Solve
4x+2y+3z=6
3x+y+2z=8
x+y+z=5
2006-11-20 08:48:23 · 5 answers · asked by Erica C 1 in Science & Mathematics ➔ Mathematics
I have gone over the problem twice. This system of linear equations has NO SOLUTION!!! This means that not even a single set of points (x,y,z) exists that can satisfy ALL three equations simultaneously. Here is my work to prove it:
(1) 4x+2y+3z=6
(2) 3x+y+2z=8
(3) x+y+z=5
(1) 4x+2y+3z=6
(2) 3x+y+2z=8 (multiply by -2, and then add equations to cancel y)
4x+2y+3z=6
-6x-2y-4z=-16
----------------------> (4) -2x-z=-10
rearrange this ---> (4) z=10-2x
(2) 3x+y+2z=8
(3) x+y+z=5 (multiply by -2, and then add equations to cancel y)
3x+y+2z=8
-2x-2y-2z= -10
----------------------> (5) x-y=-2
rearrange this ---> (5) y = 2+x
Plug (4) and (5) into (1):
4x+2(2+x)+3(10-2x)=6
4x+4+2x+30-6x=6
34 ≠ 6
Therefore, there's NO SOLUTION!!!!!!
2006-11-20 08:50:58 · answer #1 · answered by عبد الله (ドラゴン) 5 · 0⤊ 0⤋
Solve for first 2
4x+2y+3z=6
3x+y+2z=8 <--multiply by -2 to get rid of y
4x+2y+3z=6
-6x-2y-4z=-16 <--now add both
-2x - z = -10
z = -2x + 10
Now solve last 2.
3x+y+2z=8
x+y+z=5 <--multiply by -2 to get rid of z and solve for y in terms of x
3x+y+2z=8
-2x-2y-2z=-10 <-add
x - y = -2
y = x+2
Plug in y and z solutions into first equation.
Remember: y = x+2, z = -2x + 10
4x + 2y + 3z = 6
4x + 2(x+2) + 3(-2x+10) = 6
4x + 2x + 4 - 6x + 30 = 6
34 = 6 ?
No solution.
2006-11-20 17:03:03 · answer #2 · answered by bourqueno77 4 · 0⤊ 0⤋
X+Y+Z=1/4
X+Y+Z=1
X+Y+Z=5 simpilest already
2006-11-20 17:10:20 · answer #3 · answered by Paula S 1 · 0⤊ 1⤋
first of all, what grade are you in?
second of all, have you ever heard of www.cosmeo.com?
If not, go there and try their free trial. That place can help you with anything you need!!
2006-11-20 16:58:03 · answer #4 · answered by Jackie 2 · 0⤊ 1⤋
do your own homework if you ever want to get a good job!!
2006-11-20 16:56:32 · answer #5 · answered by Anonymous · 0⤊ 1⤋