How to solve this system?

Attachment 24308

Thanks a bunch!

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- Jul 18th 2012, 07:12 AMstudentttSystem of three equations and four variables
How to solve this system?

Attachment 24308

Thanks a bunch! - Jul 18th 2012, 03:47 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... - Jul 18th 2012, 09:43 PMstudentttRe: System of three equations and four variables
- Jul 18th 2012, 09:58 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 $\displaystyle x_i$'s are ordered.

- Jul 18th 2012, 11:55 PMstudentttRe: System of three equations and four variables
- Jul 19th 2012, 04:54 AMWilmerRe: System of three equations and four variables
- Jul 19th 2012, 04:57 AMWilmerRe: System of three equations and four variables
- Jul 19th 2012, 05:16 AMstudentttRe: System of three equations and four variables
- Jul 19th 2012, 05:56 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 - Jul 19th 2012, 06:13 AMWilmerRe: System of three equations and four variables
- Jul 24th 2012, 01:49 PMstudentttRe: System of three equations and four variables
Any new ideas?

- Jul 24th 2012, 04:34 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... - Jul 24th 2012, 05:58 PMrichard1234Re: System of three equations and four variables
You could set all the $\displaystyle x_i$'s equal to get

*a*solution, but you have four variables/three equations. - Jul 24th 2012, 07:48 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! - Jul 25th 2012, 12:50 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.