SOLUTION: How do you solve these 3 linear equations using elimination method 2x-3y+z=13 x+y-2z=-6 3x-y+z=8

Line up the like symbols:

eq. 1   2x - 3y +   z =  13
eq. 2    x +  y -  2z =  -6
eq. 3   3x -  y +   z =   8

I notice that the y's will cancel out if I add the middle and bottom
equations

eq. 2    x +  y -  2z =  -6
eq. 3   3x -  y +   z =   8
        ------------------
eq. 4   4x      -   z =   2

I can also make the y's cancel out if I multiply eq. 2 by 3

eq. 5   3x + 3y -  6z = -18

then add eq. 1 to it

eq. 5   3x + 3y -  6z = -18
eq. 1   2x - 3y +   z =  13
        ------------------
eq. 6   5x      -  5z =  -5

I notice that I can divide eq. 6 through by 5

eq. 7    x      -   z =  -1

Now I have two equations with y's eliminated,  

eq. 4   4x      -   z =   2  
eq. 7    x      -   z =  -1

I can make the z's cancel if I multiply eq. 7 by -1

eq. 8   -x      +   z =   1

and add it to eq. 4

eq. 4   4x      -   z =   2
eq. 8   -x      +   z =   1
       --------------------
e. 9    3x            =   3

I notice that I can divide eq. 9 through by 3 

eq. 10   x            =   1

Now that I have 1 of the letters, I can switch to
substitution. I will use eq. 7

eq. 7    x      -   z =  -1
         1      -   z =  -1
                   -z =  -2
                    z =   2
Now that I have 2 of the letters, I can substitute in one
of the original equations, say, eq. 2 

eq. 2    x +  y -  2z =  -6
         1 +  y -2(2) =  -6
         1 +  y -   4 =  -6
              y -   3 =  -6
                    y =  -3 

(x,y,z) = (1,-3,2)
                    
Edwin

2x - 3y +  z =  13 ----- eq (i)
 x +  y - 2z = - 6 ----- eq (ii)
3x -  y +  z =   8 ----- eq (iii)
      x + 2y = - 5 ----- Subtracting eq (i) from eq (iii) ----- eq (iv)

4x - 6y + 2z = 26 ----- Multiplying eq (i) by 2 ----- eq (v)
     5x - 5y = 20 ----- Adding eqs (v) & (ii) 
    5(x - y) = 5(4)
       x - y = 4 ------ eq (vi)
          3y = - 9 ---- Subtracting eq (vi) from eq (iv)
         

   x - (- 3) = 4 ------ Substituting - 3 for y in eq (vi)
       x + 3 = 4
         

3(1) - (- 3) + z = 8 ----- Substituting - 3 for y and 1 for x in eq (iii)
   3 + 3 + z = 8
       6 + z = 8