My name is cody
12yo in year 7 primary school.
perth australia

can someone explain the story
to end up with the correct answer
to this one
http://cjoint.com/data/dFlxPGB5yb.htm

I do not know how the negative arrives
in the centre of the fraction in the answer.

Is this the best place to ask my queries?
thank you
cody

Sun, 1 Apr 2007 08:35:38 +0800 from Rod :
> 
> My name is cody
> 12yo in year 7 primary school.
> perth australia
> 
> can someone explain the story
> to end up with the correct answer
> to this one
> http://cjoint.com/data/dFlxPGB5yb.htm
> 
> I do not know how the negative arrives
> in the centre of the fraction in the answer.

You had -ax on the left-hand side. What must you divide by to get x? 
Do you divide by a or by -a?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

Thank you Mr Brown.
divide by -a

Is b-c over -a
the same as negative b-c over a ?

eg -(b-c over a)
this confuses me

thank you
cody





> You had -ax on the left-hand side. What must you divide by to get x?
> Do you divide by a or by -a?
>
> --
> Stan Brown, Oak Road Systems, Tompkins County, New York, USA
>                                   http://OakRoadSystems.com/

"Rod"  wrote in message 
news:460f28d3$1_1@news.iprimus.com.au...
>
> Thank you Mr Brown.
> divide by -a
>
> Is b-c over -a
> the same as negative b-c over a ?
>
> eg -(b-c over a)
> this confuses me

Yes, or at least it is if you include the ( )  ie:


(b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )

One way to avoid the problem of dealing with the negative signs is always to 
ensure that the terms in x are positve.

eg:

-ax = b-c

add ax to both sides:


0 = b-c+ax

-b both sides

-b = - c + ax

+ b both sides

c-b = ax

/a both sides

(c-b)/a  = x

--
73
Brian
www.g8osn.org.uk

Thank you Mr Reay
then I got the answer right.
here is the question and answer given
which does not show the brackets
http://cjoint.com/data/ebitEB4dMe.htm

It is "kumon" and I have a tutor but i have to line up
with some 50 kids to get answers.

i get most right but the negative place here
looks wrong

is the book answer wrong?

also you say c-b we have to put in alphabetic b-c



"Brian Reay" <brian.reay@(spamstopper)bigfoot.com> wrote in message
news:xRHPh.232$r4.15@newsfe1-gui.ntli.net...
>
> "Rod"  wrote in message
> news:460f28d3$1_1@news.iprimus.com.au...
> >
> > Thank you Mr Brown.
> > divide by -a
> >
> > Is b-c over -a
> > the same as negative b-c over a ?
> >
> > eg -(b-c over a)
> > this confuses me
>
> Yes, or at least it is if you include the ( )  ie:
>
>
> (b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )
>
> One way to avoid the problem of dealing with the negative signs is always
to
> ensure that the terms in x are positve.
>
> eg:
>
> -ax = b-c
>
> add ax to both sides:
>
>
> 0 = b-c+ax
>
> -b both sides
>
> -b = - c + ax
>
> + b both sides
>
> c-b = ax
>
> /a both sides
>
> (c-b)/a  = x
>
> --
> 73
> Brian
> www.g8osn.org.uk
>
>
>

you made a miistake
+c both sides

> > + b both sides

"Rod"  wrote in message 
news:460f5168$1_1@news.iprimus.com.au...
> you made a miistake
> +c both sides
>
>> > + b both sides

Well spotted- it is early on a Sunday morning here and I've not had a cuppa 
yet ;-)

Good luck with your studies.

--
73
Brian
www.g8osn.org.uk

> Well spotted- it is early on a Sunday morning here and I've not had a
cuppa
> yet ;-)
>
> Good luck with your studies.

Thank you Sir.

Sun, 1 Apr 2007 11:39:11 +0800 from Rod :
> 
> Thank you Mr Brown.
> divide by -a
> 
> Is b-c over -a
> the same as negative b-c over a ?

Is 8 over -4 the same as -8 over 4?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Is 8 over -4 the same as -8 over 4?

first i think no, but each answer is -2?
so must be yes.

the answer I got is shown at the top
the answer in the book is below
http://cjoint.com/data/ebp1xokLvv.htm
I mean are the two answers the same?
i am more worried about the position of the
negative sign than the answer maybe?

thank you Mr Brown

"Rod"  wrote in message 
news:460fbb55_1@news.iprimus.com.au...
 > i am more worried about the position of the
> negative sign than the answer maybe?


Sometimes you are simply looking at a presentation convention eg

-8/2  =  8/-2    (-4 in both cases)

However,  -8/2 is often considered "better" presentation wise.

Likewise:

-b + c =  c-b

But c-b is  often considered "better" presentation wise.

There are harder examples (eg "rationalising surds") but worry about those 
when you get to secondary school.

-- 
73
Brian, G8OSN
www.g8osn.org.uk

I've added the attribution you omitted. Please show proper 
attributions in your follow-ups.
http://oakroadsystems.com/genl/unice.htm#attrib

Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown :
> > Is 8 over -4 the same as -8 over 4?
> 
> first i think no, but each answer is -2?
> so must be yes.
> 
> the answer I got is shown at the top
> the answer in the book is below
> http://cjoint.com/data/ebp1xokLvv.htm
> I mean are the two answers the same?

This HTML business is getting pretty old. It's not hard to post 
algebra. You are asking whether
     (b-c) / (-a)          and        -(b-c) / a
are the same.

Again, I'll answer that question with a question: are
     (15-7) / (-4)         and        -(15-7) / 4
the same?

(Note: It's not always safe to answer questions like these with an 
example. But if you choose your numbers so that nothing is 0 or 1, 
and maybe even try a couple of examples, it can be a helpful guide.)

> i am more worried about the position of the
> negative sign than the answer maybe?

Meaning no disrespect, I think you need to review the basics of 
division and multiplication with signed numbers. When I see a student 
confused by different ways of presenting the same thing, sometimes 
it's just a momentary perplexity but often the student just doesn't 
understand the underlying principles.

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Sometimes you are simply looking at a presentation convention eg

Thank you, Mr Reay
I notice my mistakes also.
Cody
Australia

In article ,
Stan Brown   wrote:
>Is 8 over -4 the same as -8 over 4?

	It's an interesting question, isn't it?  I can imagine
sharing an #8 loss among 4 people to get #2 loss each, but not
#8 among -4 people.  At some level, we have to say "Them's the
rules", and list what you can do to "sharings";  we have lost
the associated intuition.  Similarly, we can divide 7 by 3.5,
but not share 7 cakes among 3.5 people;  not, anyway, until
we get to "full-time equivalent" people.  As mathematicians,
we get used to "the rules", and often forget how difficult
these concepts are to "the man in the street", who knows his
maths only through the intuitive applications.

-- 
Andy Walker, School of MathSci., Univ. of Nott'm, UK.
anw@maths.nott.ac.uk

"Stan Brown"  wrote in message 
news:MPG.2079f2be9d79e39298aadb@news.individual.net...
> I've added the attribution you omitted. Please show proper
> attributions in your follow-ups.
> http://oakroadsystems.com/genl/unice.htm#attrib
>
> Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
>> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown 
>> :
>> > Is 8 over -4 the same as -8 over 4?
>>
>> first i think no, but each answer is -2?
>> so must be yes.
>>
>> the answer I got is shown at the top
>> the answer in the book is below
>> http://cjoint.com/data/ebp1xokLvv.htm
>> I mean are the two answers the same?
>
> This HTML business is getting pretty old. It's not hard to post
> algebra.

