www.mathcentre.ac.uk/.../Simple%20Linear%20Equations.mp4

In this session we're going to
be having a look at simple equations

in one variable and
the equations will be linear.

That means that there are no x-squared
terms

and no x-cube terms, just
x's and numbers. So let's have a

look at the first one.
3x plus 15 equals x plus 25.

An important thing to remember
about any equation is this equal

sign represents a balance. What
that equal sign says is that

what's on the left hand side is
exactly the same as what's on

the right hand side.

If we do anything to one side of
the equation, we have to do it

to the other side. If we don't,
the balance is disturbed.

If we can keep that in mind,

the whatever operations we
perform on either side of the

equation, so long as it's done
in exactly the same way on other side,

we should be alright.

Our first step in solving any
equation is to attempt to gather

all the x terms together and all
these free floating numbers together.

To begin with we've
got 3x on the left and and x on the right.

If we take an x away from both sides,
we take one x off the left,

and one x off the right.
That will give us 2x plus 15 equals 25.

We need to get the numbers
together. These free numbers.

We can see that if we take 15 away
from the left, and 15 away from the right

then we will have no numbers remaining
on the left, just a 2x

and taking 15 away on the right,
that gives us 10.

So, x must be equal to 5.

That was relatively
straightforward.

Lots of plus signs, no minus signs,
which we know can be complicated,

also no brackets. So let's
introduce both of those.

We will take 2x plus 3
is equal to

6 minus bracket 2x minus three bracket.

Now this right hand
side needs a little bit of dealing with.

It needs getting
into shape so we have to remove

this bracket.
2x plus 3 equals 6,

now we take away 2x, thats minus 2x
and then we're taking away minus 3.

When we take away minus 3,
that makes it plus 3.

Before we go any further, we
need to tidy this right hand

side up a little bit more
2x plus 3 is equal to 9 minus 2x

and now we're in the same
position at this line as we were

when we started there. So we
need to get the xs together,

which we can best do by adding
2x to each side.

On the right, minus 2x plus 2x.
This side gives us no x.

And on the left it's going to
give us 4X.

4X plus 3 is equal to 9
because minus 2X plus 2X gives no x.

Now I can take 3
away from each side, 4X equals 6

and so x is 6 divided by 4
and we want to write that in its

lowest terms which is 3 over 2.
Perfectly acceptable answer.

Or one and a half. So either of these
are acceptable answers.

This one (6 over 4) isn't because it's not in its
lowest terms, it must be reduced

to its lowest terms. So either
of those two are OK as answers.

With all of these equations, we
should strictly check back by

taking our answer and putting it
into the first line of the

equation and seeing if we get
the right answer.

So with this five,
3 fives are 15, and 15 is 30.
5 plus 25 is also 30

The balance that we talked about at
the beginning is maintained.

If you take this number, 3 over 2,
or one and a half.

And substitute it back into here
we should find again that

it gives us the right answer,
that both sides give the same value.

They balance.
Solet's just do that.

Let us take one and a half.
So 2 times one and a half is 3,

plus three is 6, so we've got
six on the left side.

Here we have 6, takeaway,
now let's deal with this bracket,

just this bracket on its own, nothing else.
2x takeaway, three, that's nothing,

so we are just left with 6.
We calculated this side to be 6.

Six equals six,
so again, we've got that

balance, so we know that we've
got the right answer.

Let's carry on and have a look
at one or two questions with

more brackets and this
time numbers outside those

brackets as well. So in a sense,
what we're doing is we are

increasing the complexity of the
equation, but the simple

principles that we've got so far
are going to help us out because

no matter how complicated it
gets and this does look

complicated, the same ideas work
all the time.

We begin by multiplying out the
brackets and taking care,

in particular, with any minus signs
that come up so

8 times by x is 8x and eight times by
minus three is minus 24.

