There are at least 6 different methods to solve systems of equations.
You will always get the same answer regardless of which of the methods you use.
I have no way of know which of the 6 methods you are studying.
The six different methods are

1. Substitution
2. Elimination
3. Gaussian elimination
4. Cramer's rule
5. Gauss-Jordan matrix method
6. Inverse matrix method

I used only the most basic one, substitution.  That may or may not be the
method you are studying, nor the easiest, but you can certainly understand it,
because it is very much like the substitution method when there are only
2 equations and 2 unknowns.

1. Solve the first equation for "a".
2. Substitute the expression that you get for "a" in the other two equations.
   That gives you two equations in only "b" and "c".
3. Solve one of those for b.
4. Substitute the expression that you get for "b" in the second equation.
   That gives you one equation in only "c".
5. Solve that and get the value for "c"
6. Substitute that value for "c" in one of the equations in only "b" and "c".  
7. Solve that and get the value for "b".
8. Substitute the values for "b" and"c" in one of the original equations.
9. Solve for "a".

1. a+3b+2c = 3
         a = 3-3b-2c

2.         2a-b-3c = -8
   2(3-3b-2c)-b-3c = -8
      6-6b-4c-b-3c = -8
           6-7b-7c = -8
            -7b-7c = -14   <--divide through by -7
               b+c = 2

           5a+2b+c = 9
   5(3-3b-2c)+2b+c = 9
   15-15b-10c+2b+c = 9
         15-13b-9c = 9
           -13b-9c = -6 
              

3.             b+c = 2
                 b = 2-c

4.         -13b-9c = -6
       -13(2-c)-9c = -6

5.      -26+13c-9c = -6
            -26+4c = -6
                4c = 20
                 c = 5  

6.              b+c = 2
                b-5 = 2

7.              b+5 = 2
                  b = -3

8.     a+3b+2c = 3
  a+3(-3)+2(5) = 3
        a-9+10 = 3
           a+1 = 3
             a = 2

The solution is a=2, b=-3, c=5.

Sometimes that is written (a,b,c) = (2,-3,5)     

Edwin