Hmm - it's not always easy.  In this case, the minus sign (in the final line 
of the solution) is shown in front of the dividing line, not against 
denominator or numerator.  Seeing it hand-written helps me to understand why 
the OP may take more than a moment to get his head round this.

- ( ( b - c ) / a ) may not mean a lot to someone meeting, say, negative 
denominators for the first time.

And as Andy Walker implies, the clever bit (and the crucial bit in an 
education group) is identifying exactly what is creating the problem / 
confusion / misunderstanding.

-- 
Martin

[Remove barrier to reply]

> Hmm - it's not always easy.  In this case, the minus sign (in the final
line
> of the solution) is shown in front of the dividing line, not against
> denominator or numerator.  Seeing it hand-written helps me to understand
why
> the OP may take more than a moment to get his head round this.

Thank you for your contribution Martin,
As Cody's father, I encourage him to be independent
and therefore seek his own resolution.
The problem deepens because I left school at 15 and whilst
I (think) Ican handle his work thus far, it was I, who marked his
answer as incorrect.
He thus asked me why, I can't answer it, hence my suggestion
he ask on the web.
His work comes in booklets, which parents have to mark and sign
Take for example his latest work going to a new level.

simultaneous linear equations in two levels.
http://cjoint.com/data/eddh7BHu2E.htm

The method is explained, but the mechanics or the practicalities
are not suggested, supplementary reading/info is not suggested either.
He is going this alone therefore, apart from a few minutes he can
snatch form his tutor, who he sees once a week.

I understand the regime of having the student think problems
through, but sometimes one needs an explanation to "kick start"
a process. I'm sure at the moment these are just heiroglyphs to him.
I remember he had a problem long ago with long division,
after I found the problem located in the extension of zeroes
concept, he rattled through his work.

BTW I still am unable to explain to him, why the negative
sign changes postion from the denomonator to the divisor line :)
Every question he asks, he gets a question in response.

We are probably meddling in the domain of the intelligentsia,
we need a link to be amongst others of similar progress,
but I am dashed if I can find one at the moment.
Rodney

"Rod"  wrote in message 
news:4611ad80_1@news.iprimus.com.au...
>> Hmm - it's not always easy.  In this case, the minus sign (in the final
> line
>> of the solution) is shown in front of the dividing line, not against
>> denominator or numerator.  Seeing it hand-written helps me to understand
> why
>> the OP may take more than a moment to get his head round this.
>
> Thank you for your contribution Martin,
> As Cody's father, I encourage him to be independent

That's obviously good.  And it'll help enormously if Cody can maintain 
enthusiasm and excitement at discovering new stuff.  Never be disheartened 
when facing a tough problem, and don't move on until you've cracked it.

> simultaneous linear equations in two levels.
> http://cjoint.com/data/eddh7BHu2E.htm

In the UK, the first problem (re-arranging an equation) and this 
simultaneous equation would normally not be encountered until age 14+.

> but sometimes one needs an explanation to "kick start"
> a process. I'm sure at the moment these are just heiroglyphs to him.

My approach (wherever possible) has always been to make sure the student 
understands the nature of the problem before diving into the algebra.  So 
with simultaneous equations, I would want to ensure Cody understands, and 
can sketch, a straight-line graph (hence understanding  " y = mx + c "  and 
what the m and c refer to).  Then add a second line, and recognise the 
significance of the point (i.e. co-ordinates) where the two lines cross.

I've found this kind of approach means the student understands what will be 
discovered by solviing two simul.eqs. (i.e. the only values of x and y which 
"satisfy" both equations).  And where a graphical method isn't so helpful, 
use some other kind of "real world" example.  So when introducing the idea 
of equations (before moving on to to re-arrange them) I always sketch a 
see-saw and tease out some "rules" about what can be added, subtracted etc. 
to each end to keep it balanced.  IME, a "negative sack of potatoes" soon 
starts to be a comfortable concept!

> BTW I still am unable to explain to him, why the negative
> sign changes postion from the denomonator to the divisor line :)

And it's even harder to explain in a post !  In the original post, the 
example has several "right" answers;  the "preferred" answer is partly a 
matter of opinion and partly "tidyness".  It certainly doesn't, though, 
involve putting the variable in alphabetic order, as suggested in your 3rd 
post.

> We are probably meddling in the domain of the intelligentsia,
> we need a link to be amongst others of similar progress,

Have you seen this site?
http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

If you hunt around, you'll find stuff on the basics of negative numbers, 
algebra, simul. eqs. etc.  It may help....

Just one note of caution - everyone has different ideas on how best to teach 
/ explain stuff, and if the student encounters too many different approaches 
/ styles, it can get confusing.  But rather than stick to one approach, 
understand there are many, hunt them down, then go with the method that 
works for you.

And never forget the answer in the back of the book is sometimes wrong...!!

Would be interested to know how you get on :-)

-- 
Martin

[Remove barrier to reply]

"Martin"  wrote in
message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

Cheers Martin,
I'll print out your suggestions.
I do notice another 2 books in, which he will attempt
next week, involves algebra to determine area of triangles
etc & etc, so finally he will have some meaning attached
to the characters.

> In the UK, the first problem (re-arranging an equation) and this
> simultaneous equation would normally not be encountered until age 14+.

Apparently here, he is just one year ahead, according to the Japanese
"Kumon" system we have purchased.
We felt we had to follow supplementary instruction, we had him
at school in Bangkok, in order to ingrain his Thai language first,
so his return by grade 3/4 in Aust, he was at that time unable to
speak English
He has caught up well in all his school work, although I am not
convinced he "likes" math, although he somehow knows it is important
and he enjoys helping other kids in class with math, and that is marvelous
for his self esteem.


> It certainly doesn't, though,
> involve putting the variable in alphabetic order, as suggested in your 3rd
> post.
I think Cody must have mis understood  tutor advice, it was he
who told me it had to be so.

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

Thank you very much indeed, Martin.

Best Regards,
Rodney
Perth.

"Martin" 
wrote in message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

I found that just a bit basic Martin, however, spurred on
by your graph suggestion, and a simple Google string,
Serendipity played her hand :)

http://www.themathpage.com/alg/simultaneous-equations.htm

and a graph to boot!

Thanks very much, (again)
Rodney.

My name is cody
12yo in year 7 primary school.
perth australia

can someone explain the story
to end up with the correct answer
to this one
http://cjoint.com/data/dFlxPGB5yb.htm

I do not know how the negative arrives
in the centre of the fraction in the answer.

Is this the best place to ask my queries?
thank you
cody

Sun, 1 Apr 2007 08:35:38 +0800 from Rod :
> 
> My name is cody
> 12yo in year 7 primary school.
> perth australia
> 
> can someone explain the story
> to end up with the correct answer
> to this one
> http://cjoint.com/data/dFlxPGB5yb.htm
> 
> I do not know how the negative arrives
> in the centre of the fraction in the answer.

