## www.mathcentre.ac.uk/.../Simultaneous%20Linear%20Equations%20Part%201.mp4

• 0:02 - 0:06
The purpose of this video is to
look at the solution of
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elementary simultaneous
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equations. Before we do that,
let's just have a look at a
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relatively straightforward
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single equation. Equation we're
going to look at 2X minus
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Y equals 3.
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This is a linear equation. It's
a linear equation because there
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are no terms in it that are X
squared, Y squared or X times by
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Y or indeed ex cubes. The only
terms we've got a terms in X
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terms in Y and some numbers.
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So this represents a linear
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equation. We can rearrange it so
that it says why equal
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something, so let's just do
that. We can add Y to each side
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so that we get 2X.
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Equals 3 plus Yi. Did say add
why two each side and you might
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have wondered what happened
here. Well, if I've got minus Y
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and I add why to it, I end up
with no wise at all.
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Here we've got two X equals 3
plus Y, so let's take the three
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away from each side. 2X minus
three equals Y.
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So there I've got a nice
expression for why. If I take
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any value of X. Let's say I take
X equals 1, then why will be
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equal to two times by 1 - 3,
which gives us minus one. So for
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this value of XI, get that value
of Yi can take another value of
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XX equals 2, Y will equal.
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2 * 2 - 3, which is
plus one.
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Another value of XX equals
0 Y equals 2 times by 0 -
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3 will 2 * 0 is 0 and that
gives me minus three.
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So for every value of XI can
generate a value of Y.
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I can plot these as point
so I can plot this as the
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.1 - 1 and I can plot this
one on a graph as the .21
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and this one on a graph as
the point nort minus three.
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So let's just set that up.
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Pair of axes.
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Let's mark the values of X that
we've been having a look at.
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So that was X on there and why
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there? And let's put on the
values of why that we got.
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When X was zero hour value
of why was minus three?
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So that's. There.
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When X was one hour value was
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minus one. And when X was
two hour value was one.
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Those three points lie on
a straight line.
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Y equals 2X minus three, and
that's another reason for
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calling this a linear equation.
It gives us a straight line.
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OK.
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You've got that one. Y equals 2X
minus three. Supposing we take a
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second one 3X plus two Y equals
8, a second linear equation, and
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supposing we say these two.
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Are true at the same time.
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What does that mean?
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Well, we can plot this as a
straight line. Again, it's a
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linear equation, so it's going
to give us a straight line. Now
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I don't want to have to workout
lots of points for this, So what
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I'm going to do is just sketch
it in quickly on the graph. I'm
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going to say when X is 0.
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And cover up the Exterm 2 Y is
equal to 8, so why must be
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equal to four which is going
to be up there somewhere?
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And when? Why is 03 X is equal
to 8 and so X is 8 / 3
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which gives us 2 and 2/3. So
somewhere about their two 2/3
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and we know it's a straight
line, so we can get that by
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joining up there.
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This is the equation 3X plus two
Y equals 8. So what does it mean
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for these two to be true at the
same time? Well, it must mean
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it's this point here.
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Where the two lines cross. So
when we solve a pair of
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simultaneous equations, what
we're actually looking for is
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the intersection of two
straight lines.
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Of course, it could happen that
we have one line like that.
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And apparel line.
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They would never meet.
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And one of the examples that
we're going to be looking at
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later will show what happens
in terms of the arithmetic
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when we have this particular
case. But for now, let's go
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back and think about these
two. How can we handle these
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two algebraically so that we
don't have to draw graphs? We
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don't have to rely on
sketching, we can calculate
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which is so much easier in
most cases that actually
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drawing a graph.
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So let's take these two
equations.
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And we're going to look at two
methods of solution, so I'm
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going to look at method one.
Now, let's begin with the
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original equation that we had
two X minus. Y is equal to three
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and then the one that we put
with it 3X plus two Y equals 8.
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Our first method of solution,
well, one of the things to do is
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to do what we did in the very
first case with this and
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Rearrange one of these
equations. It doesn't matter
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which one, but we'll take this
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one. So that we get Y
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equals. And we know what the
result for that one is. It's Y
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equals 2X minus three.
