SOLUTION: Can someone help me? I need to write the final answer as an ordered triple (x,y,z) Solve the system for x,y,z 2x+3y-z=4 3x-y+2z=5 x-4y+3z=1 I tried it, but ended up weird.

Algebra ->  Coordinate Systems and Linear Equations  -> Lessons -> SOLUTION: Can someone help me? I need to write the final answer as an ordered triple (x,y,z) Solve the system for x,y,z 2x+3y-z=4 3x-y+2z=5 x-4y+3z=1 I tried it, but ended up weird.       Log On


   

Question 1090977: Can someone help me?
I need to write the final answer as an ordered triple (x,y,z)
Solve the system for x,y,z
2x+3y-z=4
3x-y+2z=5
x-4y+3z=1
I tried it, but ended up weird.
Thank you.
Found 3 solutions by Alan3354, MathLover1, MathTherapy:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
I tried it, but ended up weird.
----------
What does that mean?
what did you try?

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the system for x,y,z
2x%2B3y-z=4.....eq.1
3x-y%2B2z=5.....eq.2
x-4y%2B3z=1.....eq.3
------------------------------------------
2x%2B3y-z=4.....eq.1...solve for z
z=2x%2B3y-4.....eq.(1z)
3x-y%2B2z=5.....eq.2..solve for z
2z=-3x%2By%2B5
z=-3x%2F2%2By%2F2%2B5%2F2....eq.(2z)
x-4y%2B3z=1.....eq.3..........solve for z
3z=-x%2B4y1
z+=+-x%2F3+%2B+%284+y%29%2F3+%2B+1%2F3....eq.(3z)

from eq.(1z) and eq.(2z):
2x%2B3y-4=-3x%2F2%2By%2F2%2B5%2F2...solve for x
4x%2B6y-8=-3x%2By%2B5
4x%2B3x=y-6y%2B5%2B8
7x=-5y%2B5%2B8
x=-5y%2F7%2B13%2F7.....(a)
from eq.(2z) and eq.(3z):
-3x%2F2%2By%2F2%2B5%2F2=-x%2F3+%2B+%284+y%29%2F3+%2B+1%2F3....both sides multiply by 6
-9x%2B3y%2B15=-2x+%2B+8+y+%2B+2
-8y%2B3y%2B15-2=-2x+%2B+9x
-5y%2B13=7x
x=-5y%2F7%2B13%2F7.....(b)
3x-y%2B2z=5.....eq.2....substitute x and z
3%28-5y%2F7%2B13%2F7%29-y%2B2%28-3x%2F2%2By%2F2%2B5%2F2%29=5.....eq.2
highlight%28y+=+13%2F5+-+%287+x%29%2F5%29-> real solution
z=2x%2B3y-4.....eq.(1z)......substitute y
z=2x%2B3%2813%2F5+-+%287x%29%2F5%29-4
highlight%28z++=+19%2F5+-+%2811x%29%2F5%29-> real solution
highlight%28x+=+5n+%2B+4%29, highlight%28y+=+-7n+-+3%29, highlight%28z+=+-11n+-+5%29, where n element Z
integer solutions:
(x,y,z)=(5n+%2B+4, -7n+-+3, -11n+-+5)
find some real take real solutions and equal to zero:
+y=13%2F5+-+%287+x%29%2F5
+13%2F5+-+%287+x%29%2F5=0...solve for x
+13+-+7x=0
+13+=7x
highlight%28x=13%2F7%29
plug in
z=19%2F5+-+%2811x%29%2F5%29
z=19%2F5+-+%2811%2813%2F7%29%29%2F5%29
highlight%28z=-2%2F7%29
find y:
y+=+13%2F5+-+%287%2813%2F7%29%29%2F5
highlight%28y+=+0%29

real solutions:(x,y,z)=(13%2F7,0,-2%2F7)

or
if n=0
(x,y,z)=(+4,+-+3,+-+5)
if n=1
(x,y,z)=(9, -10, -16) ...and so on


Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Can someone help me?
I need to write the final answer as an ordered triple (x,y,z)
Solve the system for x,y,z
2x+3y-z=4
3x-y+2z=5
x-4y+3z=1
I tried it, but ended up weird.
Thank you. 
2x + 3y -  z = 4 -------- eq (i)

3x -  y + 2z = 5 -------- eq (ii)

 x - 4y + 3z = 1 -------- eq (iii)



I can clearly see what you're saying!

The EASIEST method to be used here is ELIMINATION, as follows:



1) Multiply eq (i) by 2 to get: 4x + 6y - 2z = 8 ---- eq (iv)

2) Add eqs (ii) & (iv) to get: 7x + 5y = 13 --------- eq (v)		



1) Multiply eq (i) by 3 to get: 6x + 9y - 3z = 12 --- eq (vi)

2) Add eqs (iii) & (vi) to get: 7x + 5y = 13 -------- eq (vii)	



Do you notice anything? If not, it's clear that eqs (v) & (vii) are the same

Therefore, there are an INFINITE number of solutions to this system!



If you want to see what I'm talking about, substitute a value for x in eq (v) or (vii), and then solve for y. You will then have values for x and y.

Substitute these 2 values in any of the 3 ORIGINAL equations (I would choose eq 1 since that equation has a z-value with a coefficient of - 1), and then solve for z. 

You now have values for x, y, and z. 

Substitute these 3 values into all 3 ORIGINAL equations, and you will see that these values satisfy all 3. 

You can follow the same process if you so choose to use a different value for x.