You had -ax on the left-hand side. What must you divide by to get x? 
Do you divide by a or by -a?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

Thank you Mr Brown.
divide by -a

Is b-c over -a
the same as negative b-c over a ?

eg -(b-c over a)
this confuses me

thank you
cody





> You had -ax on the left-hand side. What must you divide by to get x?
> Do you divide by a or by -a?
>
> --
> Stan Brown, Oak Road Systems, Tompkins County, New York, USA
>                                   http://OakRoadSystems.com/

"Rod"  wrote in message 
news:460f28d3$1_1@news.iprimus.com.au...
>
> Thank you Mr Brown.
> divide by -a
>
> Is b-c over -a
> the same as negative b-c over a ?
>
> eg -(b-c over a)
> this confuses me

Yes, or at least it is if you include the ( )  ie:


(b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )

One way to avoid the problem of dealing with the negative signs is always to 
ensure that the terms in x are positve.

eg:

-ax = b-c

add ax to both sides:


0 = b-c+ax

-b both sides

-b = - c + ax

+ b both sides

c-b = ax

/a both sides

(c-b)/a  = x

--
73
Brian
www.g8osn.org.uk

Thank you Mr Reay
then I got the answer right.
here is the question and answer given
which does not show the brackets
http://cjoint.com/data/ebitEB4dMe.htm

It is "kumon" and I have a tutor but i have to line up
with some 50 kids to get answers.

i get most right but the negative place here
looks wrong

is the book answer wrong?

also you say c-b we have to put in alphabetic b-c



"Brian Reay" <brian.reay@(spamstopper)bigfoot.com> wrote in message
news:xRHPh.232$r4.15@newsfe1-gui.ntli.net...
>
> "Rod"  wrote in message
> news:460f28d3$1_1@news.iprimus.com.au...
> >
> > Thank you Mr Brown.
> > divide by -a
> >
> > Is b-c over -a
> > the same as negative b-c over a ?
> >
> > eg -(b-c over a)
> > this confuses me
>
> Yes, or at least it is if you include the ( )  ie:
>
>
> (b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )
>
> One way to avoid the problem of dealing with the negative signs is always
to
> ensure that the terms in x are positve.
>
> eg:
>
> -ax = b-c
>
> add ax to both sides:
>
>
> 0 = b-c+ax
>
> -b both sides
>
> -b = - c + ax
>
> + b both sides
>
> c-b = ax
>
> /a both sides
>
> (c-b)/a  = x
>
> --
> 73
> Brian
> www.g8osn.org.uk
>
>
>

you made a miistake
+c both sides

> > + b both sides

"Rod"  wrote in message 
news:460f5168$1_1@news.iprimus.com.au...
> you made a miistake
> +c both sides
>
>> > + b both sides

Well spotted- it is early on a Sunday morning here and I've not had a cuppa 
yet ;-)

Good luck with your studies.

--
73
Brian
www.g8osn.org.uk

> Well spotted- it is early on a Sunday morning here and I've not had a
cuppa
> yet ;-)
>
> Good luck with your studies.

Thank you Sir.

Sun, 1 Apr 2007 11:39:11 +0800 from Rod :
> 
> Thank you Mr Brown.
> divide by -a
> 
> Is b-c over -a
> the same as negative b-c over a ?

Is 8 over -4 the same as -8 over 4?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Is 8 over -4 the same as -8 over 4?

first i think no, but each answer is -2?
so must be yes.

the answer I got is shown at the top
the answer in the book is below
http://cjoint.com/data/ebp1xokLvv.htm
I mean are the two answers the same?
i am more worried about the position of the
negative sign than the answer maybe?

thank you Mr Brown

"Rod"  wrote in message 
news:460fbb55_1@news.iprimus.com.au...
 > i am more worried about the position of the
> negative sign than the answer maybe?


Sometimes you are simply looking at a presentation convention eg

-8/2  =  8/-2    (-4 in both cases)

However,  -8/2 is often considered "better" presentation wise.

Likewise:

-b + c =  c-b

But c-b is  often considered "better" presentation wise.

There are harder examples (eg "rationalising surds") but worry about those 
when you get to secondary school.

-- 
73
Brian, G8OSN
www.g8osn.org.uk

I've added the attribution you omitted. Please show proper 
attributions in your follow-ups.
http://oakroadsystems.com/genl/unice.htm#attrib

Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown :
> > Is 8 over -4 the same as -8 over 4?
> 
> first i think no, but each answer is -2?
> so must be yes.
> 
> the answer I got is shown at the top
> the answer in the book is below
> http://cjoint.com/data/ebp1xokLvv.htm
> I mean are the two answers the same?

This HTML business is getting pretty old. It's not hard to post 
algebra. You are asking whether
     (b-c) / (-a)          and        -(b-c) / a
are the same.

Again, I'll answer that question with a question: are
     (15-7) / (-4)         and        -(15-7) / 4
the same?

(Note: It's not always safe to answer questions like these with an 
example. But if you choose your numbers so that nothing is 0 or 1, 
and maybe even try a couple of examples, it can be a helpful guide.)

> i am more worried about the position of the
> negative sign than the answer maybe?

Meaning no disrespect, I think you need to review the basics of 
division and multiplication with signed numbers. When I see a student 
confused by different ways of presenting the same thing, sometimes 
it's just a momentary perplexity but often the student just doesn't 
understand the underlying principles.

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Sometimes you are simply looking at a presentation convention eg

Thank you, Mr Reay
I notice my mistakes also.
Cody
Australia

In article ,
Stan Brown   wrote:
>Is 8 over -4 the same as -8 over 4?

	It's an interesting question, isn't it?  I can imagine
sharing an #8 loss among 4 people to get #2 loss each, but not
#8 among -4 people.  At some level, we have to say "Them's the
rules", and list what you can do to "sharings";  we have lost
the associated intuition.  Similarly, we can divide 7 by 3.5,
but not share 7 cakes among 3.5 people;  not, anyway, until
we get to "full-time equivalent" people.  As mathematicians,
we get used to "the rules", and often forget how difficult
these concepts are to "the man in the street", who knows his
maths only through the intuitive applications.

-- 
Andy Walker, School of MathSci., Univ. of Nott'm, UK.
anw@maths.nott.ac.uk

"Stan Brown"  wrote in message 
news:MPG.2079f2be9d79e39298aadb@news.individual.net...
> I've added the attribution you omitted. Please show proper
> attributions in your follow-ups.
> http://oakroadsystems.com/genl/unice.htm#attrib
>
> Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
>> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown 
>> :
>> > Is 8 over -4 the same as -8 over 4?
>>
>> first i think no, but each answer is -2?
>> so must be yes.
>>
>> the answer I got is shown at the top
>> the answer in the book is below
>> http://cjoint.com/data/ebp1xokLvv.htm
>> I mean are the two answers the same?
>
> This HTML business is getting pretty old. It's not hard to post
> algebra.

Hmm - it's not always easy.  In this case, the minus sign (in the final line 
of the solution) is shown in front of the dividing line, not against 
denominator or numerator.  Seeing it hand-written helps me to understand why 
the OP may take more than a moment to get his head round this.

- ( ( b - c ) / a ) may not mean a lot to someone meeting, say, negative 
denominators for the first time.

And as Andy Walker implies, the clever bit (and the crucial bit in an 
education group) is identifying exactly what is creating the problem / 
confusion / misunderstanding.

-- 
Martin

[Remove barrier to reply]

> Hmm - it's not always easy.  In this case, the minus sign (in the final
line
> of the solution) is shown in front of the dividing line, not against
> denominator or numerator.  Seeing it hand-written helps me to understand
why
> the OP may take more than a moment to get his head round this.

Thank you for your contribution Martin,
As Cody's father, I encourage him to be independent
and therefore seek his own resolution.
The problem deepens because I left school at 15 and whilst
I (think) Ican handle his work thus far, it was I, who marked his
answer as incorrect.
He thus asked me why, I can't answer it, hence my suggestion
he ask on the web.
His work comes in booklets, which parents have to mark and sign
Take for example his latest work going to a new level.

simultaneous linear equations in two levels.
http://cjoint.com/data/eddh7BHu2E.htm

The method is explained, but the mechanics or the practicalities
are not suggested, supplementary reading/info is not suggested either.
He is going this alone therefore, apart from a few minutes he can
snatch form his tutor, who he sees once a week.

