first take first two equations and solve them by eliminating one variable, then take second two, and ultimately u wil get values of all three
2007-11-27 18:16:44
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answer #1
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answered by Ahmed Zia 3
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2x-3y+z=13 --(1)
x-2y-z=6 --(2)
5x+y+z=14--(3)
(2) + (3)
6x - y = 20 --(4)
(1) + (2)
3x - 5y = 19 --(5)
from (4) => y = 6x -20
by substituting this to (5) we can get
3x - 5(6x-20) = 19
3x - 30x + 100 = 19
27x = 81 ==> x=3
y = 6*3 - 20 ==> y= -2
from (2)
z = x - 2y -6
z = 3 + 4 - 6 ==> z=1
2007-11-28 02:27:32
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answer #2
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answered by Natali 2
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I must agree this is poorly worded.
I think you may wish to use Gauss-Jordan elimination on this problem.
( 1 -2 -1 | 6 )
( 2 -3 1 | 13 )
( 5 1 1 | 14 )
r2 -> r2-2r1
r3 -> r3-5r1
( 1 -2 -1 | 6 )
( 0 1 3 | 1 )
(0 11 6 | -16 )
r1 -> r1+2r2
r3 -> r3-11r2
(1 0 5 | 8 )
( 0 1 3 | 1 )
(0 0 -27 | -27 )
r3 -> r3 / -27
(1 0 5 | 8 )
( 0 1 3 | 1 )
(0 0 1 | 1 )
r1 -> r1-5r3
r2 -> r2-3r3
(1 0 0 | 3 )
( 0 1 0 | -2 )
(0 0 1 | 1 )
So
x = 3, y = -2, z = 1
2007-11-28 02:30:23
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answer #3
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answered by Ian 6
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I don't know exactly what you're asking, but here one of many ways to solve this system of equations:
2x-3y+z=13
{isolate z}
z=13-2x+3y
x-2y-z=6
{isolate z}
z=x-2y-6
{because z=z}
13-2x+3y=x-2y-6
3x-y=19
y=3x-19
{plug that back in to z=x-2y-6}
z=x-2(3x-19)-6
z=32-5x
{plug those into 5x+y+z=14}
5x+(3x+19)+(32-5x)=14
3x+13=14
3x=1
x=1/3
{plug that into z=32-5x and y=3x-19}
z=32-5(1/3)=91/3 {or 30+1/3 if you prefer)
y=3(1/3)-19=1-19= -18
so x=1/3 z=30+1/3 and y= -18
2007-11-28 02:34:41
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answer #4
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answered by Grim Strategy 2
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2x-3y+z=13 equation I
x-2y-z=6 equation II
5x+y+z=14 equation III
2x-3y+z=13 equation I
x-2y-z=6 equation II
-----------------------
3x - 5y = 19
x-2y-z=6 equation II
5x+y+z=14 equation III
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6x -y = 20
3x - 5y = 19
6x - y = 20 (5)
3x - 5y = 19
-30x + 5y = -100 (-5)
----------------------------
-27x = -81
x = 3
6x - y = 20
6(3) - y = 20
-y = 20 - 18
y = -2
x-2y-z=6
3 - 2(-2) - z = 6
3 + 4 -z = 6
-z = 6 -7
z = 1
x = 3; y = -2; z = 1
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2007-11-28 02:28:12
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answer #5
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answered by aeiou 7
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2x-3y+z=13
x-2y-z=6
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3x-5y=19
x-2y-z=6
5x+y+z=14
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6x-y=20
3x-5y=19
6x-y=20
----------------
-6x+10y=-38
6x-y=20
------------------
9y=-18
y=-2
x=3
z=1
2007-11-28 02:16:21
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answer #6
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answered by iyiogrenci 6
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