I have to try and solve this system of equations but don't know where to start
4u + v = 2y + 2z = j + k
2u - 2v = 2y - 2z = j - 2k
2u + 2v = 4y - z = j - k
That is is 6 equations in 6 variables.
First separate the equations:
(1) 4u+ v= 2y+ 2z
(2) 2u- 2v= 2y- 2z
(3) 2u+ 2v= 4y- z
(4) 2y+ 2z= j+ k
(5) 2y- 2z= j- 2k
(6) 4y- z= j- k
Add 2 times (1) to (2) to get 10u= 6y+ 2z or (7) 5u= 3y+ z
Add (2) and (3) to get (8) 6u= 6y- 3z .
Multiply (7) by 6 to get 30u= 18y+ 6z and multiply (8) by 5 to get 30u= 30y- 3z.
That gives 18y+ 6z= 30y- 3z which reduces to 12y= 9z or 4y= 3z. Replacing y by (3/4) z, 30u= (45/2)z- 3z= (39/2)z so that u= (39/30)z= (13/10)z. Replace u and y by those in (1) and solve for v as multiple of z. You can now replace y in (4), (5), and (6) by (3/4)z to get three equations in z, j, and k. Solve those three equations for the three values, then use that value of z to find y, u, and v.
thanks.
Here what i've done
(1) 4u + v= 2y + 2z
(2) 2u - 2v= 2y - 2z
(3) 2u + 2v= 4y - z
(4) 2y + 2z= j + k
(5) 2y - 2z= j - 2k
(6) 4y - z= j - k
2*(1)+(2) 10u = 6y + 2z ____ 5u = 3y + z (7)
Should that actually beAdd (2) and (3) to get (8) 6u= 6y- 3z .
(2)+(3) 4u= 6y - 3z (8)
4*(7) 20u = 12y + 4z (9)
5*(8) 20u = 30y - 15z (10)
so 12y + 4z = 30y - 15z
19y = 18z
y = (18/19)z
substituting this in (10) 20u = (540/19)z - 15z
u = (51/76)z
substituting u and y in (1) v = (23/19)zReplace u and y by those in (1)and solve for v as multiple of z.
y = (18/19)zYou can now replace y in (4), (5), and (6)by (3/4)z to get three equations in z, j, and k.
(4) 2y + 2z= j+ k _______ j + k = (74/19)z
(5) 2y - 2z= j- 2k ______ j - 2k = -(2/19)z
(6) 4y - z= j- k ________ j - k = (53/19)z
(4)+(6) j = (127/38)z
substituting j in (5) k = 131/76 z
What values of y, u and v am I solving for?Solve those three equations for the three values, then use that value of z to find y, u, and v.
I already have
y = (18/19)z
u = (51/76)z
v = (23/19)z
I know the values for j, k, u, v and y that I have are only correct in (1) but don't know what I've done wrong.
[edit]
My original problem with this system was that I thought I only had 3 equations no matter what I did with them. i.e.
(1) 4u + v= 2y + 2z
(4) 2y + 2z= j + k
isn't this just one equation, telling me that 4u + v = j + k ?
Thanks for your help
Where d'heck do these equations come from?!
(1) 4u+ v= 2y+ 2z
(2) 2u- 2v= 2y- 2z
(3) 2u+ 2v= 4y- z
An integer solution for these 3: u=15, v = 14, y=19, z=18
As far as j and k goes, these seem to have been thrown in there in a "hope for the best" manner:
in other words, there' s something wrong with them....but I can't make it out exactly...
Are you SURE there is a solution?
The last 3 equations are a dependant system. (As are the first 3)
Solutions for these three:
2y + 2z = j + k (1)
2y - 2z = j - 2k (2)
4y - z = j - k (3)
(1)-(2) 4z=3k (4)
2*(1)-(3) 5z=j+3k (5)
substituting z=(3/4)k in (5) gives j=(3/4)k so j=z
substituting these back into (1) y=(1/8)k
so one of the infinite solutions would be y=1 z=6 j=6 k=8
I think I've finally figured that it can't be figured
4u + v = 2y + 2z = j + k
2u - 2v = 2y - 2z = j - 2k
2u + 2v = 4y - z = j - k
Set 1
4u + v = 2y + 2z
2u - 2v = 2y - 2z
2u + 2v = 4y - z
valid when...
..... v = (14/15)u
..... y = (19/15)u
..... z = (6/5)u
Set 2
2y + 2z = j + k
2y - 2z = j - 2k
4y - z = j - k
valid when...
..... y = (1/8)k
..... z = (3/4)k
..... j = (3/4)k
Set 3
4u + v = j + k
2u - 2v = j - 2k
2u + 2v = j - k
valid when...
..... u = (9/8)k
..... v = (1/4)k
..... j = (15/4)k
I think this means the 3 sets cannot be correct simultaneously since we have
..... j = (3/4)k
..... j = (15/4)k
and also
..... v = (14/15)u where 14*(9/8)k <> 15*(1/4)k
Am I correct that there is no solution (apart from the obvious zero solutions)?
Thanks