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So that's equation one.
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That's equation. Two, so this
is now, let's call it equation 3
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and we got it by rearranging.
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1.
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What we're going to do with this
is if these two have to be true
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at the same time, then this
relationship must be true in
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this equation, so we can
substitute it in, so let's.
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Substitute 3
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until two so we
have 3X plus.
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Two times Y. But why is
2X minus three that's equal to
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8? And you can see that what
we've done is we've reduced.
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This. To this equation giving us
a single equation in one
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unknown, which is a simple
linear equation and we can solve
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it. Multiply out the brackets 3X
plus 4X minus 6 equals 8.
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Gather the excess together. 7X
minus 6 equals 8.
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At the six to each side, Seven X
equals 14, and so X must be 2.
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That's only given as one value.
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We need a value of Y, but up
here we've got an expression
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which says Y equals and if we
take the value of X that we've
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got and substituted in.
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Therefore, why
will be equal to 2
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* 2 - 3 gives US1?
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And so we've got a solution X
equals 2, Y equals 1.
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Are we sure it's right?
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Well, we used this equation
which came from equation one to
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generate the value of Y.
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So if we take the values of X&Y
and put them back into here,
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they should work, should give us
the right answer. So let's try
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that. X is 2.
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3 times by two is 6 plus, Y is
1 two times by one is 2 six and
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two gives us eight. Yes, this
works. This is a solution of
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that equation and of that one.
So this is our answer to the
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pair of simultaneous equations.
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Let's have a look at another one
using this particular method.
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The example we're going to use
is going to be said.
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Open X. +2
Y equals 47 and
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five X minus four
Y equals 1.
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Now. We need to make a choice.
We need to choose one of these
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two equations. And Rearrange it
so that it says Y equals or if
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we want X equals.
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The choice is entirely ours and
we have to make the choice based
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upon what we feel will be the
simplest and looking at a pair
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of equations like this often
difficult to know which is the
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simplest. Well, let's pick at
random. Let's choose this one
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and let's Rearrange Equation
too. So we'll start by getting X
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equals this time. So we say 5X
is equal to 1 and I'm going to
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add 4 Y to each side plus 4Y.
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Now I'm going to divide
throughout by the five so that I
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have. X on its own. Now I've got
to divide everything by 5.
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Everything so I had to put
that line there to show that
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I'm dividing the one and the
four Y. So this is a fraction.
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I'm sure you can tell this is
not going to be as easy as the
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previous question was. In
fact, it's going to be quite
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difficult because I have to
take this now and because it
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came from equation too.
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I'm going to have to take it and
substitute it back into equation
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one, and this isn't looking very
pretty, so let's give it a try
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sub. 3.
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Until one. So I
have 7X but X is this
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lump of algebra here 1 +
4 Y all over 5.
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+2 Y equals
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47. I can see this is
becoming quite horrific.
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Multiply throughout by 5 why?
Because we're dividing by 5. We
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want to get rid of the fraction.
The way to do that is to
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multiply everything by 5 and it
has to be everything. So if we
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multiply that by 5 because we're
dividing by 5, it's as though we
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actually do nothing to the 1 + 4
Y. That leaves a 7 * 1 + 4 Y.
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We need five times that that's
ten Y and we have to have five
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times that remember, an
equation is a balance. What
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you do to one side of the
balance you have to do to the
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other. If you don't, it's
unbalanced. So we're
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multiplying everything by 5.
So 5 * 47 five 735, five falls
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of 22135 altogether.
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Now we need to multiply out the
brackets 7 + 28 Y plus 10
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Y equals 235. So we take this
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equation. Write it down again so
that we can see it clearly.
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Now we can gather these two
together gives us 38Y.
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And we can take Seven away
from each side, which will
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give us 228.
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Exactly big numbers coming in
here 228 / 38 'cause we're
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looking for the number which
when we multiplied by 38 will
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give us 228 and that's going to
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be 6. So we've established Y is
equal to 6.
