find an algebraic expression equivalento to....

eddy2017 said:
I studied what you have written. I parsed every bit of info down. Still, I fail to see its application to the solution of the problem!. I do not see it at all.



\dfrac{y}{x} \equiv \dfrac{x}{y} * 1 \text { BECAUSE } a = a * 1 \text { for ANY number } a.xy≡yx∗1 BECAUSE a=a∗1 for ANY number a.
Furthermore

\dfrac{2y}{2y} = 1 \text { BECAUSE } \dfrac{b}{b} = 1 \text { for ANY number } b \ne 0.2y2y=1 BECAUSE bb=1 for ANY number b=0.
\therefore \dfrac{y}{x} = \dfrac{y}{x} * 1 = \dfrac{y}{x} * \dfrac{2y}{2y} = \dfrac{2y^2}{2xy}.∴xy=xy∗1=xy∗2y2y=2xy2y2.
Click to expand...
I'll track my thought process in detail

I am asked to find an equivalence between this sum [imath]\dfrac{y}{x} + \dfrac{3}{2y}[/imath] and ONE of four different fractions. But I first notice that the fractions being added do not have a common denominator so I cannot add them as they are initially AND all four proposed answers have [imath]2xy[/imath] in their denominators. Are these useful clues? The second clue suggests to me that I need 2xy in the denominator of the fractions to be added. And by the way, that would give them a PERTINENT common denominator. This is thinking in detective mode rather than any mathematical mechanics. Do you follow? I have not done a lick of mechanical work as yet. Moreover, I am now certain about what I am trying to do, at least as a first step.

How do I change the denominator of the first fraction to be added from x to 2xy without changing the fraction's value?

I know this logic:

[math]\dfrac{y}{x} \equiv \dfrac{y}{x} * 1 \equiv \dfrac{y}{x} * \dfrac{2y}{2y} \equiv \dfrac{2y^2}{2xy}.[/math]
Because I know it, I can take this short-cut

[math]\dfrac{y}{x} \equiv \dfrac{2y^2}{2xy}.[/math]
I have changed the denominator of the first fraction to be added to 2xy without changing the fraction's value.

How do I change the denominator of the second fraction to be added from 2y to 2xy without changing the fraction's value?

[math]\dfrac{3}{2y} \equiv \dfrac{3}{2y} * 1 \equiv \dfrac{3}{2y} * \dfrac{x}{x} \equiv \dfrac{3x}{2xy}.[/math]
Or, using the short-cut

[math]\dfrac{3}{2y} \equiv \dfrac{3x}{2xy}.[/math]
I have changed the denominator of the second fraction to be added to 2xy without changing the fraction's value.

I can add those two fractions into one because they have the same denominator.

[math]\dfrac{2y^2}{2xy} + \dfrac{3x}{2xy} \equiv \dfrac{2y^2 + 3x}{2xy}.[/math]
And that is one of the answers proposed so we are done.

I do not start with mechanics. I start with think about what the problem requires and what clues do I have before I deal with any mechanics. That way I know what mechanics are relevant and am not guessing.

Y todo como el diamante
Antes que luz es carbon.
 
JeffM said:
I'll track my thought process in detail

I am asked to find an equivalence between this sum [imath]\dfrac{y}{x} + \dfrac{3}{2y}[/imath] and ONE of four different fractions. But I first notice that the fractions being added do not have a common denominator so I cannot add them as they are initially AND all four proposed answers have [imath]2xy[/imath] in their denominators. Are these useful clues? The second clue suggests to me that I need 2xy in the denominator of the fractions to be added. And by the way, that would give them a PERTINENT common denominator. This is thinking in detective mode rather than any mathematical mechanics. Do you follow? I have not done a lick of mechanical work as yet. Moreover, I am now certain about what I am trying to do, at least as a first step.

How do I change the denominator of the first fraction to be added from x to 2xy without changing the fraction's value?

I know this logic:

[math]\dfrac{y}{x} \equiv \dfrac{y}{x} * 1 \equiv \dfrac{y}{x} * \dfrac{2y}{2y} \equiv \dfrac{2y^2}{2xy}.[/math]
Because I know it, I can take this short-cut

[math]\dfrac{y}{x} \equiv \dfrac{2y^2}{2xy}.[/math]
I have changed the denominator of the first fraction to be added to 2xy without changing the fraction's value.

How do I change the denominator of the second fraction to be added from 2y to 2xy without changing the fraction's value?

