Question 1054759
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1. Solve any equation for any letter.
2. Substitute what that letter equals for that letter in the 
other two equations and simplify.

[Now we'll have only a system in 2 equations and 2 letters]

3. Solve either equation for either letter.
4. Substitute what that letter equals for that letter in the 
other equation in 2 letters and simplify.
5. Substitute that in the other letter in one of the 
other equations in two letters and solve for a
second letter.
6. Finally substitute the numbers for the two letters
you have in one of the original equations and solve
for the last remaining letter:

(eq. 1)   2x -  y -  z =  15
(eq. 2)   4x + 5y + 2z =  10
(3q. 3)   -x - 4y + 3z = -20

1. Solve any equation for any letter.

I'll pick an easiest equation to solve for an easy letter.

Hmmm.  Think I'll pick eq. 1 to solve for, hmmm, y

(eq. 1)   2x -  y -  z =  15
               -y      =  15 - 2x + z
                y      = -15 + 2x - z

2. Substitute what that letter equals for that letter in the 
other two equations and simplify:

(eq. 2)   4x + 5y + 2z =  10
          4x + 5(-15 + 2x - z) + 2z = 10
          4x - 75 + 10x - 5z + 2z = 10
          14x - 3z - 75 = 10
(eq. 4)   14x - 3z = 85


(3q. 3)   -x - 4y + 3z = -20
          -x - 4(-15 + 2x - z) +3z = -20
          -x + 60 - 8x + 4z + 3z = -20
          -9x + 7z + 60 = -20
(eq. 5)   -9x + 7z = -80

Now we have only a system in 2 equations and 2 unknowns:

(eq. 4)   14x - 3z =  85
(eq. 5)   -9x + 7z = -80

3. Solve either equation for either letter.

I'll pick the easier equation to solve for the easier letter.

(eq. 4)   14x - 3z =  85
          -3z = 85 - 14x
            z = -85/3 + (14/3)x

4. Substitute what that letter equals for that letter in the 
other equation and simplify:
   
(eq. 5)   -9x + 7z = -80
          -9x + 7[-85/3 + (14/3)x] = -80
          -9x - 595/3 + (98/3)x = -80
          -27x - 595 + 98x = -240
          71x - 595 = -240
          71x = 355
          x = 5

5. Substitute that in the other letter in one of the 
other equations in two letters and solve for a
second letter:

I'll substitute x = 5 in eq. 5:

(eq. 5)   -9x + 7z = -80     
          -9(5) + 7z = -80
          -45 + 7z = -80
          7z = =35
           z = -5

6. Finally substitute the numbers for the two letters
you have in one of the original equations and solve
for the last remaining letter:

I'll pick eq. 3 to substitute x = 5 and z = -5 in

(3q. 3)   -x - 4y + 3z = -20
          -(5) - 4y + 3(-5) = -20
          -5 - 4y - 15 = -20
          -4y - 20 = -20
          -4y = 0
            y = 0

Solution: (x,y,z) = (5,0,-5) 

Follow those same steps in your second problem.

Answers are (x,y,z) = (-6,-20,8)

However your third problem is different. 

 x + 3y -  z = 12 
2x + 4y - 2z =  6 
-x - 2y +  z = -6

1. Solve any equation for any letter.

I'll pick the first to solve for x

x = 12 - 3y + z

2. Substitute what that letter equals for that letter in the 
other two equations and simplify.

2(12 - 3y + z) + 4y - 2z = 6
24 - 6y + 2z + 4y - 2z = 6
24 -2y = 6
-2y = -18
y = 9

-x - 2y +  z = -6
-(12 - 3y + z) - 2y +  z = -6
-12 + 3y - z - 2y +  z = -6
-12 + y = -6
y = 6
  
y cannot equal both 6 and 9.

So the system is inconsistent.  It has no solution.

Edwin</pre><b>