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Question

I have to solve this system:

 

3x+y+z=13

2x+2y-3z=3

3x+y-z=7

Answer 1

3X + Y + Z = 13

2X + 2Y - 3Z = 3

3X + Y   - Z   = 7 

 

 Subtract 1st and 3rd equation:

 

   3x + Y + Z = 13

-(3X + Y  - Z  = 7)

2Z = 6

 

Z = 6/2 = 3

Substitute Z = 2 in first and 2nd equation:

 

-2(3X + Y = 10)

   2X + 2y = 12

 -4x = -8

 

  X = 2 

  

  6 + Y + 3 = 13

 

     Y = 4