Thread: System of three equations and four variables

How to solve this system?

Thanks a bunch!

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...

Originally Posted by Wilmer

ONE question: are you serious?

Why not? Is this problem too easy?

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?

That's a bizarre system of equations...unfortunately it's not symmetric so you cannot assume that the 's are ordered.

Originally Posted by richard1234

That's a bizarre system of equations...unfortunately it's not symmetric so you cannot assume that the 's are ordered.

agrees... any new ideas?

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?

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...

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).

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

BUT BUT not so "interesting" is:
solve sqrt(u^2*x^2-(u-x)^2)-sqrt(u^2*w^2-(u-w)^2)-sqrt(w^2*x^2-(w-x)^2)=0 for x - Wolfram|Alpha

Any new ideas?

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...

You could set all the 's equal to get a solution, but you have four variables/three equations.

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!

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
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.