No. The theorem states that if such a system has a an infinite number of solution, then each is equal to a specific solution to the system plus any of the infinite number of solutions to the associated homogeneous system. In a situation like this, where you have more equations than unknown values, three things are possible: (1) there is no solution, (2) there is a unique solution, (3) there are an infinite number of solutions.
Here, that "associated homogenous system" is
x1+2x2+2x3 =-5
-x1+-2x2-x3=1
4x1+8x2+5x3=-8
3x1+6x2+x3=5
Here, you have four equation in three unknown values- typically, more equations than unknowns is "over determined"- there may exist values of x1, x2, and x3 that satisfy the first three but it is unlikely they will satisfy the last. In any case the only way to determine which is true is to to try to find them!
Adding the first two equations eliminates both x1 and x2: (x1+ 2x2+ 2x3)+ (-x1- 2x2- x3)= x3= -5+1= -4 so x3= -4
(you have "+-2x2" in the second equation. I am assuming you mean just "-2x2".)
Setting x3= -4, the second equation becomes -x1- 2x2= -3 and the third becomes 4x1+ 9x2= 12. Multiply the first of those equations by 4 to get -4x1- x2= -12 and add to the second equation: (-4x1- 8x2)+ (4x1+ 9x2)= x2= -12+ 12= 0. x2= 0. Setting x2= 0 and x3= 4 in the first equation, x1+ 2(0)+ 2(-4)= x1- 8= -5 so x1= 8- 5= 3.
Finally, set x1= 3, x2= 0, and x3= -4 in the fourth equation: 3(3)+ 6(0)+ (-4)= 9- 4= 5! Yes, those values satisfy all four equations.
The fact that we were able to find a solution means that (1) above, that there is NO solution, does not apply. It is still possible that there is a single, unique, solution, of that there exist an infinite number of solutions.
Now, look at
x1+2x2+2x3 = 0
-x1+-2x2-x3= 0
4x1+8x2+5x3= 0
3x1+6x2+x3= 0
Those may have an infinite number of solutions. That is, instead of solving for specific values of x1, x2, and x3, you should only be able to find, say, x1 and x2 in terms of x3. Those solutions, added to the solution above will give all solutions to this system of equations.
(If they have only a single solution, it obviously must be x1= x2= x3= 0 so that the solution above is the only solution.)