[math]\dfrac{3}{2y} \equiv \dfrac{3}{2y} * 1 \equiv \dfrac{3}{2y} * \dfrac{x}{x} \equiv \dfrac{3x}{2xy}.[/math]
Or, using the short-cut

[math]\dfrac{3}{2y} \equiv \dfrac{3x}{2xy}.[/math]
I have changed the denominator of the second fraction to be added to 2xy without changing the fraction's value.

I can add those two fractions into one because they have the same denominator.

[math]\dfrac{2y^2}{2xy} + \dfrac{3x}{2xy} \equiv \dfrac{2y^2 + 3x}{2xy}.[/math]
And that is one of the answers proposed so we are done.

I do not start with mechanics. I start with think about what the problem requires and what clues do I have before I deal with any mechanics. That way I know what mechanics are relevant and am not guessing.

Y todo como el diamante
Antes que luz es carbon.
Click to expand...
Jeff, sorry for the break I took. Reading your answer now. thank you!.

That was very good. It is clear now. Just a couple of questions so as to why I am doing something. ( I remember now someone that says that some things in math are so because they are so, and they are better to leave them alone and learn them as such without inquiring further because there is no rhyme or reason to why they are like that).
I don't think they are right that is why I am asking:

[math]y/x[/math] is equivalent to [math]y/x * 1[/math] and this is equivalent to [math]y/x * 2y/2y[/math]
what is the principle or law that permits me to do such a thing?. that is, to multiply [math]y/x * 2y/2y[/math]
Or is it arbitrary just to get [math]2y^2/2xy?[/math]
everything is clear, but I can't help but wonder. I am a guy who needs to know why is something like that, or by what power invested in me I can do that mathematical thing?
and thank you so much, Jeff, for taking your precious time to help and explain

I got it. You multiply whatever term is expedient to get to the desired denominator.
That is why you told me you were on detective mode and you had not done a lick of mathematical work yet.
 
eddy2017 said:
Jeff, sorry for the break I took. Reading your answer now. thank you!.

That was very good. It is clear now. Just a couple of questions so as to why I am doing something. ( I remember now someone that says that some things in math are so because they are so, and they are better to leave them alone and learn them as such without inquiring further because there is no rhyme or reason to why they are like that).
I don't think they are right that is why I am asking:

[math]y/x[/math] is equivalent to [math]y/x * 1[/math] and this is equivalent to [math]y/x * 2y/2y[/math]
what is the principle or law that permits me to do such a thing?. that is, to multiply [math]y/x * 2y/2y[/math]
Or is it arbitrary just to get [math]2y^2/2xy?[/math]
everything is clear, but I can't help but wonder. I am a guy who needs to know why is something like that, or by what power invested in me I can do that mathematical thing?
and thank you so much, Jeff, for taking your precious time to help and explain

I got it. You multiply whatever term is expedient to get to the desired denominator.
That is why you told me you were on detective mode and you had not done a lick of mathematical work yet.
Click to expand...
The last paragraph was perhaps the main point I was trying to make. Congratulations.

Your broader question is an excellent one. Probably my answer will annoy several true mathematicians, and they will disagree with me.

It seems plausible to me that arithmetic and geometry were first studied as experimental and observational sciences just as physics, chemistry, astronomy, and biology still are today. It was correct because experience showed it to be so. That is, mathematics showed general truths about the universe we live in, and the proof came in the form of agreement with practical
experience.

Few mathematicians have had that view for the last 150 years. Today most mathematicians believe that mathematics are interesting propositions deduced logically from a small set of propositions dependent on faith (the axioms). Mathematics need not have any connection to physical reality at all.

I suppose that we should take the mathematicians at their word and admit that mathematics is, at a formal level, ultimately based on logic applied to logically unverifiable propositions. That every number multiplied by 1 equals itself is never going to be proved by example.

However, I believe the mathematicians are being excessively hard on themselves. When we talk about arithmetic, the axioms are verifiable by experience repeated billions of time, samples sized far in excess of what satisfy physical scientists. Can you find any number that when multiplied by itself is not equal to itself?

Arithmetic and algebra are built up logically from a small set of axioms, which themselves cannot be proved logically, but that experience has shown to be without contradiction. I suggest that you look at a typical bunch of axioms used for arithmetic. Do you really have any doubt that they are true in the universe we inhabit?

 
JeffM said:
The last paragraph was perhaps the main point I was trying to make. Congratulations.

Your broader question is an excellent one. Probably my answer will annoy several true mathematicians, and they will disagree with me.

