Lesson video

Thank you for joining us for today's algebra lesson.

My name is Ms. Davis.

I'm really excited to get stuck into, what I think, is the best part of mathematics.

Simultaneous equations.

Once you get used to pronouncing those words, is one of the topics in mathematics is really fun, can help us solve lots of problems and there's loads of ways to check that our answers are right so that we can be really confident in what we're doing.

So hope you're looking forward to it as much as I am.

Let's get started.

Welcome to this lesson on solving algebra simultaneous equations by elimination.

By the end of the lesson you'll be able to solve two linear simultaneous equations algebraically using elimination.

There's a few key words there that we need to be confident with.

So, equations which represent different relationships between the same variables are called simultaneous equations.

At this stage, we're always looking at two equations, which each contain two variables.

Elimination is a technique to help solve equations simultaneously and it's where one variable in a problem is removed.

We call that eliminating one of the variables.

So we're gonna start by looking at adding or subtracting to eliminate a variable, and we're gonna be thinking about picking the right method for the right question.

When we have an equation with two variables, we can find values for both variables that satisfy the equation.

So what could x and y be for this equation? Pause the video and see if you can come up with an answer.

So we need 4xs + one y to be 16.

And there are lots of options.

You may have had some of these: x could be one and y 12, x could be three and y four, x could be five and y -4.

But of course there are infinite pairs of values which will work, especially if we're including non integer and negative numbers.

To find the solution for two variables we need two different equations linking those variables.

So there was the one we had before.

Now I'm gonna give you another equation linking those variables.

I'm just gonna number my equations so that when I'm doing my solutions I can make it really clear what I'm doing.

This is a really good habit for you to get into.

So now what we can do is we can solve these simultaneously.

That means find a value for x and y that work for both.

What we can do is we combine these equations to form a new equation.

This is only gonna be useful to us if our new equation only has one variable.

Doesn't matter if it's x or if it's y, we want our new equation to only contain one variable and then we can solve it.

We call this eliminating a variable.

The method we use to eliminate the variable and the outcome will depend on the coefficients of the variables in the equations.

You'll see what I mean in a moment.

So in this pair of simultaneous equations, you'll see that the coefficients of y are the same in both equations; we've got one y and one y.

If we subtract these equations, we will eliminate the y term from our new equation.

So if we could do equation one, subtract equation two, 4x - 2x is 2x, y - y = 0y, 16 - 12 is 4.

So our equation becomes 2x = four.

If we wanted to, we could then find a solution for x, and that is the point here.

We're looking to eliminate a variable so that we have an equation that only has one variable in it.

We have maintained a quality by subtracting the same amount from both sides.

From the left hand side we've subtracted 2x + y from 4x + y and on the right hand side we've subtracted 12 from 16.

Our second equation tells us that these things are equal.

So what we've actually done is we've subtracted 12 from both sides of equation one.

It's just the left hand side.

12 is written as 2x + y.

We know that 2x + y is 12.

It's important to understand why it is that we are allowed to do this and why this works.

So what you think will happen if we subtract these equations? Pause the video and have a think.

So what will happen is, we will eliminate the x terms because the coefficients of x are the same in both equations.

So if we do one subtract 2, 7x - 7x is 0x.

Then we've gotta be careful.

3y - negative 2y is 5y.

Just be really careful with your negative values.

3y - negative 2y is 5y.

29 - 4 is 25.

That means 5y is 25 and we could solve that now if we wanted.

Alex says, "I think I can eliminate a variable by subtracting these equations." Pause the video and see what you think.

Well in this case, to see if Alex is correct, we should probably rearrange our equations into a similar format because they're not, at the moment, in a similar format to each other, so it's hard to see what's going on.

So equation one could become 3x - 2y = 13, and it's often easier to have the variables on one side and the constant on the other side of the equation.

Two then could be written as 5x - 2y = 23.

I've just written that equality the other way around.

And now all the variables are lined up neatly so we can compare them.

Now we can see that the coefficients of y are the same in both equations, so Alex is correct.

Subtracting will eliminate the y variable.

It doesn't matter that they're both negative values.

Negative two - negative two is zero.

Let's give it a go.

3x - 5x is negative 2x.

As we said before, negative 2y - negative 2y is zero.

And 13 - 23 is negative 10.

Make sure that if you're doing equation one subtract equation two.

You always do the top row subtract the row underneath even if it gives you negative values.

That means negative 2x is negative 10.

Alex says, "Could I have done equation two, subtract equation one instead to avoid negative values?" Yes, it does not matter which equation is subtracted from which as long as you do the same thing for all your terms. It could be helpful then to rewrite the equation so that you don't make a mistake.

