
The purpose of this video is to
look at the solution of

elementary simultaneous

equations. Before we do that,
let's just have a look at a

relatively straightforward

single equation. Equation we're
going to look at 2X minus

Y equals 3.

This is a linear equation. It's
a linear equation because there

are no terms in it that are X
squared, Y squared or X times by

Y or indeed ex cubes. The only
terms we've got a terms in X

terms in Y and some numbers.

So this represents a linear

equation. We can rearrange it so
that it says why equal

something, so let's just do
that. We can add Y to each side

so that we get 2X.

Equals 3 plus Yi. Did say add
why two each side and you might

have wondered what happened
here. Well, if I've got minus Y

and I add why to it, I end up
with no wise at all.

Here we've got two X equals 3
plus Y, so let's take the three

away from each side. 2X minus
three equals Y.

So there I've got a nice
expression for why. If I take

any value of X. Let's say I take
X equals 1, then why will be

equal to two times by 1  3,
which gives us minus one. So for

this value of XI, get that value
of Yi can take another value of

XX equals 2, Y will equal.

2 * 2  3, which is
plus one.

Another value of XX equals
0 Y equals 2 times by 0 

3 will 2 * 0 is 0 and that
gives me minus three.

So for every value of XI can
generate a value of Y.

I can plot these as point
so I can plot this as the

.1  1 and I can plot this
one on a graph as the .21

and this one on a graph as
the point nort minus three.

So let's just set that up.

Pair of axes.

Let's mark the values of X that
we've been having a look at.

So that was X on there and why

there? And let's put on the
values of why that we got.

When X was zero hour value
of why was minus three?

So that's. There.

When X was one hour value was

minus one. And when X was
two hour value was one.

Those three points lie on
a straight line.

Y equals 2X minus three, and
that's another reason for

calling this a linear equation.
It gives us a straight line.

OK.

You've got that one. Y equals 2X
minus three. Supposing we take a

second one 3X plus two Y equals
8, a second linear equation, and

supposing we say these two.

Are true at the same time.

What does that mean?

Well, we can plot this as a
straight line. Again, it's a

linear equation, so it's going
to give us a straight line. Now

I don't want to have to workout
lots of points for this, So what

I'm going to do is just sketch
it in quickly on the graph. I'm

going to say when X is 0.

And cover up the Exterm 2 Y is
equal to 8, so why must be

equal to four which is going
to be up there somewhere?

And when? Why is 03 X is equal
to 8 and so X is 8 / 3

which gives us 2 and 2/3. So
somewhere about their two 2/3

and we know it's a straight
line, so we can get that by

joining up there.

This is the equation 3X plus two
Y equals 8. So what does it mean

for these two to be true at the
same time? Well, it must mean

it's this point here.

Where the two lines cross. So
when we solve a pair of

simultaneous equations, what
we're actually looking for is

the intersection of two
straight lines.

Of course, it could happen that
we have one line like that.

And apparel line.

They would never meet.

And one of the examples that
we're going to be looking at

later will show what happens
in terms of the arithmetic

when we have this particular
case. But for now, let's go

back and think about these
two. How can we handle these

two algebraically so that we
don't have to draw graphs? We

don't have to rely on
sketching, we can calculate

which is so much easier in
most cases that actually

drawing a graph.

So let's take these two
equations.

And we're going to look at two
methods of solution, so I'm

going to look at method one.
Now, let's begin with the

original equation that we had
two X minus. Y is equal to three

and then the one that we put
with it 3X plus two Y equals 8.

Our first method of solution,
well, one of the things to do is

to do what we did in the very
first case with this and

Rearrange one of these
equations. It doesn't matter

which one, but we'll take this

one. So that we get Y

equals. And we know what the
result for that one is. It's Y

equals 2X minus three.

So that's equation one.

That's equation. Two, so this
is now, let's call it equation 3

and we got it by rearranging.

1.

What we're going to do with this
is if these two have to be true

at the same time, then this
relationship must be true in

this equation, so we can
substitute it in, so let's.

Substitute 3

until two so we
have 3X plus.

Two times Y. But why is
2X minus three that's equal to

8? And you can see that what
we've done is we've reduced.

This. To this equation giving us
a single equation in one

unknown, which is a simple
linear equation and we can solve

it. Multiply out the brackets 3X
plus 4X minus 6 equals 8.

Gather the excess together. 7X
minus 6 equals 8.

At the six to each side, Seven X
equals 14, and so X must be 2.

That's only given as one value.

We need a value of Y, but up
here we've got an expression

which says Y equals and if we
take the value of X that we've

got and substituted in.

Therefore, why
will be equal to 2

* 2  3 gives US1?

And so we've got a solution X
equals 2, Y equals 1.

Are we sure it's right?