I understand the regime of having the student think problems
through, but sometimes one needs an explanation to "kick start"
a process. I'm sure at the moment these are just heiroglyphs to him.
I remember he had a problem long ago with long division,
after I found the problem located in the extension of zeroes
concept, he rattled through his work.

BTW I still am unable to explain to him, why the negative
sign changes postion from the denomonator to the divisor line :)
Every question he asks, he gets a question in response.

We are probably meddling in the domain of the intelligentsia,
we need a link to be amongst others of similar progress,
but I am dashed if I can find one at the moment.
Rodney

"Rod"  wrote in message 
news:4611ad80_1@news.iprimus.com.au...
>> Hmm - it's not always easy.  In this case, the minus sign (in the final
> line
>> of the solution) is shown in front of the dividing line, not against
>> denominator or numerator.  Seeing it hand-written helps me to understand
> why
>> the OP may take more than a moment to get his head round this.
>
> Thank you for your contribution Martin,
> As Cody's father, I encourage him to be independent

That's obviously good.  And it'll help enormously if Cody can maintain 
enthusiasm and excitement at discovering new stuff.  Never be disheartened 
when facing a tough problem, and don't move on until you've cracked it.

> simultaneous linear equations in two levels.
> http://cjoint.com/data/eddh7BHu2E.htm

In the UK, the first problem (re-arranging an equation) and this 
simultaneous equation would normally not be encountered until age 14+.

> but sometimes one needs an explanation to "kick start"
> a process. I'm sure at the moment these are just heiroglyphs to him.

My approach (wherever possible) has always been to make sure the student 
understands the nature of the problem before diving into the algebra.  So 
with simultaneous equations, I would want to ensure Cody understands, and 
can sketch, a straight-line graph (hence understanding  " y = mx + c "  and 
what the m and c refer to).  Then add a second line, and recognise the 
significance of the point (i.e. co-ordinates) where the two lines cross.

I've found this kind of approach means the student understands what will be 
discovered by solviing two simul.eqs. (i.e. the only values of x and y which 
"satisfy" both equations).  And where a graphical method isn't so helpful, 
use some other kind of "real world" example.  So when introducing the idea 
of equations (before moving on to to re-arrange them) I always sketch a 
see-saw and tease out some "rules" about what can be added, subtracted etc. 
to each end to keep it balanced.  IME, a "negative sack of potatoes" soon 
starts to be a comfortable concept!

> BTW I still am unable to explain to him, why the negative
> sign changes postion from the denomonator to the divisor line :)

And it's even harder to explain in a post !  In the original post, the 
example has several "right" answers;  the "preferred" answer is partly a 
matter of opinion and partly "tidyness".  It certainly doesn't, though, 
involve putting the variable in alphabetic order, as suggested in your 3rd 
post.

> We are probably meddling in the domain of the intelligentsia,
> we need a link to be amongst others of similar progress,

Have you seen this site?
http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

If you hunt around, you'll find stuff on the basics of negative numbers, 
algebra, simul. eqs. etc.  It may help....

Just one note of caution - everyone has different ideas on how best to teach 
/ explain stuff, and if the student encounters too many different approaches 
/ styles, it can get confusing.  But rather than stick to one approach, 
understand there are many, hunt them down, then go with the method that 
works for you.

And never forget the answer in the back of the book is sometimes wrong...!!

Would be interested to know how you get on :-)

-- 
Martin

[Remove barrier to reply]

"Martin"  wrote in
message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

Cheers Martin,
I'll print out your suggestions.
I do notice another 2 books in, which he will attempt
next week, involves algebra to determine area of triangles
etc & etc, so finally he will have some meaning attached
to the characters.

> In the UK, the first problem (re-arranging an equation) and this
> simultaneous equation would normally not be encountered until age 14+.

Apparently here, he is just one year ahead, according to the Japanese
"Kumon" system we have purchased.
We felt we had to follow supplementary instruction, we had him
at school in Bangkok, in order to ingrain his Thai language first,
so his return by grade 3/4 in Aust, he was at that time unable to
speak English
He has caught up well in all his school work, although I am not
convinced he "likes" math, although he somehow knows it is important
and he enjoys helping other kids in class with math, and that is marvelous
for his self esteem.


> It certainly doesn't, though,
> involve putting the variable in alphabetic order, as suggested in your 3rd
> post.
I think Cody must have mis understood  tutor advice, it was he
who told me it had to be so.

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

Thank you very much indeed, Martin.

Best Regards,
Rodney
Perth.

"Martin" 
wrote in message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

I found that just a bit basic Martin, however, spurred on
by your graph suggestion, and a simple Google string,
Serendipity played her hand :)

http://www.themathpage.com/alg/simultaneous-equations.htm

and a graph to boot!

Thanks very much, (again)
Rodney.

My name is cody
12yo in year 7 primary school.
perth australia

can someone explain the story
to end up with the correct answer
to this one
http://cjoint.com/data/dFlxPGB5yb.htm

I do not know how the negative arrives
in the centre of the fraction in the answer.

Is this the best place to ask my queries?
thank you
cody

Sun, 1 Apr 2007 08:35:38 +0800 from Rod :
> 
> My name is cody
> 12yo in year 7 primary school.
> perth australia
> 
> can someone explain the story
> to end up with the correct answer
> to this one
> http://cjoint.com/data/dFlxPGB5yb.htm
> 
> I do not know how the negative arrives
> in the centre of the fraction in the answer.

You had -ax on the left-hand side. What must you divide by to get x? 
Do you divide by a or by -a?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

Thank you Mr Brown.
divide by -a

Is b-c over -a
the same as negative b-c over a ?

eg -(b-c over a)
this confuses me

thank you
cody





> You had -ax on the left-hand side. What must you divide by to get x?
> Do you divide by a or by -a?
>
> --
> Stan Brown, Oak Road Systems, Tompkins County, New York, USA
>                                   http://OakRoadSystems.com/

"Rod"  wrote in message 
news:460f28d3$1_1@news.iprimus.com.au...
>
> Thank you Mr Brown.
> divide by -a
>
> Is b-c over -a
> the same as negative b-c over a ?
>
> eg -(b-c over a)
> this confuses me

Yes, or at least it is if you include the ( )  ie:


(b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )

One way to avoid the problem of dealing with the negative signs is always to 
ensure that the terms in x are positve.

eg:

-ax = b-c

add ax to both sides:


0 = b-c+ax

-b both sides

-b = - c + ax

+ b both sides

c-b = ax

/a both sides

(c-b)/a  = x

--
73
Brian
www.g8osn.org.uk

Thank you Mr Reay
then I got the answer right.
here is the question and answer given
which does not show the brackets
http://cjoint.com/data/ebitEB4dMe.htm

It is "kumon" and I have a tutor but i have to line up
with some 50 kids to get answers.

i get most right but the negative place here
looks wrong

is the book answer wrong?

also you say c-b we have to put in alphabetic b-c



"Brian Reay" <brian.reay@(spamstopper)bigfoot.com> wrote in message
news:xRHPh.232$r4.15@newsfe1-gui.ntli.net...
>
> "Rod"  wrote in message
> news:460f28d3$1_1@news.iprimus.com.au...
> >
> > Thank you Mr Brown.
> > divide by -a
> >
> > Is b-c over -a
> > the same as negative b-c over a ?
> >
> > eg -(b-c over a)
> > this confuses me
>
> Yes, or at least it is if you include the ( )  ie:
>
>
> (b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )
>
> One way to avoid the problem of dealing with the negative signs is always
to
> ensure that the terms in x are positve.
>
> eg:
>
> -ax = b-c
>
> add ax to both sides:
>
>
> 0 = b-c+ax
>
> -b both sides
>
> -b = - c + ax
>
> + b both sides
>
> c-b = ax
>
> /a both sides
>
> (c-b)/a  = x
>
> --
> 73
> Brian
> www.g8osn.org.uk
>
>
>

you made a miistake
+c both sides

> > + b both sides

"Rod"  wrote in message 
news:460f5168$1_1@news.iprimus.com.au...
> you made a miistake
> +c both sides
>
>> > + b both sides

Well spotted- it is early on a Sunday morning here and I've not had a cuppa 
yet ;-)

Good luck with your studies.

