How to solve this system?

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Thanks a bunch!

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- July 18th 2012, 08:12 AMstudentttSystem of three equations and four variables
How to solve this system?

Attachment 24308

Thanks a bunch! - July 18th 2012, 04:47 PMWilmerRe: System of three equations and four variables
ONE question: are you serious?

Well, I'd attack that (if attackable!) this way:

you have terms that contain :

x1x2; let a = that term

x1x3; let b...

x1x4; let c...

x2x3; let d...

x2x4; let e...

x3x4; let f...

So the equations can be rewritten:

sqrt(c) - sqrt(b) = sqrt(f) [1]

sqrt(c) - sqrt(a) = sqrt(e) [2]

sqrt(b) - sqrt(a) = sqrt(d) [3]

Put those 3 equations through the elimination process...and see what happens... - July 18th 2012, 10:43 PMstudentttRe: System of three equations and four variables
- July 18th 2012, 10:58 PMrichard1234Re: System of three equations and four variables
That's a bizarre system of equations...unfortunately it's not symmetric so you cannot assume that the 's are ordered.

- July 19th 2012, 12:55 AMstudentttRe: System of three equations and four variables
- July 19th 2012, 05:54 AMWilmerRe: System of three equations and four variables
- July 19th 2012, 05:57 AMWilmerRe: System of three equations and four variables
- July 19th 2012, 06:16 AMstudentttRe: System of three equations and four variables
- July 19th 2012, 06:56 AMWilmerRe: System of three equations and four variables
Don't give up! I'm playing with it now; getting interesting stuff!

I'll let you know later...

Have a look here; that's why I say "interesting":

solve sqrt(a^2-b^2)-sqrt(c^2-d^2)-sqrt(e^2-f^2)=0 for a - Wolfram|Alpha - July 19th 2012, 07:13 AMWilmerRe: System of three equations and four variables
- July 24th 2012, 02:49 PMstudentttRe: System of three equations and four variables
Any new ideas?

- July 24th 2012, 05:34 PMWilmerRe: System of three equations and four variables
Not from me!

I make that a "lost cause": 3 equations, 4 variables; you need minimum of 4 equations.

Also, almost impossible to manipulate even without the square roots.

And all of the "math heavyweights" at this site have (wisely!) stayed away from this mess... - July 24th 2012, 06:58 PMrichard1234Re: System of three equations and four variables
You could set all the 's equal to get

*a*solution, but you have four variables/three equations. - July 24th 2012, 08:48 PMWilmerRe: System of three equations and four variables
Or one way is to set each term equal to a square, and get:

c^2 = a^2 + b^2

c^2 = d^2 + e^2

b^2 = d^2 + f^2

...in other words, that's 3 right triangles, possibly:

1073^2 = 448^2 + 975^2

1073^2 = 495^2 + 952^2

975^2 = 495^2 + 840^2

Then, using my c^2 (as example):

x1^2 x4^2 / 4 - (x1 - x4)^2 = 1073^4

But that also turns out as ugly as the 7 witches of MacBeth! - July 25th 2012, 01:50 AMstudentttRe: System of three equations and four variables
Do you think we really

**need minimum of 4 equations**? I know a lot of examples when it is possible to solve a system of eqns where a number of eqns is less than a number of variables. E.g.:

1) One eq and 2 vars: (x-2)^2+(y-3)^2=0. Answer: x=2, y=3

2) System of two eqns and three variables:

(x-2)^2+(y-3)^2+(z-4)^2=0

x+y+z=9

Answer: x=2, y=3, z=4.

Sometimes (a case when a number of eqns is less than a number of variables) we just have more than one solution. E.g.: system of two eqns and three variables:

x+y+z=5

x+y-z=3

Answer: (x,y,z) may take the form (x,4-x,1) so we get many roots.