It seems plausible to me that arithmetic and geometry were first studied as experimental and observational sciences just as physics, chemistry, astronomy, and biology still are today. It was correct because experience showed it to be so. That is, mathematics showed general truths about the universe we live in, and the proof came in the form of agreement with practical
experience.

Few mathematicians have had that view for the last 150 years. Today most mathematicians believe that mathematics are interesting propositions deduced logically from a small set of propositions dependent on faith (the axioms). Mathematics need not have any connection to physical reality at all.

I suppose that we should take the mathematicians at their word and admit that mathematics is, at a formal level, ultimately based on logic applied to logically unverifiable propositions. That every number multiplied by 1 equals itself is never going to be proved by example.

However, I believe the mathematicians are being excessively hard on themselves. When we talk about arithmetic, the axioms are verifiable by experience repeated billions of time, samples sized far in excess of what satisfy physical scientists. Can you find any number that when multiplied by itself is not equal to itself?

Arithmetic and algebra are built up logically from a small set of axioms, which themselves cannot be proved logically, but that experience has shown to be without contradiction. I suggest that you look at a typical bunch of axioms used for arithmetic. Do you really have any doubt that they are true in the universe we inhabit?

Click to expand...
Thank you so much, Jeff. It is all I can say. I'm speechless at your understanding not only of Math but of the laws of the Universe and of the immensely finite knowledge we have of everything.
 
eddy2017 said:
Thank you so much, Jeff. It is all I can say. I'm speechless at your understanding not only of Math but of the laws of the Universe and of the immensely finite knowledge we have of everything.
Click to expand...
Thank you, but we all stand on the shoulders of giants.
 
eddy2017 said:
I went the easy road or at least what for me was the easy way. Harry the cat's way.
I saw that a common denominator to both fractions was"
CD=2xy
then I cross-multiplied
[math]y*2y =2y^2 and 3*x =3x=[/math]so,
[math](2y^2 +3x )/2xy[/math]
Click to expand...
I thought I'd share my thinking behind what is going on. It is not "cross-multiplying" that you are doing. "Cross-multiplying" is done across an = sign.

When simplifying \(\displaystyle \frac{y}{x}+\frac{3}{2y}\), first decide what common denominator you are going to use. The simplest (but not always best) way to get a common denominator, is to multiply the existing denominators together.

So, here, we'll use the common denominator \(\displaystyle x*2y\) or \(\displaystyle 2xy\).

Next, we have to change each fraction into an equivalent fraction with that common denominator. Equivalent fractions are obtained by multiplying the numerator and denominator of a fraction by the same thing.

So we want to change \(\displaystyle \frac{y}{x}\) into \(\displaystyle \frac {??}{2xy}\). How to determine \(\displaystyle ??\).
Well, what did you multiply the original denominator \(\displaystyle x\) by to get the new denominator \(\displaystyle 2xy\)? The answer is \(\displaystyle 2y\).
So, now you need to multiply the original numerator \(\displaystyle y\) by \(\displaystyle 2y\) as well and you get the new numerator of \(\displaystyle 2y^2\).

Go through a similar process with the second fraction.

And then add together.
 
Harry_the_cat said:
I thought I'd share my thinking behind what is going on. It is not "cross-multiplying" that you are doing. "Cross-multiplying" is done across an = sign.

When simplifying \(\displaystyle \frac{y}{x}+\frac{3}{2y}\), first decide what common denominator you are going to use. The simplest (but not always best) way to get a common denominator, is to multiply the existing denominators together.

So, here, we'll use the common denominator \(\displaystyle x*2y\) or \(\displaystyle 2xy\).

Next, we have to change each fraction into an equivalent fraction with that common denominator. Equivalent fractions are obtained by multiplying the numerator and denominator of a fraction by the same thing.

So we want to change \(\displaystyle \frac{y}{x}\) into \(\displaystyle \frac {??}{2xy}\). How to determine \(\displaystyle ??\).
Well, what did you multiply the original denominator \(\displaystyle x\) by to get the new denominator \(\displaystyle 2xy\)? The answer is \(\displaystyle 2y\).
So, now you need to multiply the original numerator \(\displaystyle y\) by \(\displaystyle 2y\) as well and you get the new numerator of \(\displaystyle 2y^2\).

Go through a similar process with the second fraction.

And then add together.
Click to expand...
Thank you. Very simple and to the point. I loved it.

I have to thank you all for the good advice and explanation
I have learned a few things out of this exercise. That is the magic of learning and you're making possible. Thank you all!
 
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