# System of three equations and four variables

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• Jul 18th 2012, 07:12 AM
studenttt
System of three equations and four variables
How to solve this system?
Attachment 24308
Thanks a bunch!
• Jul 18th 2012, 03:47 PM
Wilmer
Re: 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 PM
studenttt
Re: System of three equations and four variables
Quote:

Originally Posted by Wilmer
ONE question: are you serious?

Why not? Is this problem too easy?

Quote:

Originally Posted by Wilmer
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...

Three equations (square after elimination process) with 6 variables (instead of 4 initial variables) and new roots (after being squared) gives nothing good :(
Any new ideas?
• Jul 18th 2012, 09:58 PM
richard1234
Re: 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 PM
studenttt
Re: System of three equations and four variables
Quote:

Originally Posted by richard1234
That's a bizarre system of equations...unfortunately it's not symmetric so you cannot assume that the \$\displaystyle x_i\$'s are ordered.

agrees... any new ideas?
• Jul 19th 2012, 04:54 AM
Wilmer
Re: System of three equations and four variables
Quote:

Originally Posted by studenttt
Why not? Is this problem too easy?

No; the other way: it appears impossible, with 4 unknowns and 3 equations.
Do you have any background...like, where does it come from?
• Jul 19th 2012, 04:57 AM
Wilmer
Re: System of three equations and four variables
Quote:

Originally Posted by studenttt
Three equations (square after elimination process) with 6 variables (instead of 4 initial variables) and new roots (after being squared) gives nothing good :(
Any new ideas?

Nope. And it'll only get worse...
• Jul 19th 2012, 05:16 AM
studenttt
Re: System of three equations and four variables
Quote:

Originally Posted by Wilmer
Do you have any background...like, where does it come from?

No background... teacher took it from his book (wrapped, so I do not know what book it was).
• Jul 19th 2012, 05:56 AM
Wilmer
Re: 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&#40;a&#94;2-b&#94;2&#41;-sqrt&#40;c&#94;2-d&#94;2&#41;-sqrt&#40;e&#94;2-f&#94;2&#41;&#61;0 for a - Wolfram|Alpha
• Jul 19th 2012, 06:13 AM
Wilmer
Re: System of three equations and four variables
• Jul 24th 2012, 01:49 PM
studenttt
Re: System of three equations and four variables
Any new ideas?
• Jul 24th 2012, 04:34 PM
Wilmer
Re: 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 PM
richard1234
Re: 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 PM
Wilmer
Re: 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 AM
studenttt
Re: System of three equations and four variables
Quote:

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

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