--
73
Brian
www.g8osn.org.uk

> Well spotted- it is early on a Sunday morning here and I've not had a
cuppa
> yet ;-)
>
> Good luck with your studies.

Thank you Sir.

Sun, 1 Apr 2007 11:39:11 +0800 from Rod :
> 
> Thank you Mr Brown.
> divide by -a
> 
> Is b-c over -a
> the same as negative b-c over a ?

Is 8 over -4 the same as -8 over 4?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Is 8 over -4 the same as -8 over 4?

first i think no, but each answer is -2?
so must be yes.

the answer I got is shown at the top
the answer in the book is below
http://cjoint.com/data/ebp1xokLvv.htm
I mean are the two answers the same?
i am more worried about the position of the
negative sign than the answer maybe?

thank you Mr Brown

"Rod"  wrote in message 
news:460fbb55_1@news.iprimus.com.au...
 > i am more worried about the position of the
> negative sign than the answer maybe?


Sometimes you are simply looking at a presentation convention eg

-8/2  =  8/-2    (-4 in both cases)

However,  -8/2 is often considered "better" presentation wise.

Likewise:

-b + c =  c-b

But c-b is  often considered "better" presentation wise.

There are harder examples (eg "rationalising surds") but worry about those 
when you get to secondary school.

-- 
73
Brian, G8OSN
www.g8osn.org.uk

I've added the attribution you omitted. Please show proper 
attributions in your follow-ups.
http://oakroadsystems.com/genl/unice.htm#attrib

Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown :
> > Is 8 over -4 the same as -8 over 4?
> 
> first i think no, but each answer is -2?
> so must be yes.
> 
> the answer I got is shown at the top
> the answer in the book is below
> http://cjoint.com/data/ebp1xokLvv.htm
> I mean are the two answers the same?

This HTML business is getting pretty old. It's not hard to post 
algebra. You are asking whether
     (b-c) / (-a)          and        -(b-c) / a
are the same.

Again, I'll answer that question with a question: are
     (15-7) / (-4)         and        -(15-7) / 4
the same?

(Note: It's not always safe to answer questions like these with an 
example. But if you choose your numbers so that nothing is 0 or 1, 
and maybe even try a couple of examples, it can be a helpful guide.)

> i am more worried about the position of the
> negative sign than the answer maybe?

Meaning no disrespect, I think you need to review the basics of 
division and multiplication with signed numbers. When I see a student 
confused by different ways of presenting the same thing, sometimes 
it's just a momentary perplexity but often the student just doesn't 
understand the underlying principles.

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Sometimes you are simply looking at a presentation convention eg

Thank you, Mr Reay
I notice my mistakes also.
Cody
Australia

In article ,
Stan Brown   wrote:
>Is 8 over -4 the same as -8 over 4?

	It's an interesting question, isn't it?  I can imagine
sharing an #8 loss among 4 people to get #2 loss each, but not
#8 among -4 people.  At some level, we have to say "Them's the
rules", and list what you can do to "sharings";  we have lost
the associated intuition.  Similarly, we can divide 7 by 3.5,
but not share 7 cakes among 3.5 people;  not, anyway, until
we get to "full-time equivalent" people.  As mathematicians,
we get used to "the rules", and often forget how difficult
these concepts are to "the man in the street", who knows his
maths only through the intuitive applications.

-- 
Andy Walker, School of MathSci., Univ. of Nott'm, UK.
anw@maths.nott.ac.uk

"Stan Brown"  wrote in message 
news:MPG.2079f2be9d79e39298aadb@news.individual.net...
> I've added the attribution you omitted. Please show proper
> attributions in your follow-ups.
> http://oakroadsystems.com/genl/unice.htm#attrib
>
> Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
>> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown 
>> :
>> > Is 8 over -4 the same as -8 over 4?
>>
>> first i think no, but each answer is -2?
>> so must be yes.
>>
>> the answer I got is shown at the top
>> the answer in the book is below
>> http://cjoint.com/data/ebp1xokLvv.htm
>> I mean are the two answers the same?
>
> This HTML business is getting pretty old. It's not hard to post
> algebra.

Hmm - it's not always easy.  In this case, the minus sign (in the final line 
of the solution) is shown in front of the dividing line, not against 
denominator or numerator.  Seeing it hand-written helps me to understand why 
the OP may take more than a moment to get his head round this.

- ( ( b - c ) / a ) may not mean a lot to someone meeting, say, negative 
denominators for the first time.

And as Andy Walker implies, the clever bit (and the crucial bit in an 
education group) is identifying exactly what is creating the problem / 
confusion / misunderstanding.

-- 
Martin

[Remove barrier to reply]

> Hmm - it's not always easy.  In this case, the minus sign (in the final
line
> of the solution) is shown in front of the dividing line, not against
> denominator or numerator.  Seeing it hand-written helps me to understand
why
> the OP may take more than a moment to get his head round this.

Thank you for your contribution Martin,
As Cody's father, I encourage him to be independent
and therefore seek his own resolution.
The problem deepens because I left school at 15 and whilst
I (think) Ican handle his work thus far, it was I, who marked his
answer as incorrect.
He thus asked me why, I can't answer it, hence my suggestion
he ask on the web.
His work comes in booklets, which parents have to mark and sign
Take for example his latest work going to a new level.

simultaneous linear equations in two levels.
http://cjoint.com/data/eddh7BHu2E.htm

The method is explained, but the mechanics or the practicalities
are not suggested, supplementary reading/info is not suggested either.
He is going this alone therefore, apart from a few minutes he can
snatch form his tutor, who he sees once a week.

I understand the regime of having the student think problems
through, but sometimes one needs an explanation to "kick start"
a process. I'm sure at the moment these are just heiroglyphs to him.
I remember he had a problem long ago with long division,
after I found the problem located in the extension of zeroes
concept, he rattled through his work.

BTW I still am unable to explain to him, why the negative
sign changes postion from the denomonator to the divisor line :)
Every question he asks, he gets a question in response.

We are probably meddling in the domain of the intelligentsia,
we need a link to be amongst others of similar progress,
but I am dashed if I can find one at the moment.
Rodney

"Rod"  wrote in message 
news:4611ad80_1@news.iprimus.com.au...
>> Hmm - it's not always easy.  In this case, the minus sign (in the final
> line
>> of the solution) is shown in front of the dividing line, not against
>> denominator or numerator.  Seeing it hand-written helps me to understand
> why
>> the OP may take more than a moment to get his head round this.
>
> Thank you for your contribution Martin,
> As Cody's father, I encourage him to be independent

That's obviously good.  And it'll help enormously if Cody can maintain 
enthusiasm and excitement at discovering new stuff.  Never be disheartened 
when facing a tough problem, and don't move on until you've cracked it.

> simultaneous linear equations in two levels.
> http://cjoint.com/data/eddh7BHu2E.htm

In the UK, the first problem (re-arranging an equation) and this 
simultaneous equation would normally not be encountered until age 14+.

> but sometimes one needs an explanation to "kick start"
> a process. I'm sure at the moment these are just heiroglyphs to him.