Well, we used this equation
which came from equation one to

generate the value of Y.

So if we take the values of X&Y
and put them back into here,

they should work, should give us
the right answer. So let's try

that. X is 2.

3 times by two is 6 plus, Y is
1 two times by one is 2 six and

two gives us eight. Yes, this
works. This is a solution of

that equation and of that one.
So this is our answer to the

pair of simultaneous equations.

Let's have a look at another one
using this particular method.

The example we're going to use
is going to be said.

Open X. +2
Y equals 47 and

five X minus four
Y equals 1.

Now. We need to make a choice.
We need to choose one of these

two equations. And Rearrange it
so that it says Y equals or if

we want X equals.

The choice is entirely ours and
we have to make the choice based

upon what we feel will be the
simplest and looking at a pair

of equations like this often
difficult to know which is the

simplest. Well, let's pick at
random. Let's choose this one

and let's Rearrange Equation
too. So we'll start by getting X

equals this time. So we say 5X
is equal to 1 and I'm going to

add 4 Y to each side plus 4Y.

Now I'm going to divide
throughout by the five so that I

have. X on its own. Now I've got
to divide everything by 5.

Everything so I had to put
that line there to show that

I'm dividing the one and the
four Y. So this is a fraction.

I'm sure you can tell this is
not going to be as easy as the

previous question was. In
fact, it's going to be quite

difficult because I have to
take this now and because it

came from equation too.

I'm going to have to take it and
substitute it back into equation

one, and this isn't looking very
pretty, so let's give it a try

sub. 3.

Until one. So I
have 7X but X is this

lump of algebra here 1 +
4 Y all over 5.

+2 Y equals

47. I can see this is
becoming quite horrific.

Multiply throughout by 5 why?
Because we're dividing by 5. We

want to get rid of the fraction.
The way to do that is to

multiply everything by 5 and it
has to be everything. So if we

multiply that by 5 because we're
dividing by 5, it's as though we

actually do nothing to the 1 + 4
Y. That leaves a 7 * 1 + 4 Y.

We need five times that that's
ten Y and we have to have five

times that remember, an
equation is a balance. What

you do to one side of the
balance you have to do to the

other. If you don't, it's
unbalanced. So we're

multiplying everything by 5.
So 5 * 47 five 735, five falls

of 22135 altogether.

Now we need to multiply out the
brackets 7 + 28 Y plus 10

Y equals 235. So we take this

equation. Write it down again so
that we can see it clearly.

Now we can gather these two
together gives us 38Y.

And we can take Seven away
from each side, which will

give us 228.

Exactly big numbers coming in
here 228 / 38 'cause we're

looking for the number which
when we multiplied by 38 will

give us 228 and that's going to

be 6. So we've established Y is
equal to 6.

Having done that, we can take it
and we can substitute it back

into the equation that we first
had for X. So remember that for

that we had X was equal to.

And what we had for that was 1
+ 4 Y all over five. We

substitute in the six, so we
have 1 + 24 or over 5 and

quickly we can see that's 25 /
5. So we have X equals 5. So

again we've got our pair of
values. Our answer to the pair

of simultaneous equations. We
haven't checked it though.

Now remember that this came from
the second equation, so really

to check it we've got to go back
to the very first equation that

we had written down that one. If
you remember was Seven X +2, Y

is equal to 47.

So let's just check 7 *
5. That gives us 35 +

2 * 6.

That gives us 12, so we 35 + 12
equals 47. And yes, that is what

we wanted, so we now know that
this is correct, but I just stop

and think about it.

We got all those fractions to
work with. We got this lump of

algebra to carry around with us.
Is there not an easier way of

doing these? Yes there is. It's
useful to have seen the method

that we have got simply because
we will need it again when we

look at the second video of
simultaneous equations, but.

That is a simple way of handling
these, so let's go on now and

have a look at method 2.

Now this method is sometimes
called elimination and we can

see why it gets that name and
this is the method that you

really do need to practice and
become accustomed to. So let's

start with the same equations
that we had last time.

And see.

How it works and how much easier
it actually is?

OK method of elimination. What
do we do?

What we do is we seek
to make the

coefficients in front
of the wise.

Or in front of the axes.

The same.

Once we've gotten the same,
then we can either add the

two equations together or
subtract them according to

the signs that are there.

By doing that, we will get rid
of that particular unknown, the

one that we chose.

To make the coefficients
numerically the same.

So. This one what would we do?

Well, if we look at this and

this. Here we have two Y and
here we have minus four Y.

So if I were to double that, I'd
have four. Why there? And it's

plus four Y Ana minus four Y
there, and that seems are pretty

good thing to do, because then
they're both for Y. One of them

is plus and one of them is
minus. And if I add them

together they will disappear. So
let me just number the equations

one and two.