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Having done that, we can take it
and we can substitute it back
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into the equation that we first
had for X. So remember that for
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that we had X was equal to.
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And what we had for that was 1
+ 4 Y all over five. We
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substitute in the six, so we
have 1 + 24 or over 5 and
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quickly we can see that's 25 /
5. So we have X equals 5. So
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again we've got our pair of
values. Our answer to the pair
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of simultaneous equations. We
haven't checked it though.
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Now remember that this came from
the second equation, so really
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to check it we've got to go back
to the very first equation that
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we had written down that one. If
you remember was Seven X +2, Y
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is equal to 47.
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So let's just check 7 *
5. That gives us 35 +
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2 * 6.
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That gives us 12, so we 35 + 12
equals 47. And yes, that is what
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we wanted, so we now know that
this is correct, but I just stop
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and think about it.
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We got all those fractions to
work with. We got this lump of
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algebra to carry around with us.
Is there not an easier way of
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doing these? Yes there is. It's
useful to have seen the method
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that we have got simply because
we will need it again when we
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look at the second video of
simultaneous equations, but.
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That is a simple way of handling
these, so let's go on now and
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have a look at method 2.
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Now this method is sometimes
called elimination and we can
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see why it gets that name and
this is the method that you
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really do need to practice and
become accustomed to. So let's
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that we had last time.
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And see.
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How it works and how much easier
it actually is?
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OK method of elimination. What
do we do?
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What we do is we seek
to make the
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coefficients in front
of the wise.
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Or in front of the axes.
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The same.
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Once we've gotten the same,
then we can either add the
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two equations together or
subtract them according to
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the signs that are there.
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By doing that, we will get rid
of that particular unknown, the
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one that we chose.
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To make the coefficients
numerically the same.
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So. This one what would we do?
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Well, if we look at this and
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this. Here we have two Y and
here we have minus four Y.
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So if I were to double that, I'd
have four. Why there? And it's
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plus four Y Ana minus four Y
there, and that seems are pretty
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good thing to do, because then
they're both for Y. One of them
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is plus and one of them is
minus. And if I add them
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together they will disappear. So
let me just number the equations
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one and two.
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And then I can keep a record of
what I'm doing. So I'm going to
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multiply the first equation by
two and that's going to lead
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Maine to a new equation 3.
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So let's do that.
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2 * 7 X is 14X
plus two times, that is 4
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Y equals 2 times that, and
2 * 47 is 94.
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Now equation two. I'm leaving as
it is not going to touch it.
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Now I've got two equations.
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This is plus four Y and this
is minus four Y. So if I
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add the two equations together,
what happens? I get 14X plus 5X.
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That's 19 X.
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No whys. 'cause I've plus four Y
add it to minus four Y know wise
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at all equals.
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95
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and so X is 95 over 19, which
gives me 5 which if you
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remember, is the answer we have
to the last question.
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Now we need to take this and
substitute it back. Doesn't
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matter which equation we choose
to substitute it back into.
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Let's take this one.
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X is 5, so five times
by 5 - 4 Y equals
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1.
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And so we have 25 -
4 Y equals 1.
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Take the four way over
to that side by adding
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four Y to each side.
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So that will give us 25 is equal
to four Y plus one. Take the one
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away from its side, 24 is 4 Y
and so why is equal to 6 and
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we've got exactly the same
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let's just look.
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How much simpler that is? How
much quicker that answer came
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out. One thing to notice. Well,
two things in actual fact. First
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of all, I try to keep the equal
signs underneath each other.
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This is not only makes it look
neat, it enables you to see what
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it is you're doing.
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Keep the equations together so
the setting out of this work
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actually helps you to be able to
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check it. Second thing to notice
is down this side. I've kept a
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record of exactly what I've done
multiplying the equation by two,
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adding the two equations
together. That's very helpful
• 20:40 - 20:45
when you want to check your
work. What did I do? How did I
• 20:45 - 20:49
actually work this out? By
having this record down the
• 20:49 - 20:54
side, you don't have to work it
out again. You can see exactly
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what it is that you did. Now
let's take a third example and
• 20:59 - 21:03
again. Will solve it by means of
the method of elimination. Just
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so we've got a second example of
that method to look at three X
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+7 Y is 27 and 5X.