My approach (wherever possible) has always been to make sure the student 
understands the nature of the problem before diving into the algebra.  So 
with simultaneous equations, I would want to ensure Cody understands, and 
can sketch, a straight-line graph (hence understanding  " y = mx + c "  and 
what the m and c refer to).  Then add a second line, and recognise the 
significance of the point (i.e. co-ordinates) where the two lines cross.

I've found this kind of approach means the student understands what will be 
discovered by solviing two simul.eqs. (i.e. the only values of x and y which 
"satisfy" both equations).  And where a graphical method isn't so helpful, 
use some other kind of "real world" example.  So when introducing the idea 
of equations (before moving on to to re-arrange them) I always sketch a 
see-saw and tease out some "rules" about what can be added, subtracted etc. 
to each end to keep it balanced.  IME, a "negative sack of potatoes" soon 
starts to be a comfortable concept!

> BTW I still am unable to explain to him, why the negative
> sign changes postion from the denomonator to the divisor line :)

And it's even harder to explain in a post !  In the original post, the 
example has several "right" answers;  the "preferred" answer is partly a 
matter of opinion and partly "tidyness".  It certainly doesn't, though, 
involve putting the variable in alphabetic order, as suggested in your 3rd 
post.

> We are probably meddling in the domain of the intelligentsia,
> we need a link to be amongst others of similar progress,

Have you seen this site?
http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

If you hunt around, you'll find stuff on the basics of negative numbers, 
algebra, simul. eqs. etc.  It may help....

Just one note of caution - everyone has different ideas on how best to teach 
/ explain stuff, and if the student encounters too many different approaches 
/ styles, it can get confusing.  But rather than stick to one approach, 
understand there are many, hunt them down, then go with the method that 
works for you.

And never forget the answer in the back of the book is sometimes wrong...!!

Would be interested to know how you get on :-)

-- 
Martin

[Remove barrier to reply]

"Martin"  wrote in
message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

Cheers Martin,
I'll print out your suggestions.
I do notice another 2 books in, which he will attempt
next week, involves algebra to determine area of triangles
etc & etc, so finally he will have some meaning attached
to the characters.

> In the UK, the first problem (re-arranging an equation) and this
> simultaneous equation would normally not be encountered until age 14+.

Apparently here, he is just one year ahead, according to the Japanese
"Kumon" system we have purchased.
We felt we had to follow supplementary instruction, we had him
at school in Bangkok, in order to ingrain his Thai language first,
so his return by grade 3/4 in Aust, he was at that time unable to
speak English
He has caught up well in all his school work, although I am not
convinced he "likes" math, although he somehow knows it is important
and he enjoys helping other kids in class with math, and that is marvelous
for his self esteem.


> It certainly doesn't, though,
> involve putting the variable in alphabetic order, as suggested in your 3rd
> post.
I think Cody must have mis understood  tutor advice, it was he
who told me it had to be so.

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

Thank you very much indeed, Martin.

Best Regards,
Rodney
Perth.

"Martin" 
wrote in message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

I found that just a bit basic Martin, however, spurred on
by your graph suggestion, and a simple Google string,
Serendipity played her hand :)

http://www.themathpage.com/alg/simultaneous-equations.htm

and a graph to boot!

Thanks very much, (again)
Rodney.

Sun, 1 Apr 2007 08:35:38 +0800 from Rod :
> 
> My name is cody
> 12yo in year 7 primary school.
> perth australia
> 
> can someone explain the story
> to end up with the correct answer
> to this one
> http://cjoint.com/data/dFlxPGB5yb.htm
> 
> I do not know how the negative arrives
> in the centre of the fraction in the answer.

You had -ax on the left-hand side. What must you divide by to get x? 
Do you divide by a or by -a?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

Thank you Mr Brown.
divide by -a

Is b-c over -a
the same as negative b-c over a ?

eg -(b-c over a)
this confuses me

thank you
cody





> You had -ax on the left-hand side. What must you divide by to get x?
> Do you divide by a or by -a?
>
> --
> Stan Brown, Oak Road Systems, Tompkins County, New York, USA
>                                   http://OakRoadSystems.com/

"Rod"  wrote in message 
news:460f28d3$1_1@news.iprimus.com.au...
>
> Thank you Mr Brown.
> divide by -a
>
> Is b-c over -a
> the same as negative b-c over a ?
>
> eg -(b-c over a)
> this confuses me

Yes, or at least it is if you include the ( )  ie:


(b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )

One way to avoid the problem of dealing with the negative signs is always to 
ensure that the terms in x are positve.

eg:

-ax = b-c

add ax to both sides:


0 = b-c+ax

-b both sides

-b = - c + ax

+ b both sides

c-b = ax

/a both sides

(c-b)/a  = x

--
73
Brian
www.g8osn.org.uk

Thank you Mr Reay
then I got the answer right.
here is the question and answer given
which does not show the brackets
http://cjoint.com/data/ebitEB4dMe.htm

It is "kumon" and I have a tutor but i have to line up
with some 50 kids to get answers.

i get most right but the negative place here
looks wrong

is the book answer wrong?

also you say c-b we have to put in alphabetic b-c



"Brian Reay" <brian.reay@(spamstopper)bigfoot.com> wrote in message
news:xRHPh.232$r4.15@newsfe1-gui.ntli.net...
>
> "Rod"  wrote in message
> news:460f28d3$1_1@news.iprimus.com.au...
> >
> > Thank you Mr Brown.
> > divide by -a
> >
> > Is b-c over -a
> > the same as negative b-c over a ?
> >
> > eg -(b-c over a)
> > this confuses me
>
> Yes, or at least it is if you include the ( )  ie:
>
>
> (b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )
>
> One way to avoid the problem of dealing with the negative signs is always
to
> ensure that the terms in x are positve.
>
> eg:
>
> -ax = b-c
>
> add ax to both sides:
>
>
> 0 = b-c+ax
>
> -b both sides
>
> -b = - c + ax
>
> + b both sides
>
> c-b = ax
>
> /a both sides
>
> (c-b)/a  = x
>
> --
> 73
> Brian
> www.g8osn.org.uk
>
>
>

you made a miistake
+c both sides

> > + b both sides

"Rod"  wrote in message 
news:460f5168$1_1@news.iprimus.com.au...
> you made a miistake
> +c both sides
>
>> > + b both sides

Well spotted- it is early on a Sunday morning here and I've not had a cuppa 
yet ;-)

Good luck with your studies.

--
73
Brian
www.g8osn.org.uk

> Well spotted- it is early on a Sunday morning here and I've not had a
cuppa
> yet ;-)
>
> Good luck with your studies.

Thank you Sir.

Sun, 1 Apr 2007 11:39:11 +0800 from Rod :
> 
> Thank you Mr Brown.
> divide by -a
> 
> Is b-c over -a
> the same as negative b-c over a ?

Is 8 over -4 the same as -8 over 4?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Is 8 over -4 the same as -8 over 4?

first i think no, but each answer is -2?
so must be yes.

the answer I got is shown at the top
the answer in the book is below
http://cjoint.com/data/ebp1xokLvv.htm
I mean are the two answers the same?
i am more worried about the position of the
negative sign than the answer maybe?

thank you Mr Brown

"Rod"  wrote in message 
news:460fbb55_1@news.iprimus.com.au...
 > i am more worried about the position of the
> negative sign than the answer maybe?


Sometimes you are simply looking at a presentation convention eg

-8/2  =  8/-2    (-4 in both cases)

However,  -8/2 is often considered "better" presentation wise.

Likewise:

-b + c =  c-b

But c-b is  often considered "better" presentation wise.

There are harder examples (eg "rationalising surds") but worry about those 
when you get to secondary school.

