Question 146538
This is a set of linear equation in x,y,z.
1. x - y + z = 5     ..........(1)
  2y + 3z = 14.............(2)
 -3y + 2z =5...............(3)

eq: (2 ) & (3) contains two variables y & z, make any one of the variable's coefficient same & eliminate.
  (2) x 3 & (3) x 2, we get
  2y x 3 + 3z x 3 = 14 x 3      ...........(4)
  -3y x2 +2z x 2 = 5 x 2...................(5)

i.e. 6y +9z = 42 ........(4)
    -6y + 4z = 10 .........(5)
(4) +(5) ,when you add the like terms of this two eq: y term gets eliminated. i.e.  13z = 52 ; z = 52 / 13 = 4. now substitute z=4  in (2) & get the value of y.
i.e 2y + 3 x 4 =14.
  2y +12 = 14 
  2y = 14 - 12 = 2;  y = 2/2 =1.
substitute  z=4 &y=1 in eq: (1)
  x - 1 +4 = 5
 x + 3 = 5 
  x = 5-3 =2. 
the solution ; x= 2, y = 1 & z= 4.
( you can cross check your answer by substituting all three values in any of the given eq: i.e. from (1)  x - y +z= 2 - 1 + 4 = 1 + 4 = 5 , so answer is correct)

2. 2x + 2y + z = -5.  .........(1)
  2x + y + 3z = 7 .......(2)
  -4x -2y -6z = -14. ...........(3)

solve (1) & (2), since coefficients of x are same by subtracting eliminate x
  (1) - (2),
y -2z = -12; y = -12 +2z ..........(4)
now take eq: (1)& (3), coefficients of y are same, eliminate y
    - 2x -5z = -19
       -2x = -19 + 5z
     x= 19- 5z/2  .........(5)substituting (4) & (5) in (1) & solve
  2x + 2y + z = -5
  2 ( 19- 5z/2) + 2 ( -12 +2z) + z = -5.
   19 - 5z  - 24 +4z +z = -5.
  -5z +4z +z + 19 -24 = -5
       -5z+ 5z - 5 = -5
    0= 0. , this shows that the equations has infinite solutions.
here eq: (2) & (3) are one and same line .( multiply (2) with -2 , we get eq: (3), i.e. these two equations are same).  such  linear equations has infinite solutions.
( before solving system of equation , verify whether given lines are same, if not then proceed for solution. above steps are for clearing doubts, no need to do it if lines are same)