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+2 Y is 60.
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OK, we've got a choice to make.
We can make either the
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coefficients in front of the axe
numerically the same, or the
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coefficients in front of the
wise. Well, in order to do that,
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I'd have to multiply the Y.
Certainly have to multiply this
• 21:31 - 21:35
equation by two to give me 14
there and this one by 7:00 to
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give me 14 there.
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How do I make that choice? Well?
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Fairly clearly 2 times by 7 is
14, so 1 by 1, one by the other.
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But I don't really like
multiplying by 7 difficult
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number. I prefer to multiply by
three and five, so my choices
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actually governed by how well I
think I can handle the
• 21:58 - 22:03
arithmetic. So let's multiply
this one by 5 and this one by
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three will give us 15X an 15X
number. The equations one.
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2. And I'll take equation one
and I will multiply it by 5 and
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that will give me a new equation
• 22:18 - 22:24
3. So multiplying
it by 5:15
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X plus 35
Y is equal
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to 135. And then
equation two, I will multiply by
• 22:36 - 22:41
three and that will give me a
new equation for.
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Oh, here we go. Multiplying this
by three 15X.
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Plus six Y is
equal to 48.
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These are now both 15X and
they're both plus 15X.
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So if I take this equation away
from that equation, I'll have
• 23:06 - 23:13
15X minus 15X no X is at all.
I live eliminative, the X, I'll
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just have the Wise left, so
let's do that equation 3 minus
• 23:19 - 23:25
equation 4. 15X takeaway
15X no axis 35 Y
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takeaway six Y that gives
us 29 Y.
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And then 135 takeaway 48?
And that's going to give us
• 23:39 - 23:45
A7 their 87 altogether. And
so why is 87 over 29,
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which gives us 3?
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Having got that, I need to know
the value of X so I can take
• 23:55 - 24:01
Y equals 3 and substituted back
let's say into equation one. So
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I have 3X plus Seven times Y 7
threes are 21 is equal to 27 and
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so 3X is 6 taking 21 away from
each side and access 2.
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Check this in here 5 twos are
ten 2306 ten and six gives me 16
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which is what I want so I know
that this is my answer. My
• 24:28 - 24:31
solution to this pair of
simultaneous equations and again
• 24:31 - 24:36
look how straightforward that
is. Much, much easier than the
• 24:36 - 24:40
first method that we saw. Also
think about using letters as
• 24:40 - 24:45
well. If we've got letters to
use instead of coefficients the
• 24:45 - 24:50
numbers here. So we might have a
X plus BY. Again, this is a much
• 24:50 - 24:53
better method to use. Again,
notice the setting down keeping
• 24:53 - 24:57
it compact, keeping the equal
signs under each other and
• 24:57 - 25:01
keeping a record of what we've
done. So do something comes out
• 25:01 - 25:04
wrong, we can check it, see what
we are doing.
• 25:05 - 25:09
Now all the examples that we've
looked at so far of all had
• 25:09 - 25:13
whole number coefficients. They
might have been, plus they might
• 25:13 - 25:16
be minus, but they've been whole
number, and everything that
• 25:16 - 25:20
we've looked at as being in this
sort of form XY number XY
• 25:20 - 25:23
number. Well, not all equations
come like that, so let's just
• 25:23 - 25:26
have a look at a couple of
• 25:26 - 25:29
examples that. Don't look like
the ones we've just done.
• 25:31 - 25:36
First of all, let's have this
One X equals 3 Y.
• 25:37 - 25:43
And X over 3 minus Y equals
34 pair of simultaneous
• 25:43 - 25:47
equations. Linear simultaneous
equations again 'cause they both
• 25:47 - 25:54
got just X&Y in an numbers,
nothing else, no X squared's now
• 25:54 - 25:55
ex wise etc.
