Help‽ Solving algebraic equation in 2 variables‽

[Turtle]

No...

 

For example, if "otherstuff" = 20, and "a" = 2,

 

(20 - (2+4)) = 14

 

((20 - 2) +4) = 22

 

moo

 

Ahhhh....OK. Now I see the difference and I know how to get to the answer.

Take the original equation, drop the constant, let a=3 and b=4 and then solve. Whichever bracketing gives the answer 10 is correct.

 

((b/2)*((a-2)*;)-a+4)=?

 

Thanks for hangin' in moo!

[Turtle]

Ahhhh....OK. Now I see the difference and I know how to get to the answer.

Take the original equation, drop the constant, let a=3 and b=4 and then solve. Whichever bracketing gives the answer 10 is correct.

 

((b/2)*((a-2)*;)-a+4)=?

 

Thanks for hangin' in moo!

 

OK

((4/2)*((3-2)*4)-3+4)=

(2)*((1*4)-3+4)=

(2)* (4-3+4)=

(2)*(5)=

10

 

Is that helpful?

 

PS 3 am here; must get sleeeepp...Thanks for the help and I'll pick up tomorrow. Guten nacht.

[Jay-qu]

I updated post 13.. it all looks good to me, unless there is a flaw in the original equation i used, point it out and Ill re-do it if so ;)

[moo]

Well, with all brackets (hopefully) in place, the problem is:

 

((b/2) * ((((a - 2) * ;) - a) + 4)) = 68,921

 

And using these values, it adds up to 10:

a = 3

b = 4

 

((b/2) * ((((a - 2) * :hihi: - a) + 4)) = ?

((4/2) * ((((3 - 2) * 4) - 3) + 4)) = ?

( 2 * ((( 1 * 4) - 3) + 4)) = ?

( 2 * (( 4 - 3) + 4)) = ?

( 2 * (1 + 4)) = ?

( 2 * 5) = ?

(10) = ?

 

moo

[moo]

Hmmm... unless I (or my app) made a boo-boo, it seems to works with either of these:

 

a = 2

b = 68921

 

OR

 

a = 68921

b = 2

 

moo ;)

[moo]

But this may be more like the answer you're looking for. ;)

 

a = 86

b = 41

 

moo

[CraigD]

[math](b/2)*((a -2)*b) -a +4 = 68921[/math]

can rearrange to

…

[math]b=\frac{0+-\sqrt{-4(a-2)(-2a-137836)}}{2(a-2)}[/math]

…

[math]a=\frac{-413488+-\sqrt{170992174528}}{12}[/math]

which does not result in integers..

I agree. :thumbs_up The step I hinted at in post #4,

Next step, I’d substitute

c = 2a^2 +137830a -275668

isn’t necessary or desirable. (note that the only difference between Jay-qu’s solution for b and mine is that I reduce the fraction by canceling [math]\frac{\sqrt4}2[/math] – we both have the same number)

 

It’s interesting to turn this question around by generalizing its simplified form,

(a -2)*b^2 -2a = 137834

into

(a -2)*b^2 -2a = c

and asking “for what values of c are there integer solutions a and b?”

[CraigD]

[salivating]So your program...MUMPS? [/salivating] We may have more use of it as I mean to get at a general case of the equation …

For a “well-behaved” function like this one, an exhaustive search program is pretty simple. Here’s the one I used in, as Turtle guessed :), MUMPS:

f a=3:1 f b=3:1 s c=(b/2)*((a-2)*B)-a+4 w:c=68921 a," ",b,! q:c'<68921

It’s easy to “generalize by changing the constant”, and rerun – changing 68921 to 68920 gives a solution a=4056, b=6, changing it to 68922 doesn’t appear to give a solution, to 68923 gives a=43, b=58, etc.

[pgrmdave]

I'm just curious - why this particular equation? Is it just a random equation that you came up with, or does it have a real-world application?

[moo]

Tap tap tap... Is this thing on? B)

 

Thought I posted 3 different integer combinations that will work...

((b/2) * ((((a - 2) * :) - a) + 4)) = 68,921

 

a = 86

b = 41

 

OR

 

a = 2

b = 68921

 

OR

 

a = 68921

b = 2

moo

[CraigD]

I'm just curious - why this particular equation? Is it just a random equation that you came up with, or does it have a real-world application?

I’ve no idea how Turtle came up with this particular equation, but find it personally interesting because I think considering the general case of it and others like it lead, ultimately, to a deep/hard number theory problem that’s preoccupied me for years: an algorithm for finding the prime factorization of the sum of 2 numbers given their prime factorization, without simply evaluating, adding, and factoring the result.

 

I’ve hinted at this in several threads, especially 3313 and outright asked it in ”Re: How to Use The 4th Dimension to find your soulmate - IN ONE DAY!” in the same-named unlikely thread, but haven’t worked up the nerve to throw it into a thread of its own.

[pgrmdave]

Those solutions seem to work fine, but I think that Turtle also wants to see how you arrived at the solutions, with the full algebraic work done so that he can solve similar equations in the future.

[CraigD]

Thought I posted 3 different integer combinations that will work...

((b/2) * ((((a - 2) * B) - a) + 4)) = 68,921

You’re using a different equation than the rest of us. We’re using:

((b/2) *   ((a - 2) * :)   -a   + 4)  = 68921

moo’s using:

((b/2) * ((((a - 2) * B)   - a) + 4)) = 68921

[moo]

I think that Turtle also wants to see how you arrived at the solutions, with the full algebraic work done so that he can solve similar equations in the future.

True, but I thought perhaps knowing the answer might help to work out the method to get there (maybe not?). I just used a quick hack to run them down.

 

moo

[moo]

You’re using a different equation than the rest of us. We’re using:

((b/2) *   ((a - 2) * B)   -a   + 4)  = 68921

moo’s using:

((b/2) * ((((a - 2) * :)   - a) + 4)) = 68921

 

If you check my first post in this thread, I discussed with Turtle how the "-a+4" part should be framed. B)

 

moo

[Qfwfq]

The two equations Craig posted are equivalent. The two that Moo posted aren't.

 

N - (a + 4) is like N - a -4

[CraigD]

If you check my first post in this thread, I discussed with Turtle how the "-a+4" part should be framed. ;)

Indeed you did, and I can see how you arrived at ((b/2)*((((a -2)*B) -a) +4)) = 68921 from that discussion.

 

However, the fact remains that the original equation

((b/2)*((a-2)*:)-a+4)=68921

Isn’t equivalent to

((b/2)*((((a-2)*B)-a)+4))=68921.

 

This post (red added by me)

OK

((4/2)*((3-2)*4)-3+4)=

(2)*((1*4)-3+4)=

(2)* (4-3+4)=

(2)*(5)=

10

is incorrect. The red parenthesis are not matching, so shouldn’t have been removed. It should read something like:

((4/2)*((3-2)*4)-3+4) =

((2)*((1)*4)-3+4) =

(2*(1*4)-3+4) =

(2*(4)-3+4) =

(2*4-3+4) =

(8-3+4) =

9

 

The impact of changing the equation in this way is more than just that of changing a constant so that a solution exists

((b/2)*((((a-2)*B)-a)+4))=68921 simplifies to

(a -2)b^2 -ab -137838 = 0

which is importantly different than ((b/2)*((a-2)*B)-a+4)=68921, which simplifies to

(a -2)*b^2 +0b -(2a +137834) =0

 

It’s still an interesting equation, but not the same as the original in post #3.