-- 
73
Brian, G8OSN
www.g8osn.org.uk

I've added the attribution you omitted. Please show proper 
attributions in your follow-ups.
http://oakroadsystems.com/genl/unice.htm#attrib

Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown :
> > Is 8 over -4 the same as -8 over 4?
> 
> first i think no, but each answer is -2?
> so must be yes.
> 
> the answer I got is shown at the top
> the answer in the book is below
> http://cjoint.com/data/ebp1xokLvv.htm
> I mean are the two answers the same?

This HTML business is getting pretty old. It's not hard to post 
algebra. You are asking whether
     (b-c) / (-a)          and        -(b-c) / a
are the same.

Again, I'll answer that question with a question: are
     (15-7) / (-4)         and        -(15-7) / 4
the same?

(Note: It's not always safe to answer questions like these with an 
example. But if you choose your numbers so that nothing is 0 or 1, 
and maybe even try a couple of examples, it can be a helpful guide.)

> i am more worried about the position of the
> negative sign than the answer maybe?

Meaning no disrespect, I think you need to review the basics of 
division and multiplication with signed numbers. When I see a student 
confused by different ways of presenting the same thing, sometimes 
it's just a momentary perplexity but often the student just doesn't 
understand the underlying principles.

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Sometimes you are simply looking at a presentation convention eg

Thank you, Mr Reay
I notice my mistakes also.
Cody
Australia

In article ,
Stan Brown   wrote:
>Is 8 over -4 the same as -8 over 4?

	It's an interesting question, isn't it?  I can imagine
sharing an #8 loss among 4 people to get #2 loss each, but not
#8 among -4 people.  At some level, we have to say "Them's the
rules", and list what you can do to "sharings";  we have lost
the associated intuition.  Similarly, we can divide 7 by 3.5,
but not share 7 cakes among 3.5 people;  not, anyway, until
we get to "full-time equivalent" people.  As mathematicians,
we get used to "the rules", and often forget how difficult
these concepts are to "the man in the street", who knows his
maths only through the intuitive applications.

-- 
Andy Walker, School of MathSci., Univ. of Nott'm, UK.
anw@maths.nott.ac.uk

"Stan Brown"  wrote in message 
news:MPG.2079f2be9d79e39298aadb@news.individual.net...
> I've added the attribution you omitted. Please show proper
> attributions in your follow-ups.
> http://oakroadsystems.com/genl/unice.htm#attrib
>
> Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
>> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown 
>> :
>> > Is 8 over -4 the same as -8 over 4?
>>
>> first i think no, but each answer is -2?
>> so must be yes.
>>
>> the answer I got is shown at the top
>> the answer in the book is below
>> http://cjoint.com/data/ebp1xokLvv.htm
>> I mean are the two answers the same?
>
> This HTML business is getting pretty old. It's not hard to post
> algebra.

Hmm - it's not always easy.  In this case, the minus sign (in the final line 
of the solution) is shown in front of the dividing line, not against 
denominator or numerator.  Seeing it hand-written helps me to understand why 
the OP may take more than a moment to get his head round this.

- ( ( b - c ) / a ) may not mean a lot to someone meeting, say, negative 
denominators for the first time.

And as Andy Walker implies, the clever bit (and the crucial bit in an 
education group) is identifying exactly what is creating the problem / 
confusion / misunderstanding.

-- 
Martin

[Remove barrier to reply]

> Hmm - it's not always easy.  In this case, the minus sign (in the final
line
> of the solution) is shown in front of the dividing line, not against
> denominator or numerator.  Seeing it hand-written helps me to understand
why
> the OP may take more than a moment to get his head round this.

Thank you for your contribution Martin,
As Cody's father, I encourage him to be independent
and therefore seek his own resolution.
The problem deepens because I left school at 15 and whilst
I (think) Ican handle his work thus far, it was I, who marked his
answer as incorrect.
He thus asked me why, I can't answer it, hence my suggestion
he ask on the web.
His work comes in booklets, which parents have to mark and sign
Take for example his latest work going to a new level.

simultaneous linear equations in two levels.
http://cjoint.com/data/eddh7BHu2E.htm

The method is explained, but the mechanics or the practicalities
are not suggested, supplementary reading/info is not suggested either.
He is going this alone therefore, apart from a few minutes he can
snatch form his tutor, who he sees once a week.

I understand the regime of having the student think problems
through, but sometimes one needs an explanation to "kick start"
a process. I'm sure at the moment these are just heiroglyphs to him.
I remember he had a problem long ago with long division,
after I found the problem located in the extension of zeroes
concept, he rattled through his work.

BTW I still am unable to explain to him, why the negative
sign changes postion from the denomonator to the divisor line :)
Every question he asks, he gets a question in response.

We are probably meddling in the domain of the intelligentsia,
we need a link to be amongst others of similar progress,
but I am dashed if I can find one at the moment.
Rodney

"Rod"  wrote in message 
news:4611ad80_1@news.iprimus.com.au...
>> Hmm - it's not always easy.  In this case, the minus sign (in the final
> line
>> of the solution) is shown in front of the dividing line, not against
>> denominator or numerator.  Seeing it hand-written helps me to understand
> why
>> the OP may take more than a moment to get his head round this.
>
> Thank you for your contribution Martin,
> As Cody's father, I encourage him to be independent

That's obviously good.  And it'll help enormously if Cody can maintain 
enthusiasm and excitement at discovering new stuff.  Never be disheartened 
when facing a tough problem, and don't move on until you've cracked it.

> simultaneous linear equations in two levels.
> http://cjoint.com/data/eddh7BHu2E.htm

In the UK, the first problem (re-arranging an equation) and this 
simultaneous equation would normally not be encountered until age 14+.

> but sometimes one needs an explanation to "kick start"
> a process. I'm sure at the moment these are just heiroglyphs to him.

My approach (wherever possible) has always been to make sure the student 
understands the nature of the problem before diving into the algebra.  So 
with simultaneous equations, I would want to ensure Cody understands, and 
can sketch, a straight-line graph (hence understanding  " y = mx + c "  and 
what the m and c refer to).  Then add a second line, and recognise the 
significance of the point (i.e. co-ordinates) where the two lines cross.

I've found this kind of approach means the student understands what will be 
discovered by solviing two simul.eqs. (i.e. the only values of x and y which 
"satisfy" both equations).  And where a graphical method isn't so helpful, 
use some other kind of "real world" example.  So when introducing the idea 
of equations (before moving on to to re-arrange them) I always sketch a 
see-saw and tease out some "rules" about what can be added, subtracted etc. 
to each end to keep it balanced.  IME, a "negative sack of potatoes" soon 
starts to be a comfortable concept!

> BTW I still am unable to explain to him, why the negative
> sign changes postion from the denomonator to the divisor line :)

And it's even harder to explain in a post !  In the original post, the 
example has several "right" answers;  the "preferred" answer is partly a 
matter of opinion and partly "tidyness".  It certainly doesn't, though, 
involve putting the variable in alphabetic order, as suggested in your 3rd 
post.

> We are probably meddling in the domain of the intelligentsia,
> we need a link to be amongst others of similar progress,

Have you seen this site?
http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

If you hunt around, you'll find stuff on the basics of negative numbers, 
algebra, simul. eqs. etc.  It may help....

Just one note of caution - everyone has different ideas on how best to teach 
/ explain stuff, and if the student encounters too many different approaches 
/ styles, it can get confusing.  But rather than stick to one approach, 
understand there are many, hunt them down, then go with the method that 
works for you.

And never forget the answer in the back of the book is sometimes wrong...!!