We've multiplied everything
inside that bracket by what's

outside, so we've 8 times by x
and we've 8 times by minus 3

Now we remove this
bracket were taking away 6,

that's minus six and we're
taking away minus 2x,

a minus minus gives us a
plus. So that's plus two X

equals, two times x here,
is 2x, 2 times by 2 is 4.

and now we have to multiply by
minus five, so we've got

minus five times by 5. That's minus 25.
And minus five times by minus x.

So that's plus 5x.

Each side needs tidying up. We
need to look at this side and

gather x terms and numbers together.
so with 8x plus 2x that's 10x.

Take away 24, takeaway six,
that's taking away 30 altogether.

2x and 5x gives us 7x.

Add on 4 takeaway 25. Now
that's the equivalent of taking

away 21. Let's get the xs together.
We can take 7x away from each side

so we would have 3x minus 30 there
equals just

minus 21 because we've taken that
7x away.

Add the 30 to each side,
because minus 30 + 30 gives us

nothing. And add it on over
here. So we get three x equals,

30 added to minus 21, just
the same as 30 takeaway 21,

the answer is 9.

And three times by something
to give us 9 means the

something, the x, must be 3.
Again we ought to be able

to take that 3, put it back
into this line and see that

in fact we've got the correct
answer. So let's just try.

3 minus 3 is zero.
8 times by 0 is still 0,

so we can forget about that term.

2 times by 3three is 6, so six
takeaway six is nothing, so

we've actually got nothing
there, zero, and nothing there, zero.

so there's nothing on the left side of the equation.

The right hand side should come out to
0 as well. Let's see that it does.

3 plus 2 is 5 and 2 fives are 10.
5 takeaway 3 is 2,

and 5 twos are 10.
So we've got 10 takeaway 10,

this gives us nothing again,
so we've got nothing on the right.

The equation balances with this value,

so this is our one and only solution.

We take a final
example of this kind with

x plus 1 times by
2x plus 1.

Equals x plus 3
times by 2x plus 3

- 14.

Now, I did say that these were
linear equations that there

would be no x-squared terms in them.

But when we start
to multiply out these

brackets, we are going to
get some x-squared terms.

Let's have a look, to show you what
actually does happen.

We do x times by two x.
That's two x-squared.

x times by one, that is plus x.

One times two x, that is
plus 2x.

1 times by 1

+ 1.
Equals...

x times by two x,
that's two x-squared.

x times by three is 3x..

3 times by 2x is 6x.

3 times by 3 is 9
and finally takeaway 14.

Now we need to tidy up
both sides here.

Two x-squared plus 3x
plus one equals

2 x-squared plus 9x,
(taking these two terms together)

Plus 9, minus 14.

And this needs a little bit more
tidying, but before we do that,

let's just have a look at what
we've got.

We've got 2 x-squared there, and
2 x-squared there. They are both

positive, so I can take two x-sqaured
away from both sides.

That means that the two
x-squared vanishes from both

sides, so let's do that. Take
two x-squared from each side,

and that leaves us 3x plus one,
is equal to 9x, and now we can

do this bit 9 takeaway 14,

Well, that's going to be
takeaway five. I still have the

five to subtract. Now we're back
to where we used to be getting

the xs together, getting the
numbers together. We can

subtract 3x from both sides, so
that gives me one equals 6x minus 5

We can add the five
to both sides so that

6 equals 6X, and so 1 is equal to x.

And again, we can check that
this works. We can substitute it back in.

1 plus 1 is 2,
two ones are two, and one is 3.

So effectively we've got 3 times
by two at this side, which gives 6.

Here 1 + 3 is 4
two times by one is 2, + 3 is 5

so we've got four times by 5.
So altogether there there's 20.

Takeaway 14 is again 6, so
again, this equation is balanced

exactly when we take x to be
equal to 1.

Now, those equations that we've
just looked at have really been

about whole numbers. The
coefficients have been whole

numbers. Everything's been in
terms of integers. What happens

when we start to get some
fractions in there? Some

rational numbers? How do we deal
with that? So again, we're going

to add in more complexity, but
again, the rules are the same.

