3x + 2y + z = 9
3 - 4y - z = 5
5x + y + 3z = 24
2007-12-09 16:22:58 · 5 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
X= 2
Y=-1
Z=5
2007-12-09 16:31:27 · answer #1 · answered by dark_x2000 2 · 0⤊ 0⤋
Assume:-
3x + 2y + z = 9
3x - 4y - z = 5----ADD
6x - 2y = 14
9x - 12y - 3z = 15
5x + y + 3z = 24------ADD
14x - 11y = 39
6x - 2y = 14
14x - 11y = 39
66x - 22y = 154
-28x + 22y = - 78----ADD
38x = 76
x = 2
12 - 2y = 14
- 2y = 2
y = - 1
6 - 2 + z = 9
z = 5
x = 2 , y = - 1 , z = 5
2007-12-09 21:21:36 · answer #2 · answered by Como 7 · 2⤊ 0⤋
there are many ways to solve this problem, but i think we should use matrices. but first, it must be in the common form.
the only one that is not in this form is the second one.
3-4y-z=5
-4y-z=5-3=2
0x-4y-z=2
so now we can use matrices
{3,2,1} [x]
{0,-4,-1} [y]
{5,1,3} [z]
that whole thing=
(9)
(5)
(24)
inverse of the first matrice is:
.48,.22,-.09
.22,-.17,-.13
-.87,-.3,.52
this matrice times the third matrice results to:
3.2599995
-1.9899999999...
3.1499996
x=the first number
y=the second
z=the third
bye
i think i messed up somewher cuz that dosnt make sense
sry
but i hope this answer is rite i didn't try inputing the values
2007-12-09 16:47:17 · answer #3 · answered by Harris 6 · 0⤊ 0⤋
The second equation. Is the first term '3' or '3x'? As is, your equation is 4y+z= -2
2007-12-09 16:51:50 · answer #4 · answered by vcs7578 5 · 0⤊ 0⤋
what are you solving for? all of the variables or just one?
2007-12-09 16:30:09 · answer #5 · answered by tucey003 1 · 0⤊ 0⤋