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Please write in Algebraically method?
2x-3y+z=13
x-2y-z=6
5x+y+z=14

2007-11-27 18:07:28 · 6 answers · asked by I Rock! 1 in Science & Mathematics Mathematics

6 answers

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 · answer #1 · answered by Ahmed Zia 3 · 0 0

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 · answer #2 · answered by Natali 2 · 0 0

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 · answer #3 · answered by Ian 6 · 0 0

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 · answer #4 · answered by Grim Strategy 2 · 0 0

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
---------------------
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 · answer #5 · answered by aeiou 7 · 0 0

2x-3y+z=13
x-2y-z=6
------------------------
3x-5y=19



x-2y-z=6
5x+y+z=14
---------------------
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 · answer #6 · answered by iyiogrenci 6 · 1 0

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