So this time I've written equation two above equation one and now I'm gonna do 2 - 1 which gives me 2x = 10.

The y's are still being eliminated.

Consider the pair of simultaneous equations below.

What do you notice about the coefficients of the variables? Pause the video and have a look at this one.

You might have said something like, the coefficients of b are not the same but they are a zero pair.

Let's think carefully about that language of zero pair.

Three and negative three are not the same number, they are a zero pair.

This time we can eliminate the variable b by adding the equations.

If you think about it, 3 + negative 3 is zero.

Two values are a zero pair.

That means they sum to zero.

So let's try adding them.

2a + 5a is 7a.

3b + negative 3b is 0b.

That was the point.

22 + 13 is 35.

Make sure that if you are adding, you are adding all the terms. Again, we've maintained equality by adding the same amount to both sides of the equation 'cause we've added 5a - 3b to the left hand side and 13 to the right hand side and those two things are equal.

So whether we eliminate a variable by adding or subtracting will depend on the coefficients of the variables.

In general, if one term is the same in both equations, then subtracting will eliminate that variable.

If one term in each equation make a zero pair, then adding will eliminate that variable.

In both cases you'll be left with an equation in one variable which can then be solved.

So I'd like you to match the pairs of simultaneous equations with the correct method to eliminate a variable.

Once you're happy with your matchup, come up for the answers.

So A, subtracting will eliminate r because the coefficients of r are the same.

B, adding will eliminate t because the coefficients of t are a zero pair, C, neither adding or subtracting will eliminate either variables because none of the coefficients of like terms are either the same or a zero pair.

And D, adding will eliminate r because the coefficients of r are a zero pair.

You might have also spotted that subtracting would eliminate t because the coefficient of t are the same.

That wasn't one of our options here, but if you were gonna solve this equation you would have that choice as to whether you wanted to add, therefore eliminate r, or subtract and therefore eliminate t.

Time to put this into practise.

I would like you to sort the pairs of simultaneous equations into the correct parts of the table.

It is possible that some may go in more than one section.

I want you to think carefully about how you justify where you put each pair of simultaneous equations.

When you're happy come back for the answers.

Okay, G and H we're gonna come back to.

Just gonna have a look at the first few.

So A, subtracting eliminates a, there's 4.

5 a's in both.

B, subtracting eliminates b, there's negative b in both.

C, adding eliminates b because there's a positive 2b and a negative 2b and there are zero pair.

D, subtracting eliminates b 'cause there's 9b in both.

E, adding eliminates a.

Doesn't matter that I wrote the terms the other way round in the equation, one's got 3a, one's got negative 3a, they are a zero pair.

Adding will eliminate A.

And F, adding will eliminate a again.

Negative 5a + positive 5a is zero.

Now G and H will depend on how you decide to rearrange these.

So, for G, if you rearrange them as 8a - 8b = negative 16 and 8a + 3b = 39, then subtracting would eliminate A.

However, it's possible that you rearrange them in a different way.

For example, if you had the top one as 8b - 8a = 16, and then the bottom one is 8a + 3b = 39, adding would eliminate A as your coefficient of A are a zero pair.

So that does depend how you chose to rearrange.

For h, again, it's gonna depend how you rearranged.

So you may have rearranged it as b subtract a is nine and then negative b - a is seven.

This case you actually have two choices.

You could subtract to eliminate A, 'cause you've got negative A in both, or add to eliminate b because you've got positive b and negative b.

Alternatively, you might have written the second equation as b + a = negative 7, in which case subtracting would eliminate b, 'cause you've got B in both, and adding would eliminate A because you've got a negative A and a positive A.

Just take your time to check your happy with those.

We're gonna practise actually solving simultaneous equations in a moment, but just be aware that in H once you'd rearranged and there was still a choice of how to eliminate, you do have to make a choice.

You can't do both.

You can't subtract and add, combine the two ideas together.

You have to decide which one you're going to do and make sure you do the same for every term.

Of course you could then go back and check by doing it the other way.

That would be fine as well.

Okay, we're gonna put that into practise now and actually solve some simultaneous equations.

We've done all the hard work already.

Once we've eliminated a variable we can solve for that variable.

This then allows us to find the value of the other variable, too.

Let's try this pair of simultaneous equations.

So the first thing I want to spot is the coefficients of r are the same, so subtracting will eliminate r.

You may wish to rewrite these to avoid as many negative terms as possible.

That's absolutely fine.

I've done that so I'm gonna do 2 - 1.

So 2r - 2r is zero.

5t - 3t is 2t.

