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.