using these three variable can anyone solve this 3x3 system using linear inequalities???
2007-01-10 13:58:39 · 5 answers · asked by princess 1 in Science & Mathematics ➔ Mathematics
x= 25/28
y= -102/28 or -51/14
z= -39/28
2007-01-10 14:22:31 · answer #1 · answered by Arash J 2 · 0⤊ 0⤋
x - 2y + 3z = 4
2x + 4y - 2z = -10
5x - 3y + z = 14
x - 2y + 3z = 4
2x + 4y - 2z = -10
Multiply top by 2
2x - 4y + 6z = 8
2x + 4y - 2z = -10
4x + 4z = -2
2x + 2z = -1
2x + 4y - 2z = -10
5x - 3y + z = 14
Multiply top by 3 and bottom by 4
6x + 12y - 6z = -30
20x - 12y + 4z = 56
26x - 2z = 26
13x - z = 13
-z = -13x + 13
z = 13x - 13
4x + 4z = -2
x + z = (-1/2)
z = -x - (1/2)
13x - 13 = -x - (1/2)
26x - 26 = -2x - 1
28x = 25
x = (25/28)
z = -x - (1/2)
z = (-25/28) - (1/2)
z = (-25 - 14)/28
z = (-39/28)
x - 2y + 3z = 4
(25/28) - 2y + 3(-39/28) = 4
(25/28) - 2y - (117/28) = 4
-2y - (92/28) = 4
-2y = 4 + (92/28)
-2y = (112 + 92)/28
-2y = (204/28)
-2y = (51/7)
y = (-51/14)
ANS :
x = (25/28)
y = (-51/14)
z = (-39/28)
2007-01-11 00:46:28 · answer #2 · answered by Sherman81 6 · 0⤊ 0⤋
x=-59/4
y=-51/7
z=-39/28
2007-01-10 22:23:40 · answer #3 · answered by dark magician girl 1 · 0⤊ 0⤋
Same answer as Arash J.
2007-01-10 22:33:13 · answer #4 · answered by Anonymous · 0⤊ 0⤋
Sorry, i can´t help! I´m just about to learn something like that stuff in school!
2007-01-10 22:03:16 · answer #5 · answered by wanna b me 1 · 0⤊ 0⤋