What we do to one side of the
equation we must do to the other

side in order to preserve that

particular balance. So let's
introduce some fractions

along with some brackets.

4 bracket X plus 2, all over 5
is equal to

7 plus 5x over 13.

We've got some fractions here.

Numbers in the denominator 5 and
30. We want rid of those we want

to be able to work with whole
numbers with integers, so we

have to find a way of getting
rid of them. That means we

have to multiply everything
because what we do to one side,

we must do to the other.

The common denominator for five and
13 is 65, that's five times by 13.

So let's do that. Let's
multiply everything by 65 and

I'll write it down in full so
that we can see it happening.

we have 65 times 4, brackets x+2 over 5.
Then we have 65 times by 7.

Remember, I said we have to
multiply everything, so it's not

just these fraction bits, it's
any spare numbers that there are

around as well.

Plus 65 times by
5x over 30.

Now let's look at each term and
make it simpler. Tidy it up.

For a start, five will divide
into 65, so 5 into five goes one

and five into 65 goes 13 times.

And then four times by 13?
Well, that's 52, so we have

52 times by x +2 equals.

That's in a nice familiar form
where used to that sort of

former. We've arrived at it by
choosing to multiply everything

by this common denominator.
Let's tidy this side up.

7 times by 65. Well, that's
pretty tall order. Let's try it.

Seven 5s are 35. Seven 6s are 42 and
three is 45.

Here with 65 times 5 x over 13.
13 goes into 13 once,

and 13 goes into 65 five times.
So we five times by 5x is 25x.

You may say 'what happened to
these ones?'. Well if I divide by

one it stays unchanged so I
don't have to write them down.

And now we left with an
equation that were used to

handling. We've met these kind
before, so let's multiply out

the brackets, get the xs
together and solve the equation.

So we multiply out the brackets
52 times 2 is 104

equals 455 plus 25x.
Take the 25x away from each side,

gives me 27x there.

And no x is there.
Take the 104 away

from each side,
which gives me 351.

And now lots of big numbers,
really, that shouldn't be a

problem to us. 351 divided by 27 will
certainly go once

(there's one 27 in the 35 and eight
over and 27s into 8

go three times,
so answer is x equals 13 and we

should go back and check it
and make sure that it's right.

So let's do that.

13 + 2? Well, that's 15

15 divided 5 is 3 and
now times by 4, so that's 12.

Hang on to that number 12
at that side.

5 times 13 divided by 13.
So the answer is just five

plus the 7 is again 12.
The same as this side.

So the answer is correct.
It balances.

Let's practice that one again
and have a look at another example.

We take X plus 5 over 6

minus x plus one over 9

is equal to x plus three over 4.

We haven't got any brackets.

Does that make any
difference? The thing you have

to remember is that this line.

Not only acts as a division
sign, but it acts as a bracket.

It means that all of x plus 5
is divided by 6.

So it might be as well if we
kept that in mind and put

brackets around these terms so
that we're clear that

we've written down that these
are to be kept together and are

all divided by 6. These two are
to be kept together and all

divided by 9. And similarly here
they are all divided by 4.

Next step we need a common
denominator. We need a number

into which all of these will
divide exactly. Now we could

multiply them altogether and
we'd be certain.

But the arithmetic would be
horrendous. 6 times 9 times 4 is very big.

Can we find a smaller
number into which six, nine, and

four will all divide?

Well, a candidate for that is 36.
36 will divide by 6l

36 will divide by 9 and 36 will divide
by 4, so lets multiply

throughout by that number 36.
We have 36, because we've put the

brackets in we are quite clear
that we're multiplying

everything over that 6 by the 36.
Minus 36 times x plus one.

All over 9 equals 36 times by x,
plus three all over 4, so we

made it quite clear by using the
brackets what this 36 is

multiplying . 6 into six goes
once and six into 36 goes 6 times.

9 into nine goes once and 9 into 36
goes 4.