And then I can keep a record of
what I'm doing. So I'm going to

multiply the first equation by
two and that's going to lead

Maine to a new equation 3.

So let's do that.

2 * 7 X is 14X
plus two times, that is 4

Y equals 2 times that, and
2 * 47 is 94.

Now equation two. I'm leaving as
it is not going to touch it.

Now I've got two equations.

This is plus four Y and this
is minus four Y. So if I

add the two equations together,
what happens? I get 14X plus 5X.

That's 19 X.

No whys. 'cause I've plus four Y
add it to minus four Y know wise

at all equals.

95

and so X is 95 over 19, which
gives me 5 which if you

remember, is the answer we have
to the last question.

Now we need to take this and
substitute it back. Doesn't

matter which equation we choose
to substitute it back into.

Let's take this one.

X is 5, so five times
by 5  4 Y equals

1.

And so we have 25 
4 Y equals 1.

Take the four way over
to that side by adding

four Y to each side.

So that will give us 25 is equal
to four Y plus one. Take the one

away from its side, 24 is 4 Y
and so why is equal to 6 and

we've got exactly the same
answer as we had before. And

let's just look.

How much simpler that is? How
much quicker that answer came

out. One thing to notice. Well,
two things in actual fact. First

of all, I try to keep the equal
signs underneath each other.

This is not only makes it look
neat, it enables you to see what

it is you're doing.

Keep the equations together so
the setting out of this work

actually helps you to be able to

check it. Second thing to notice
is down this side. I've kept a

record of exactly what I've done
multiplying the equation by two,

adding the two equations
together. That's very helpful

when you want to check your
work. What did I do? How did I

actually work this out? By
having this record down the

side, you don't have to work it
out again. You can see exactly

what it is that you did. Now
let's take a third example and

again. Will solve it by means of
the method of elimination. Just

so we've got a second example of
that method to look at three X

+7 Y is 27 and 5X.

+2 Y is 60.

OK, we've got a choice to make.
We can make either the

coefficients in front of the axe
numerically the same, or the

coefficients in front of the
wise. Well, in order to do that,

I'd have to multiply the Y.
Certainly have to multiply this

equation by two to give me 14
there and this one by 7:00 to

give me 14 there.

How do I make that choice? Well?

Fairly clearly 2 times by 7 is
14, so 1 by 1, one by the other.

But I don't really like
multiplying by 7 difficult

number. I prefer to multiply by
three and five, so my choices

actually governed by how well I
think I can handle the

arithmetic. So let's multiply
this one by 5 and this one by

three will give us 15X an 15X
number. The equations one.

2. And I'll take equation one
and I will multiply it by 5 and

that will give me a new equation

3. So multiplying
it by 5:15

X plus 35
Y is equal

to 135. And then
equation two, I will multiply by

three and that will give me a
new equation for.

Oh, here we go. Multiplying this
by three 15X.

Plus six Y is
equal to 48.

These are now both 15X and
they're both plus 15X.

So if I take this equation away
from that equation, I'll have

15X minus 15X no X is at all.
I live eliminative, the X, I'll

just have the Wise left, so
let's do that equation 3 minus

equation 4. 15X takeaway
15X no axis 35 Y

takeaway six Y that gives
us 29 Y.

And then 135 takeaway 48?
And that's going to give us

A7 their 87 altogether. And
so why is 87 over 29,

which gives us 3?

Having got that, I need to know
the value of X so I can take

Y equals 3 and substituted back
let's say into equation one. So

I have 3X plus Seven times Y 7
threes are 21 is equal to 27 and

so 3X is 6 taking 21 away from
each side and access 2.

Check this in here 5 twos are
ten 2306 ten and six gives me 16

which is what I want so I know
that this is my answer. My

solution to this pair of
simultaneous equations and again

look how straightforward that
is. Much, much easier than the

first method that we saw. Also
think about using letters as

well. If we've got letters to
use instead of coefficients the

numbers here. So we might have a
X plus BY. Again, this is a much

better method to use. Again,
notice the setting down keeping

it compact, keeping the equal
signs under each other and

keeping a record of what we've
done. So do something comes out

wrong, we can check it, see what
we are doing.

Now all the examples that we've
looked at so far of all had

whole number coefficients. They
might have been, plus they might

be minus, but they've been whole
number, and everything that

we've looked at as being in this
sort of form XY number XY

number. Well, not all equations
come like that, so let's just

have a look at a couple of

examples that. Don't look like
the ones we've just done.

First of all, let's have this
One X equals 3 Y.

And X over 3 minus Y equals
34 pair of simultaneous

equations. Linear simultaneous
equations again 'cause they both

got just X&Y in an numbers,
nothing else, no X squared's now

ex wise etc.

