Solve the system by elimination -2x+2y+3z=0 -2x-y+z=-3 2x+3y+3z=5

 -2x+2y+3z=0 
-2x-y+z=-3
2x+3y+3z=5

Adding the third equation to the second and first, we get 
5y+6z=5
2y+4z=2

Multiplying the second equation by -6/4 and adding, we get 
   2y=2 and y=1 by dividing both sides by 2. Plugging y=1 into 2y+4z=2, we get 2+4z=2 and by subtracting 2 from both sides we get 4z=0=z. PLugging y=1 and z=0 into -2x-y+z=-3, we get -2x-1+0=-3 and by adding 1 to both sides, we get -2x=-2 and by dividing both sides by -2, we get x=1

The answer would be (c)