And 8 - 4 is 4.

If 2t is 4, I know that t is two.

We have solved for one variable but we haven't finished.

We also need to find the value of r.

But now we know the value of t; that's easy.

You can pick either of your original equations and then just substitute the value for t.

So I've picked equation one.

And then instead of writing t, I can now write two.

So this becomes 2r + 3 lots of two = four.

So 2r + 6 = 4.

And then I can solve it as a linear equation.

Subtracting six from both sides gives me 2r = negative two, therefore r is negative one.

Lucas says, "The best thing about simultaneous equations is that it is easy to check your answers!" And I absolutely agree.

We can substitute the values for t and r back into the equations and check it works.

So let's try the top equation; 2 lots of negative one + 5 lots of 2 is 8.

Is that true? Yes, it is.

Let's try the second equation 'cause it needs to satisfy both.

2 lots of negative one + 3 lots of 2, is that four? Yes, it is.

So we know that our values must be correct.

Alex and Lucas are solving these equations simultaneously.

Alex says, "I think we should add the equations." Lucas says, "I think we should subtract the equations." Have a look at that question.

Who do you agree with? Well let's rearrange the equations so that the x terms are lined up and the y terms are lined up.

Both methods are going to work.

This doesn't happen very often, but it does happen where you do have a choice for methods.

So subtracting would eliminate y, but adding would eliminate x.

Either choice is fine as long as you do the same thing to all the terms. Let's look at Alex's method then.

So 2.

2y + 2.

2y is 4.

4y.

The x's will cancel when you add and then make sure you add the constants, 25 + negative 3 is 22, 4.

4y is 22, so y is five.

Now we need to substitute five back into our equation.

You can pick either equation.

I've gone with equation one.

So 2.

2 lots of 5 + 3.

5x = 25.

That means 11 + 3.

5x is 25, 3.

5x is 14, so x is four.

Now it's up to you how much you're gonna challenge yourself with your decimal skills, and what bits you're gonna allow yourself to check using a calculator.

However, there's elements of these values that will help us with some trickier divisions.

Now when you're doing 22 divided by 4.

4, for example, it's useful to spot that 22 divided by 2.

2 is 10.

Therefore if we're doing 22 divided by 4.

4, that'll be five.

Equally, when we were doing 14 divided by 3.

5, it's helpful to remember that 14 divided by 7 is two.

So if you are doing 14 divided by 3.

5, which is half of seven, that's four.

So all those number skills that you have, you can bring in here to help you with these.

Of course, we know that we can check our answers to see if we are correct.

So x is four and y is five.

We can substitute them back in to check and that will allow us to back up our decimal skills.

Of course, you may want to use a calculator for certain bits as well to support you.

So let's look at Lucas', which was subtracting the equations.

So 3.

5x - negative 3.

5x, and this is where you need to be careful.

If you've got 3.

5 and you're subtracting negative 3.

5, that's the same as adding 3.

5, so you get 7x.

Then 25 - negative 3 is 28.

It is important that you're thinking carefully about those negative values.

Obviously, we'll check our answers at the end to see if they work.

So substitute x is 4 into one of those equations and solve.

By adding the equations we get the same solutions as we did by subtracting the equations.

You might want to think about which method you preferred.

Both methods had their advantages.

Most of the time you will not have a choice.

There'll be an obvious, easier way to do it by looking at your coefficient.

Okay, time for a check.

I'm gonna show you one example on the left hand side and then you're gonna give it a go.

I'm gonna rewrite the equation so the A terms and the B terms are lined up.

I can see that the A terms are a zero pair, so I'm gonna add to eliminate, 4b + negative 5b is -b, 7 + negative 5 is 2.

so negative b is 2, which means that b is negative 2.

Then I can substitute into either equation.

I've gone with equation one.

Four lots of negative 2 - 5a is 7.

So negative 8 - 5a is 7.

And negative 5a is 15, and A is negative 3.

I'm absolutely gonna substitute those back in to check.

If you look at the first equation, we've got 4 lots of negative 2 - 5 lots of negative 3.

And I want to see if that is seven.

So that is negative 8 - negative 15 or negative 8 + 15, which is 7.

And then you still need to check it works for the second equation.

So 5 lots of negative 3 - 5 lots of negative 2, so that is negative 15.

Subtract negative 10, or negative 15 add to 10 which is negative five.

So those work for both equations.

They must be our solutions.

Time for you to have a go.

Have a go on the right hand side.

Try and do it without a calculator, but you might wanna check with a calculator once you're done.

So, setting up our equations again, we're gonna want the A's and the B's lined up.

