Substitute that back in and we find that b is negative 0.

28.

Question four was a tricky but beautiful problem.

We could write the equation two a plus b, plus three c equals 57 and a plus b equals 13 and then five a plus c equals 52 from the information that we're given.

But from here it's not so clear how we could use elimination, so we won't.

We'll use substitution.

Rearrange equation two and we get b in terms of a.

Rearrange equation three, we get c in terms of a.

And then when we look at equation one, we can substitute in in the b position our 13 minus a and in the c position our 52 minus five a.

From there, I need to expand those brackets and simplify and I find that a equals eight.

Substituting a equals eight back into one of our original equations tells me that b equals five, substituting those in, I find that c equals 12.

We need to put those into context.

A bag of apples contains eight pieces of fruit.

A bag of bananas contains five pieces of fruit and a bag of clementines contains 12 pieces of fruit.

I hope you enjoyed that problem as much as I did.

Sadly, that's the end of our lesson now, but we've learned that we can use our knowledge of forming and solving pairs of simultaneous equations to solve problems in a variety of contexts.

That's all for today.

I look forward to welcoming you back to maths very soon.

Goodbye for now.