• 25:56 - 26:02
We need to get them into a form
that we can use and that would
• 26:02 - 26:08
be nice to have XY number. So
let's do that with this One X
• 26:08 - 26:12
equals 3 Y, so will have X minus
three Y equals 0.
• 26:14 - 26:19
This one got a fraction in it.
Fractions we don't like, can't
• 26:19 - 26:23
handle fractions. Let's get rid
of the three by multiplying
• 26:23 - 26:27
everything in this equation by
three. So will do that three
• 26:27 - 26:30
times X over 3 just leaves us
• 26:30 - 26:37
with X. Three times the Y
minus three Y equals 3 times.
• 26:37 - 26:40
This going to be 102.
• 26:42 - 26:43
Problem.
• 26:45 - 26:49
These two bits here are exactly
• 26:49 - 26:52
the same. But these two
bits are different.
• 26:54 - 26:56
What's going to happen?
• 26:57 - 27:00
Well, clearly if we subtract
these two equations one from
• 27:00 - 27:04
the other, there won't be
anything left this side when
• 27:04 - 27:07
we've done the subtraction X
from X, no access.
• 27:08 - 27:12
Minus 3 Y takeaway minus three
Y know why is left, and yet
• 27:12 - 27:15
we're going to have 0 - 102
equals minus 102 at this side.
• 27:15 - 27:18
In other words, we're gonna
end up with that.
• 27:21 - 27:23
Which is a wee bit strange.
• 27:24 - 27:27
What's the problem? What's the
difficulty? Remember right back
• 27:27 - 27:31
at the beginning when we drew a
couple of graphs?
• 27:32 - 27:37
In the first case we had two
lines that actually crossed, but
• 27:37 - 27:42
in the second case I drew 2
lines that were parallel.
• 27:43 - 27:48
And that's exactly what we've
got here. We have got 2 lines
• 27:48 - 27:52
that are parallel because
they've got this same form. They
• 27:52 - 27:57
are parallel lines so they don't
meet. And what this is telling
• 27:57 - 28:02
us is there in fact is no
solution to this pair of
• 28:02 - 28:07
equations because they come from
2 parallel lines that do not
• 28:07 - 28:11
meet no solution. There isn't
one fixed point, so we would
• 28:11 - 28:12
write that down.
• 28:13 - 28:15
Simply say no solutions.
• 28:18 - 28:20
And it's important to keep
an eye out for that.
• 28:21 - 28:25
Check back, make sure the
arithmetic's correct yes, but do
• 28:25 - 28:26
remember that can happen.
• 28:27 - 28:34
Let's take just one more final
example, X over 5.
• 28:35 - 28:39
Minus Y over 4 equals 0.
• 28:40 - 28:48
3X plus 1/2 Y equals
70. Now for this one.
• 28:49 - 28:52
We've got fractions with
dominators five and four, and we
• 28:52 - 28:56
need to get rid of those. So we
need a common denominator.
• 28:57 - 29:01
With which we can multiply
everything in the equation and
• 29:01 - 29:07
those get rid of the five in the
fall. The obvious one to choose
• 29:07 - 29:13
is 20, because 20 is 5 times by
4. Let us write that down in
• 29:13 - 29:19
falls 20 times X over 5 - 20
times Y over 4 equals 0
• 29:19 - 29:22
be'cause. 20 * 0 is still 0.
• 29:23 - 29:29
Little bit of counseling 5
into 20 goes 4.
• 29:30 - 29:34
4 into 20 goes 5.
• 29:35 - 29:40
So we have 4X minus five
Y equals 0.
• 29:42 - 29:47
So that was our first equation
that was our second equation.
• 29:47 - 29:52
This one is now become our third
equation. So equation one has
• 29:52 - 29:54
gone to equation 3.
• 29:54 - 29:57
Let's look at equation two. Now
that we need to deal with it,
• 29:57 - 29:59
it's got a half way in it. So if
• 29:59 - 30:04
I multiply every. Anything by
two. This will become just why?
• 30:04 - 30:06
So we have 6X.