And you'll see that the A's are a zero pair, so add them to eliminate.

We get 3b = 15.

B then is five.

Substituting into either equation, 2 lots of 5 - 6a is 16, so that's 10 - 6a is 16, so negative 6a is 6, and A is negative one.

If you substituted into equation two instead you would've got 5 + 6a = negative 1.

So 6a = negative 6 and a is negative one.

You get the same answer either way, of course.

So A is negative one and B is five.

And, of course, we can double check that by substituting back into the two original equations.

All right, time for a practise then.

For question one, I'd like you to solve these simultaneous equations to find the value of each variable.

You need to make sure you write what A equals and then what E equals, then for B, what G equals, then what I equals.

The reason being there's a code in question two.

So for question 2 you need to use the values of the letters you found out to crack that code.

So for example, if you found out that a = 8, then everywhere in the code that there was the number eight, you would write A.

Of course there isn't any numbers eight in our code, so A cannot be eight.

If you do it correctly, it should spell a mystery word.

Give that a go.

So for question three, I'd like you to solve these simultaneous equations for X and Y.

Make sure you're laying out your working clearly.

Feel free to rearrange any equations as necessary.

Give it a go.

The same for d, e, and f, don't forget, you can check your answers work before you then come back to find out the answers from me.

Off you go.

Okay, I would like you to have a go at spotting the mistakes in these questions.

If someone's had a go at solving these equations is absolutely fine to have answers that are fractions, decimals, negatives, that happens quite a lot.

However, there are some mistakes in here.

See if you can find them.

Let's have a look then.

So you should have A is five and E is four.

G is seven and I is one.

L is three and N is negative three.

N is negative one and T is two.

If you've got those correct, you have found the mystery word is 'eliminating'.

All right, we're gonna go through our solutions a little bit more for this one.

So I'd like you to pause the videos and check your lines of working.

Remember, where I've chosen to substitute into an equation you may have substituted into the other one, so check your final answers.

If you didn't get the correct final answer, see if you can use my working to see where you went wrong.

The same for d, e, and f.

For F, there was lots of options with how to solve it.

It all depended on how you rearranged the original equations and then you still add a choice as to whether to add or subtract.

If you rearranged in the same way that I have, then you've got a choice; you could subtract to eliminate the x values, and if you do 2 - 1 that would give you 6y = 18, so y is three.

Or you could have added to eliminate the y values, which would've given you 5x = 60, so x is 12.

So either would be a good first step and then you can substitute in to get the other value.

Of course, it's possible that you rearrange in a different way to start.

Check your answers and then we'll have a look at the last bit.

Okay, let's have a look at these mistakes then.

So we got, in a model, with negative values in the first one.

The main problem is that we should have added not subtracted.

The coefficients of b are a zero pair, so adding will have eliminated.

What has happened is this person has subtracted the As but then not actually subtracted the Bs because negative 5b - 5b should still give you negative 10b.

So although they subtracted the As and they subtracted the constants, if they had subtracted the Bs, they would've still had negative 10b.

So they've done a mixture of adding and subtracting, which does not work.

If you add them, you'd get 12a = 0, which gives you a = 0.

That's fine, zero could be a solution.

If A is 0 then you get B as three fifths.

For question B.

This time they have made the right choice that we're gonna subtract.

They've written in their working that they're gonna do 2 - 1.

That's fine, 6a - 3a is 3a, but 3.

8 - 8.

3 is negative 4.

5.

If you've chosen to do 2 - 1, you need to do that for all the terms. So 3a should be negative 4.

5 and therefore A would be negative 1.

5.

If that wasn't enough, they've also made a mistake when they substituted.

Once they found that A was 1.

5, which is incorrect anyway, they needed to do three lots of 1.

5 + 4b is 8.

3.

Of course we know that A is actually negative 1.

5, so we do three lots of negative 1.

5 + 4b = 8.

3.

And then we get b as 3.

2.

So that's two mistakes that you can watch out for in your own working.

Fantastic work today.

We have looked at the fact that if we have two equations linking the same two variables, we can add or subtract the equations to create a third valid equation.

Where the coefficients of one variable are the same in both equations, subtracting eliminates the variable.

Where the coefficients of one variable are a zero pair, adding eliminates the variable.

Once we've eliminated a variable we can solve for the remaining variable.

Substituting that back into the original equation will allow you to find the value of the other unknown.

Don't forget to find the value of both unknowns and show you're working really clearly.

Don't forget Lucas' point that simultaneous equations are fantastic because you can always check your answers work.

And if they're not right, you can go back through your working and correct it.

Thank you for joining us for that lesson and I look forward to seeing you again.