4 into 4 goes
once and four into 36 goes 9.

So now I have this bracket to
multiply by 6 this bracket to

multiply by 4 and this bracket
to multiply by 9. I don't have

to worry about the ones because
I'm dividing by them so they

leave everything unchanged. So
let's multiply out six times by

6X plus 30 (6 times by 5).
This is a minus four I'm

multiplied by, so I need to be a

bit careful. Minus four times x
is minus 4x,

Minus 4 times 1 is minus 4

Equals 9X
and 9 threes

are 27. So now
we need to simplify this side

6x takeaway 4x. That's just two
X30 takeaway, four is 26, and

that's equal to 9X plus 27.

Let me take 2X away from each
side, so I have 26 equals 7X

plus 27 and now I'll take the
Seven away from each side and

I'll have minus one is equal to
7X and so now I need to divide

both sides by 7 and so I get
minus 7th for my answer. Don't

worry that this is a fraction,
sometimes they workout like

that. Don't worry, that is the
negative number. Sometimes they

workout like that. Let's have a
look at another one 'cause this

is a process that you're going
to have to be able to do quite

complicated questions. So we'll
take 4 - 5 X.

All over 6.

Minus 1 - 2 X all over

3. Equals
13 over 42.

What are we going to do? First
of all, let's remind ourselves

that this line.

Not only means divide.

Divide 4 - 5 X by 6 but it means
divide all of 4 - 5 X by 6. So

let's put it in a bracket to
remind ourselves and let's do

the same there.

Now we need a common
denominator, six and three and

40. Two, well, six goes into 42
and three goes into 42 as well.

So let's choose 42 as our
denominator, an multiply

everything by 42. So will have
42 * 4 - 5 X or

over 6 - 42 * 1 -
2 X all over 3.

Equals 42 *
13 over 42.

Six goes into six once and six
goes into 42 Seven times.

Three goes into three once and
free goes into 4214 times.

42 goes into itself once and

again once. So now I need to
multiply out these brackets.

And simplify this side.

So 7 times by 4 gives us
28 Seven times. My minus five

gives us minus 35 X. Now here
we have a minus sign and the

14, so it's minus 14 times by
1 - 14 and then it's minus

14 times by minus 2X. So it's
plus 28X. Remember we've got to

take extra care when we've got
those minus signs.

One times by 13 is just
13. Now we need to tidy

this side up 28 takeaway 14
is just 1435 - 35 X

plus 28X. Or what's the
difference there? It is 7 so

it's minus Seven X equals 30.
Take the 14 away from each

side. We've minus Seven X equals

minus one. And divide both
sides by minus 7 - 1 divided

by minus Seven is just a 7th
again fractional answer, but

not to worry.

When we looked at these, now we
want to have a look at a type

of equation which occasionally
causes problems. This particular

equation or kind of equation
looks relatively

straightforward. Translate
numbers get rather more

difficult. It can cause
difficulties, so we got three

over 5 equals 6 over X

straightforward. But what do we
do? Let's think about it. First

of all, in terms of fractions,
this is a fraction 3/5, and it

is equal to another fraction

which is 6. Well family
obviously 3/5 is the same

fraction as 6/10 and so
therefore X has got to be equal

to 10. That's not going to
happen with every question. It's

not going to be as easy as that.

We're going to have to juggle
with the numbers, so how do we

do that? Well, again, we need a
common denominator. We need a

number that will be divisible by
5 and a number that will be

divisible by X. And the obvious

choice is 5X. So let us

multiply. Both sides by 5X.
So we 5X times by

3/5 equals 5X times by
6 over X and now

five goes into five once
and five goes into five

once there. X goes into X
once an X goes into X. Once

there, let's remember that we're
multiplying by these, so we have

one times X times three. That's
3X and divided by one, so it's

still three X equals 5 1 6,
which is just 30 and divided by

one, so it's still 30.