Would be interested to know how you get on :-)

-- 
Martin

[Remove barrier to reply]

"Martin"  wrote in
message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

Cheers Martin,
I'll print out your suggestions.
I do notice another 2 books in, which he will attempt
next week, involves algebra to determine area of triangles
etc & etc, so finally he will have some meaning attached
to the characters.

> In the UK, the first problem (re-arranging an equation) and this
> simultaneous equation would normally not be encountered until age 14+.

Apparently here, he is just one year ahead, according to the Japanese
"Kumon" system we have purchased.
We felt we had to follow supplementary instruction, we had him
at school in Bangkok, in order to ingrain his Thai language first,
so his return by grade 3/4 in Aust, he was at that time unable to
speak English
He has caught up well in all his school work, although I am not
convinced he "likes" math, although he somehow knows it is important
and he enjoys helping other kids in class with math, and that is marvelous
for his self esteem.


> It certainly doesn't, though,
> involve putting the variable in alphabetic order, as suggested in your 3rd
> post.
I think Cody must have mis understood  tutor advice, it was he
who told me it had to be so.

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

Thank you very much indeed, Martin.

Best Regards,
Rodney
Perth.

"Martin" 
wrote in message news:oUpQh.4553$e9.319@newsfe6-gui.ntli.net...

> Have you seen this site?
> http://www.bbc.co.uk/schools/gcsebitesize/maths/index.shtml

I found that just a bit basic Martin, however, spurred on
by your graph suggestion, and a simple Google string,
Serendipity played her hand :)

http://www.themathpage.com/alg/simultaneous-equations.htm

and a graph to boot!

Thanks very much, (again)
Rodney.

Sun, 1 Apr 2007 08:35:38 +0800 from Rod :
> 
> My name is cody
> 12yo in year 7 primary school.
> perth australia
> 
> can someone explain the story
> to end up with the correct answer
> to this one
> http://cjoint.com/data/dFlxPGB5yb.htm
> 
> I do not know how the negative arrives
> in the centre of the fraction in the answer.

You had -ax on the left-hand side. What must you divide by to get x? 
Do you divide by a or by -a?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

Thank you Mr Brown.
divide by -a

Is b-c over -a
the same as negative b-c over a ?

eg -(b-c over a)
this confuses me

thank you
cody





> You had -ax on the left-hand side. What must you divide by to get x?
> Do you divide by a or by -a?
>
> --
> Stan Brown, Oak Road Systems, Tompkins County, New York, USA
>                                   http://OakRoadSystems.com/

"Rod"  wrote in message 
news:460f28d3$1_1@news.iprimus.com.au...
>
> Thank you Mr Brown.
> divide by -a
>
> Is b-c over -a
> the same as negative b-c over a ?
>
> eg -(b-c over a)
> this confuses me

Yes, or at least it is if you include the ( )  ie:


(b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )

One way to avoid the problem of dealing with the negative signs is always to 
ensure that the terms in x are positve.

eg:

-ax = b-c

add ax to both sides:


0 = b-c+ax

-b both sides

-b = - c + ax

+ b both sides

c-b = ax

/a both sides

(c-b)/a  = x

--
73
Brian
www.g8osn.org.uk

Thank you Mr Reay
then I got the answer right.
here is the question and answer given
which does not show the brackets
http://cjoint.com/data/ebitEB4dMe.htm

It is "kumon" and I have a tutor but i have to line up
with some 50 kids to get answers.

i get most right but the negative place here
looks wrong

is the book answer wrong?

also you say c-b we have to put in alphabetic b-c



"Brian Reay" <brian.reay@(spamstopper)bigfoot.com> wrote in message
news:xRHPh.232$r4.15@newsfe1-gui.ntli.net...
>
> "Rod"  wrote in message
> news:460f28d3$1_1@news.iprimus.com.au...
> >
> > Thank you Mr Brown.
> > divide by -a
> >
> > Is b-c over -a
> > the same as negative b-c over a ?
> >
> > eg -(b-c over a)
> > this confuses me
>
> Yes, or at least it is if you include the ( )  ie:
>
>
> (b-c)/-a   =  -(b-c)/a  (also = (c-b)/a )
>
> One way to avoid the problem of dealing with the negative signs is always
to
> ensure that the terms in x are positve.
>
> eg:
>
> -ax = b-c
>
> add ax to both sides:
>
>
> 0 = b-c+ax
>
> -b both sides
>
> -b = - c + ax
>
> + b both sides
>
> c-b = ax
>
> /a both sides
>
> (c-b)/a  = x
>
> --
> 73
> Brian
> www.g8osn.org.uk
>
>
>

you made a miistake
+c both sides

> > + b both sides

"Rod"  wrote in message 
news:460f5168$1_1@news.iprimus.com.au...
> you made a miistake
> +c both sides
>
>> > + b both sides

Well spotted- it is early on a Sunday morning here and I've not had a cuppa 
yet ;-)

Good luck with your studies.

--
73
Brian
www.g8osn.org.uk

> Well spotted- it is early on a Sunday morning here and I've not had a
cuppa
> yet ;-)
>
> Good luck with your studies.

Thank you Sir.

Sun, 1 Apr 2007 11:39:11 +0800 from Rod :
> 
> Thank you Mr Brown.
> divide by -a
> 
> Is b-c over -a
> the same as negative b-c over a ?

Is 8 over -4 the same as -8 over 4?

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/

> Is 8 over -4 the same as -8 over 4?

first i think no, but each answer is -2?
so must be yes.

the answer I got is shown at the top
the answer in the book is below
http://cjoint.com/data/ebp1xokLvv.htm
I mean are the two answers the same?
i am more worried about the position of the
negative sign than the answer maybe?

thank you Mr Brown

"Rod"  wrote in message 
news:460fbb55_1@news.iprimus.com.au...
 > i am more worried about the position of the
> negative sign than the answer maybe?


Sometimes you are simply looking at a presentation convention eg

-8/2  =  8/-2    (-4 in both cases)

However,  -8/2 is often considered "better" presentation wise.

Likewise:

-b + c =  c-b

But c-b is  often considered "better" presentation wise.

There are harder examples (eg "rationalising surds") but worry about those 
when you get to secondary school.

-- 
73
Brian, G8OSN
www.g8osn.org.uk

I've added the attribution you omitted. Please show proper 
attributions in your follow-ups.
http://oakroadsystems.com/genl/unice.htm#attrib

Sun, 1 Apr 2007 22:04:16 +0800 from Rod :
> Sun, 1 Apr 2007 09:24:51 -0400 from Stan Brown :
> > Is 8 over -4 the same as -8 over 4?
> 
> first i think no, but each answer is -2?
> so must be yes.
> 
> the answer I got is shown at the top
> the answer in the book is below
> http://cjoint.com/data/ebp1xokLvv.htm
> I mean are the two answers the same?

This HTML business is getting pretty old. It's not hard to post 
algebra. You are asking whether
     (b-c) / (-a)          and        -(b-c) / a
are the same.

Again, I'll answer that question with a question: are
     (15-7) / (-4)         and        -(15-7) / 4
the same?

(Note: It's not always safe to answer questions like these with an 
example. But if you choose your numbers so that nothing is 0 or 1, 
and maybe even try a couple of examples, it can be a helpful guide.)

> i am more worried about the position of the
> negative sign than the answer maybe?

Meaning no disrespect, I think you need to review the basics of 
division and multiplication with signed numbers. When I see a student 
confused by different ways of presenting the same thing, sometimes 
it's just a momentary perplexity but often the student just doesn't 
understand the underlying principles.

-- 
Stan Brown, Oak Road Systems, Tompkins County, New York, USA
                                  http://OakRoadSystems.com/