Can't remember how to solve equation with two variables

[tim9000]

Umm from memory I used to use...that triangle:
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Fibonachii was it? Pathetic I can't even remember the name.
To factorise...or was it expand...polynomials...anyway, I don't think that's elevant here.
My question is; I had an equation and it boild down to:
3R^2 + A*R - y*B - C*y^2 = 0

where A, B and C are constants.

Did I have to do completing the squares or some factorisation to find how R is proportional to y?

Thanks

 

[phinds]

You cannot solve a single equation with two variables, you have to have two equations but to get how one variable relates to the other, just move everything but the variable of interest to the other side of the equation by doing the appropriate add, subtract, multiply, divide.

For example, if you had Ax + By = 0, and you wanted to know how y was related to x, you would subtract off By then divide by A and get

Ax = -By

x = -(B/A)y

 

[tim9000]

You cannot solve a single equation with two variables, you have to have two equations but to get how one variable relates to the other, just move everything but the variable of interest to the other side of the equation by doing the appropriate add, subtract, multiply, divide.

For example, if you had Ax + By = 0, and you wanted to know how y was related to x, you would subtract off By then divide by A and get

Ax = -By

x = -(B/A)y

That wouldn't work in this case, the best I could do would be

3R^2 + A*R = y*B + C*y^2

isn't it?

Ah, ok, so bottom line, I need to use simultanious eqations?

Thanks

 

[Mark44]

That wouldn't work in this case, the best I could do would be

3R^2 + A*R = y*B + C*y^2

isn't it?

You could solve the equation above for R in terms of y, by using the Quadratic Formula. Start by moving all terms to the left side.
$3R^2 + AR - yB - Cy^2 = 0$

In the Quadratic Formula, a = 3R², b = A, and c = -yB - Cy².

In post #1 you said you were trying to show that R is proportional to y -- that's not the case in the equation above. If two variables are proportional, then either one is some constant multiple of the other.

Ah, ok, so bottom line, I need to use simultanious eqations?

Thanks

 

[phinds]

Ah. I missed the y^2. Just use the quadratic equation, as Mark suggested.

 

[Svein]

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Fibonachii was it?

Pascal.

 

[Mark44]

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Fibonachii was it?

Pascal.

And for the record, Fibonacci is the correct spelling, not Fibonachii. In Italian, 'c' followed immediately by 'i' or 'e' has the ch sound. If followed by 'a', 'o', or 'u', it has the k sound.

 

[tim9000]

And for the record, Fibonacci is the correct spelling, not Fibonachii. In Italian, 'c' followed immediately by 'i' or 'e' has the ch sound. If followed by 'a', 'o', or 'u', it has the k sound.

Aha, thanks for the tip, I'm not great with English, let alone Latin.

 

[tim9000]

Pascal.

AAH yep, that's it.

 

[tim9000]

You could solve the equation above for R in terms of y, by using the Quadratic Formula. Start by moving all terms to the left side.
$3R^2 + AR - yB - Cy^2 = 0$

In the Quadratic Formula, a = 3R², b = A, and c = -yB - Cy².

In post #1 you said you were trying to show that R is proportional to y -- that's not the case in the equation above. If two variables are proportional, then either one is some constant multiple of the other.

That would be helpful to solve for R or y using the quadratic formula.
So are you saying this relation regardless of anything, will never be proportional, and the best I can do is solving for something like this:

R = √((12(yB+Cy^2)/36)
or
R = - √((2A^2 + 12(yB+Cy^2)/36)

?
Thanks

 

[SteamKing]

Aha, thanks for the tip, I'm not great with English, let alone Latin.

Fibonacci is not Latin, it's Italian.

https://en.wikipedia.org/wiki/Fibonacci

 

[tim9000]

Fibonacci is not Latin, it's Italian.

https://en.wikipedia.org/wiki/Fibonacci

Isn't Italian just modern Latin?

Would it have been ok if I had of said
'I'm not great with English, let alone Latin derivatives'?

 

[SteamKing]

Isn't Italian just modern Latin?

No, not really. If that were true, you could say the same about French or Spanish or Portuguese, all of which derive from Latin like Italian, but are distinct from it in their own separate ways. This group of languages is called the Romance Languages, because they each derive from the language of the Romans (Latin).

 

[tim9000]

No, not really. If that were true, you could say the same about French or Spanish or Portuguese, all of which derive from Latin like Italian, but are distinct from it in their own separate ways. This group of languages is called the Romance Languages, because they each derive from the language of the Romans (Latin).

Ah I have heard that term actually. I wonder when at what point each started to diverge into what you'd distinguish as a separate entity, ah well I don't want to get too far off topic. (though I doubt there are near as many things taken from Latin verbally, discounting nouns, in French, as there would be in Italian, or even Spanish)

 

[Mark44]

That would be helpful to solve for R or y using the quadratic formula.
So are you saying this relation regardless of anything, will never be proportional, and the best I can do is solving for something like this:

R = √((12(yB+Cy^2)/36)
or
R = - √((2A^2 + 12(yB+Cy^2)/36)

?
Thanks

Yes, that's what I'm saying. In the equation you posted, R and y are not proportional.

 

[tim9000]

Yes, that's what I'm saying. In the equation you posted, R and y are not proportional.

Ok, I'll see if I can come up with another equation to solve it simultaniously.
Thank you

 

[Mark44]

Ah I have heard that term actually. I wonder when at what point each started to diverge into what you'd distinguish as a separate entity, ah well I don't want to get too far off topic. (though I doubt there are near as many things taken from Latin verbally, discounting nouns, in French, as there would be in Italian, or even Spanish)

I would guess that French has about as many words that are recognizably of Latin origin as Italian, Spanish, or Portuguese. Although Spanish and Italian share words that are identical or close (casa means house in Italian and Spanish, but the s is pronounced differently; cortar (Sp.) and cortare (It.) -- to cut), there are many words that are very different, due to the influence on Spanish by the Moors for 700 years.