3X is equal to 30, so X must be
equal to 10, which is what we

had before. Is there another way
of looking at this equation?

Well, yes there is.

If two fractions are equal that
way up there also equal the

other way up.

This makes it easier still
because all that we need to do

now is multiply by the common
denominator and we can see what

that common denominator is. It's
quite clearly 6 because six

divides into six and three
divides into 6.

So we had five over three is
equal to X over 6, and we're

going to multiply by this common
denominator of six. So six goes

into six once on each occasion,
three goes into three once and

into six twice, so again X is
equal to 10 two times by 5,

one times by X.

Whichever way you use.

Doesn't matter. They should come
out the same. There's no reason

why they shouldn't, but you do
have to be careful. The number

work can be a bit tricky
sometimes. Let's have a look at

just a couple more.

Five over 3X
is equal to

25 over 27.

OK. What I think I'm going to
do with these is flip them over

3X over 5 is equal to 27 over
25. I can see straight away.

I've got a common denominator
here of 25. Five goes into 25

exactly and so does 25. So if I
do that multiplication 25 * 3 X

over 5 is equal to 25 * 27 over

25. 25 goes into itself
once on each occasion.

Five goes into itself once and
five goes into 25 five times. So

I have 15X5 times by three. X is
equal to 27, and so X is 27 over

15. Dividing both sides by 15
and there is here a common

factor between top and bottom of
three, which gives me 9 over 5,

so that's an acceptable answer
because it's in its lowest

forms. Or I could write it as

one. And four fifths, which is
also an acceptable answer.

Now, some of you may not like
what I did there when I flipped

it over, and we might want to
think, well, how would I do it

if I had to start from there. So
let's tackle that in another

way. So again we five over 3X is
equal to 25 over 27 this time.

We're not going to flip it over.
Let's look for a common

denominator here between these
two, so we want something that

3X will divide into exactly, and
something that 27 will divide

into exactly, well. Three will
divide into 27, so the 27, so to

speak ought to be a part of our
answer. What we need is an X,

because if we had 27 X, 3X would
divide into it 9 times and the

27 would just divide into it.

X times, so that's going to
be our common denominator. The

thing that we are going to
multiply both sides by 27 X.

So we can look at this and we
can see that X divides into X

one St X device into X. Once
there we can also see that three

divides into three and three
divides into 27 nine times and

over here 27 goes into 27 once
each time. So I have 9 * 1

* 5 and 95 S 45.

Divided by 1 * 1, which is one,
so it's still 45 equals 1 times

by 25. And times by the X
there 25 X. Now I need to divide

both sides by 25 so I have 45
over 25 equals X.

This is not in its lowest
terms. I can divide top on

bottom by 5, giving me 9 over
5 again, which again I can

write as one and four fifths.

Let's take one final example.

This time, let's look at some
fractions, but this time,

mysteriously, the X is already
on the top. That's really good

for us. All we need to do is
look at what's our common

denominator, well, 7.

And 49 what number divides
exactly by both of these and it

will be 49. So we just need
to do 49 times by 19 X

over 7 equals 49 * 57 over
4949 goes into 49 once each

time. And Seven goes into
49 Seven times.

And so I have 7 times by 19 X
equals 57 and you might say,

well, hang on a minute,
shouldn't you have multiplied

out that first? Well, I didn't
want to. Why didn't I want to?

Well, sometimes when you play
darts, your arithmetic improves

and triple 19 on a dartboard is
57. So 19 divides into 57 three

times, and I don't want to lose
that relationship. So I'm going

to divide each side by 19 so 7X.

Is equal to three, which means X
must be 3 over 7.

Playing darts does help with
arithmetic. We finished there

with simple linear equations.
The important thing in dealing

with these kinds of equations
and any kind of equation is to

remember that the equal sign is
a balance. What it tells you is

that what's on the left hand
side is exactly equal to what's

on the right hand side. So
whatever you do to one side, you

have to do to the other side,
and you must follow.

The rules of arithmetic
when you do it.