SOLUTION: Find the solution to the system of equations: (x - 2y + z = -3 |y + 2z = -5 |x + y + 3z = -6 (

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Question 1175220: Find the solution to the system of equations:
(x - 2y + z = -3
|y + 2z = -5
|x + y + 3z = -6
(

Found 3 solutions by greenestamps, MathLover1, MathTherapy:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


There are many methods -- elimination, Cramer's rule, matrices with Gauss-Jordan....

Using Cramer's rule is a well-defined process; if you want to solve the problem that way, just do it.

Using Gauss-Jordan elimination is also a well-defined process, but there are multiple possible paths to the solution, and the process is prone to simple arithmetic errors.

Given no instruction on what method to use, I would solve the problem algebraically using elimination.

Since one of the equations involves only two of the three variables, I would solve that equation for one variable in terms of the other and then use substitution to reduce the problem to two equations in two unknowns.

Presumably if you are working on a problem like this you know how to solve a system of two linear equations; so I leave the last part of the work to you.

y%2B2z+=+-5
y+=+-5-2z

Then

x-2%28-5-2z%29%2Bz+=+-3
x%2B5z+=+-13 [1]

and

x%2B%28-5-2z%29%2B3z+=+-6
x%2Bz+=+-1 [2]

Solve [1] and [2] for x and z; then find y from one of the given equations.

Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
x+-2y+%2B+z+=+-3 .....eq.1
y+%2B+2z+=+-5.....eq.2
x+%2B+y+%2B+3z+=+-6.....eq.3
-------------------------------------

y+%2B+2z+=+-5.....eq.2, solve for y
y++=-+2z+-5.........a)

go to
x+-2y+%2B+z+=+-3 .....eq.1....substitute y
x+-2%28-+2z+-5%29+%2B+z+=+-3
x+%2B4z+%2B10+%2B+z+=+-3
x+=+-5z-13........b)

go to
x+%2B+y+%2B+3z+=+-6.....eq.3...substitute y
x+-+2z+-5+%2B+3z+=+-6
x++-5+%2B+z+=+-6
x++=+5-6-z
x++=+-1-z........c)

from b) and c) we have
+-5z-13=-1-z.....solve for z
+1-13=5z-z
+-12=4z
+z=-12%2F4
+z=-3

go to
y++=-+2z+-5.........a)..substitute z
y++=-+2%2A%28-3%29+-5
y++=6+-5
y++=1
go to
x++=+-1-z........c)..substitute z
x++=+-1-%28-3%29
x++=+-1%2B3
x++=+2


solution set:
x+=+2
+y+=+1
+z+=+-3

Answer by MathTherapy(10552) About Me  (Show Source):
You can put this solution on YOUR website!
Find the solution to the system of equations:
(x - 2y + z = -3
|y + 2z = -5
|x + y + 3z = -6
This is NOT as COMPLEX as the other "respondents" make it seem! 
EASISET/QUICKEST/LEAST TIME-CONSUMING method, as follows: 

 x - 2y +  z = - 3 ------ eq (i)
      y + 2z = - 5 ------ eq (ii)
 x +  y + 3z = - 6 ------ eq (iii)
     3y + 2z = - 3 ------ Subtracting eq (i) from eq (iii) ------ eq (iv)
2y = 2 ------- Subtracting eq (ii) from eq (iv)


1 + 2z = - 5 ------- Substituting 1 for y in eq (ii)
2z = - 5 - 1
2z = - 6


x - 2(1) + - 3 = - 3 ------- Substituting 1 for y in eq (ii)
x - 2 - 3 = - 3
x - 5 = - 3