Since you have a system of 4 equations in 2 variables, I am going to find the solution for a system of 2 of them.  I will then check the solution in the other 2 equations.  If the solution does not work for all 4 equations, then there is no solution.

3x = y - 3                     Equation 1

2y = 6 - 3x                   Equation 2

2y = 6 - (y - 3)             Substitute the value of 3x in equation 1 for the 3x in equation 2

2y = 6 - y + 3               Distributive property (distribute the negative)

2y = 9 - y                     Simplify

2y + y = 9 - y + y          Add y to each side

3y = 9                          Simplify

3y/3 = 9/3                    Divide each side by 3

y = 3                            Simplify

3x= y - 3                       Equation 2

3x = 3 - 3                      Substitute the value of y

3x = 0                           Simplify

3x/3 = 0/3                     Divide each side by 3

x = 0                             Simplify

Now that we have the solution to the first two equations (0,3), we need to check the solution in the third and fourth equations.

5x = 14 - y                    Equation 3

5(0) = 14 - 3                 Substitution

0 = 11                          Simplify

The solution to the first two equations does not work in the third, so there is no solution for the system of 3 equations.  If the first three equations do not have a solution, then there will not be a solution for all 4.

There is no solution to this system of 4 equations.