• 30:07 - 30:14
Plus Y equals 34 and so
equation two has become. Now
• 30:14 - 30:19
equation for. We want to
eliminate one of the variables
• 30:19 - 30:24
OK, which one well I'd have to
do quite a bit of multiplication
• 30:24 - 30:30
by 6:00 AM by 4. If it was, the
ex is that I wanted to get rid
• 30:30 - 30:35
of look, there's a minus five
here and one there, so to speak.
• 30:35 - 30:40
So if we multiply this one by
five, will get these two the
• 30:40 - 30:45
same. So let's do that 4X minus
five Y equals 0, and then times
• 30:45 - 30:47
in this by 5.
• 30:47 - 30:54
30X plus five Y equals and then
we do this by 5, five, 420, not
• 30:54 - 31:01
down and two to carry 5 threes
are 15 and the two is 17, so
• 31:01 - 31:08
that gives us 170 and now we can
just add these two together. So
• 31:08 - 31:14
equation three state as it was
equation for we multiplied by 5.
• 31:14 - 31:17
So that's gone to equation 5 and
• 31:17 - 31:22
now. Finally, we're going to
add together equations three
• 31:22 - 31:29
and five, and so we have 34
X equals 170 and wise have
• 31:29 - 31:30
been illuminated.
• 31:34 - 31:42
34 X is 170 and so
X is 170 / 34 and
• 31:42 - 31:44
that gives us 5.
• 31:45 - 31:50
We need to go back and
substituting to one of our two
• 31:50 - 31:54
equations. It's just
have a look which one?
• 31:55 - 31:58
Doesn't really matter, I
think. Actually choose to go
• 31:58 - 32:02
for that one. Why? because I
can see that five over 5 gives
• 32:02 - 32:06
me one, and that's a very
simple number. Might make the
• 32:06 - 32:07
arithmetic so much easier.
• 32:08 - 32:15
So we'll have X over 5 minus
Y. Over 4 equals 0. Take the
• 32:15 - 32:17
Five and substituted in.
• 32:22 - 32:28
5 over 5. That's just one, and
so I have one takeaway Y over 4
• 32:28 - 32:34
equals 0, so one must be equal
to Y over 4. If I multiply
• 32:34 - 32:39
everything by 4I end up with
four equals Y. So there's my
• 32:39 - 32:44
pair of answers X equals 5, Y
equals 4 and I really should
• 32:44 - 32:49
just check by looking at the
second equation now, remember.
• 32:49 - 32:54
2nd equation was 3X
plus 1/2 Y equals 17.
• 32:56 - 32:59
3X also, half Y
• 32:59 - 33:06
equals 17. So let's substitute
these in. X is 5, three X is.
• 33:06 - 33:12
Therefore AR15, three fives plus
1/2 of Y. But why is 4 so 1/2
• 33:12 - 33:18
of it is 2. That gives me 17,
which is what I want. Yes, this
• 33:18 - 33:24
is correct. Let's just recap for
a moment. Apparel simultaneous
• 33:24 - 33:29
equations. They represent two
straight lines in effect when we
• 33:29 - 33:35
solve them together, we are
looking for the point where the
• 33:35 - 33:37
two straight lines intersect.
• 33:39 - 33:43
The method of elimination is
much, much better to use than
• 33:43 - 33:45
the first method that we saw.
• 33:46 - 33:50
Remember also in the way that
we've set this one out. Keep a
• 33:50 - 33:52
record of what it is that you
• 33:52 - 33:57
do. Set you workout so that the
equal signs come under each
• 33:57 - 34:00
other and so that at a glance
you can look at what you've
• 34:00 - 34:01
done. Check your working.
• 34:02 - 34:06
Finally, remember the answer
that you get can always be
• 34:06 - 34:11
checked by substituting the pair
of values into the equations
• 34:11 - 34:16
that you began with. That means
strictly you should never get
• 34:16 - 34:19
one of these wrong. However,
mistakes do happen.
Title:
www.mathcentre.ac.uk/.../Simultaneous%20Linear%20Equations%20Part%201.mp4
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