We need to get them into a form
that we can use and that would

be nice to have XY number. So
let's do that with this One X

equals 3 Y, so will have X minus
three Y equals 0.

This one got a fraction in it.
Fractions we don't like, can't

handle fractions. Let's get rid
of the three by multiplying

everything in this equation by
three. So will do that three

times X over 3 just leaves us

with X. Three times the Y
minus three Y equals 3 times.

This going to be 102.

Problem.

These two bits here are exactly

the same. But these two
bits are different.

What's going to happen?

Well, clearly if we subtract
these two equations one from

the other, there won't be
anything left this side when

we've done the subtraction X
from X, no access.

Minus 3 Y takeaway minus three
Y know why is left, and yet

we're going to have 0  102
equals minus 102 at this side.

In other words, we're gonna
end up with that.

Which is a wee bit strange.

What's the problem? What's the
difficulty? Remember right back

at the beginning when we drew a
couple of graphs?

In the first case we had two
lines that actually crossed, but

in the second case I drew 2
lines that were parallel.

And that's exactly what we've
got here. We have got 2 lines

that are parallel because
they've got this same form. They

are parallel lines so they don't
meet. And what this is telling

us is there in fact is no
solution to this pair of

equations because they come from
2 parallel lines that do not

meet no solution. There isn't
one fixed point, so we would

write that down.

Simply say no solutions.

And it's important to keep
an eye out for that.

Check back, make sure the
arithmetic's correct yes, but do

remember that can happen.

Let's take just one more final
example, X over 5.

Minus Y over 4 equals 0.

3X plus 1/2 Y equals
70. Now for this one.

We've got fractions with
dominators five and four, and we

need to get rid of those. So we
need a common denominator.

With which we can multiply
everything in the equation and

those get rid of the five in the
fall. The obvious one to choose

is 20, because 20 is 5 times by
4. Let us write that down in

falls 20 times X over 5  20
times Y over 4 equals 0

be'cause. 20 * 0 is still 0.

Little bit of counseling 5
into 20 goes 4.

4 into 20 goes 5.

So we have 4X minus five
Y equals 0.

So that was our first equation
that was our second equation.

This one is now become our third
equation. So equation one has

gone to equation 3.

Let's look at equation two. Now
that we need to deal with it,

it's got a half way in it. So if

I multiply every. Anything by
two. This will become just why?

So we have 6X.

Plus Y equals 34 and so
equation two has become. Now

equation for. We want to
eliminate one of the variables

OK, which one well I'd have to
do quite a bit of multiplication

by 6:00 AM by 4. If it was, the
ex is that I wanted to get rid

of look, there's a minus five
here and one there, so to speak.

So if we multiply this one by
five, will get these two the

same. So let's do that 4X minus
five Y equals 0, and then times

in this by 5.

30X plus five Y equals and then
we do this by 5, five, 420, not

down and two to carry 5 threes
are 15 and the two is 17, so

that gives us 170 and now we can
just add these two together. So

equation three state as it was
equation for we multiplied by 5.

So that's gone to equation 5 and

now. Finally, we're going to
add together equations three

and five, and so we have 34
X equals 170 and wise have

been illuminated.

34 X is 170 and so
X is 170 / 34 and

that gives us 5.

We need to go back and
substituting to one of our two

equations. It's just
have a look which one?

Doesn't really matter, I
think. Actually choose to go

for that one. Why? because I
can see that five over 5 gives

me one, and that's a very
simple number. Might make the

arithmetic so much easier.

So we'll have X over 5 minus
Y. Over 4 equals 0. Take the

Five and substituted in.

5 over 5. That's just one, and
so I have one takeaway Y over 4

equals 0, so one must be equal
to Y over 4. If I multiply

everything by 4I end up with
four equals Y. So there's my

pair of answers X equals 5, Y
equals 4 and I really should

just check by looking at the
second equation now, remember.

2nd equation was 3X
plus 1/2 Y equals 17.

3X also, half Y

equals 17. So let's substitute
these in. X is 5, three X is.

Therefore AR15, three fives plus
1/2 of Y. But why is 4 so 1/2

of it is 2. That gives me 17,
which is what I want. Yes, this

is correct. Let's just recap for
a moment. Apparel simultaneous

equations. They represent two
straight lines in effect when we

solve them together, we are
looking for the point where the

two straight lines intersect.

The method of elimination is
much, much better to use than

the first method that we saw.

Remember also in the way that
we've set this one out. Keep a

record of what it is that you

do. Set you workout so that the
equal signs come under each

other and so that at a glance
you can look at what you've

done. Check your working.

Finally, remember the answer
that you get can always be

checked by substituting the pair
of values into the equations

that you began with. That means
strictly you should never get

one of these wrong. However,
mistakes do happen.