Is a system of equations consistent?

[SticksandStones]

Homework Statement

Tell if the given system is consistent or not:
$$X_{1} -2X_{4} = -3$$
$$2X_{2}+2X_{3} = 0$$
$$X_{3}+3X_{4} = 1$$
$$-2X_{1}+3X_{2}+2X_{3}+X_{4} = 5$$

Homework Equations

Once put into Reduced Row Echelon Form, the system becomes:
$$X_{1}-2X_{4} = -3$$
$$X_{2}-3X_{4} = -1$$
$$X_{3}+3X_{4} = 1$$

The Attempt at a Solution

I put the system into reduced row echelon form (see above) and can not see a way to find an exact solution to the system so I want to say that it is inconsistent. However, I'm not entirely sure on this matter.

Is it inconsistent?

 

[Vid]

It's consistent, but with infinitely many solutions. The 4th row became 0=0 meaning that X4 is allowed to take on any value it wants and the system will still be consistent. The rest of the solutions depends on what value of x4 is chosen. X4 = 0 gives the solution's {-3,-1,1,0}. Since there is one degree of freedom (ie. one variable that is allowed to do whatever it pleases), the solutions all lie on a line in R^4. If there had been two degrees of freedom, all of the solutions would be on a plane in R^4

 

[SticksandStones]

It's consistent, but with infinitely many solutions. The 4th row became 0=0 meaning that X4 is allowed to take on any value it wants and the system will still be consistent. The rest of the solutions depends on what value of x4 is chosen. X4 = 0 gives the solutions {-3,-1,1,0}. Since there is one degree of freedom (ie. one variable that is allowed to do whatever is pleases), the solutions all lie on a line in R^4. If there had been two degrees of freedom, all of the solutions would be on a plane in R^4

Thank you. That makes far more sense.