/sci/ - Math problem

Name: Anonymous 2007-02-26 18:13 ID:o+9UCkUh

Hay guys how do I solve for this? a,b,c,d are constants and I have to solve for v,x,y,z.

v (x + y + z) = a
x (v + y + z) = b
y (v + x + z) = c
z (v + x + y) = d

Name: Anonymous 2007-02-26 18:33 ID:fqioNTT7

STEP ONE
vx + vy + vz = a
xv + xy + xz = b
yv + yx + yz = c
zv + zx + zy = d

Name: Anonymous 2007-02-26 19:11 ID:S9j8eDwH

Solve for one variable, substitute into the next, until you have an expression for every individual variable.

Name: Anonymous 2007-02-26 20:31 ID:o+9UCkUh

Okay I tried working it out and eventually got to

a = v/2 { 3(a+v) +- sqrt[(a+v)^2 - 4b] +- sqrt[(a+v)^2 - 4c] +- sqrt[(a+v)^2 - 4d] }

How do I solved for v? I am pretty sure this is the wrong way to do it though...

>>2
That makes it so you have 6 variables (m = xy, n = vy and so on) but only 4 equations.

Name: Anonymous 2007-02-26 20:33 ID:o+9UCkUh

BTW I got the equation from this:

v + x + y + z = a/v + v = b/x + x = c/y + y = d/z + z

I just set x,y,z in terms of v and then inserted that into v (x + y + z) = a.

Name: Anonymous 2007-02-27 9:42 ID:AEHHVCfT

Halp

Name: Anonymous 2007-02-27 11:33 ID:F6bMbtxR

vxyz(v+x+y+z) = axyz + v^2xyz = bvyz + vx^2yz = cvxz + vxy^2z = dvxy + vxyz^2

v^2xyz + vx^2yz + vxy^2z + vxyz^2 = axyz + v^2xyz

vx^2yz + vxy^2z + vxyz^2 = axyz

a = vx + vy + vz

Using a similar method, you can see b = xv + xy + xz, c = yv + yx + yz, d = zv + zx + zy.

Name: Anonymous 2007-02-27 11:51 ID:Heaven

>>7
Wow I'm really stupid. Solved for the constants instead of the variables. Never mind ;x

Name: Anonymous 2007-02-27 11:53 ID:F6bMbtxR

>>8
Ok, real solution here:

vx^2yz + vxy^2z + vxyz^2 = axyz
v(x^2yz + xy^2z + xyz^2) = axyz
v(xyz(x + y + z)) = axyz
v = a/(x+y+z).

Similarly, x = b/(v+y+z), y = c/(v+x+z), z = d/(v+x+y).

Name: Anonymous 2007-02-27 12:22 ID:Heaven

>>9
i lol'd

Name: Anonymous 2007-02-27 15:41 ID:AEHHVCfT

>>9
You are not supposed to have variables in terms of other variables. You are supposed to have the variables in terms of the constants.

Name: Anonymous 2007-02-27 16:46 ID:7jGZv00m

>>4

multiplication creates variables?

Name: Anonymous 2007-02-27 23:30 ID:AEHHVCfT

>>12
xy is not going to be constant certainly. I just meant that you can't solve the equations using linear algebra techniques.

Name: Anonymous 2007-02-28 15:03 ID:b7+yBL5Q

equation := {v(x+y+z) = a, x(v+y+z) = b, y(v+x+z) = c, z(v+x+y) = d}

{v(x + y + z) = a, x(v + y + z) = b, y(v + x + z) = c, z(v + x + y) = d}

solve(equations, {v,x,y,z})

piecewise([equations <> 0, {}], [equations = 0, {[v = z4, x = z1, y = z2, z = z3]}])

mupad tells me that it can't be solved through normal stuff for x,v,c,z != 0

Name: Anonymous 2007-02-28 15:57 ID:qewDUfSh

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483 Name: Anonymous : 2007-02-28 12:57 ID:zH5nuAqs

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Name: Anonymous 2007-03-01 4:42 ID:lb1giXHj

HATE NIGGERS?

Name: Anonymous 2007-03-01 19:58 ID:IkB/yKp4

simotaneous equations bro
L2 do math....

Name: Anonymous 2007-03-01 23:11 ID:t1F9uL6V

Well, you have 4 unknowns (n) and 4 equations (m), so m=n, and you should be able to solve it using linear calculus. Seek the inverse triangle...

Name: Anonymous 2007-03-02 9:22 ID:eeRp4OYr

>>18
Why don't you try doing it fag, if you think you are so smart. PROTIP: These equations are not linear. So you have to some weird fucking methods to solve it.

Name: Anonymous 2007-03-02 9:38 ID:SuxS0WQv

Just make a matrix. Reduce to row echelon form. and then solve simultaneously. easy? I think so.

Name: Anonymous 2007-03-02 10:35 ID:Heaven

>>20
Hint: It's called linear algebra because it works with simultaneous LINEAR equations.

Name: Anonymous 2007-03-02 19:40 ID:VVCENZt2

>>20
Lol. GTFO /sci/

Math problem

1 Name: Anonymous 2007-02-26 18:13 ID:o+9UCkUh

2 Name: Anonymous 2007-02-26 18:33 ID:fqioNTT7

3 Name: Anonymous 2007-02-26 19:11 ID:S9j8eDwH

4 Name: Anonymous 2007-02-26 20:31 ID:o+9UCkUh

5 Name: Anonymous 2007-02-26 20:33 ID:o+9UCkUh

6 Name: Anonymous 2007-02-27 9:42 ID:AEHHVCfT

7 Name: Anonymous 2007-02-27 11:33 ID:F6bMbtxR

8 Name: Anonymous 2007-02-27 11:51 ID:Heaven

9 Name: Anonymous 2007-02-27 11:53 ID:F6bMbtxR

10 Name: Anonymous 2007-02-27 12:22 ID:Heaven

11 Name: Anonymous 2007-02-27 15:41 ID:AEHHVCfT

12 Name: Anonymous 2007-02-27 16:46 ID:7jGZv00m

13 Name: Anonymous 2007-02-27 23:30 ID:AEHHVCfT

14 Name: Anonymous 2007-02-28 15:03 ID:b7+yBL5Q

15 Name: Anonymous 2007-02-28 15:57 ID:qewDUfSh

16 Name: Anonymous 2007-03-01 4:42 ID:lb1giXHj

17 Name: Anonymous 2007-03-01 19:58 ID:IkB/yKp4

18 Name: Anonymous 2007-03-01 23:11 ID:t1F9uL6V

19 Name: Anonymous 2007-03-02 9:22 ID:eeRp4OYr

20 Name: Anonymous 2007-03-02 9:38 ID:SuxS0WQv

21 Name: Anonymous 2007-03-02 10:35 ID:Heaven

22 Name: Anonymous 2007-03-02 19:40 ID:VVCENZt2