sqrt(x)-sqrt(3x-3)=1

(sqrt(x))^2 = (1+sqrt(3x-3))^2

x = (1+sqrt(3x-3))(1+sqrt(3x-3))

x = 1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2

x = 3x-2+2(sqrt(3x-3))

(-2x+2) = 2(sqrt(3x-3))

(-2x+2)/2 = 2(sqrt(3x-3))/2     <<<---===  after this line, your writing was incorrect,
                                           so I REPLACED IT. Below is MY writing.

(-x+1)^2 = (sqrt(3x-3))^2

x^2 - 2x + 1 = 3x-3

x^2 - 5x + 4 = 0

(x-4)*(x-1) = 0


So, the two candidates are  x= 4  and  x= 1.
Substitute them into the original equation and make sure
that x= 4 does not work, while x= 1 works.


ANSWER.  The only solution to the original equation is x= 1.

Question: sqrt(x)-sqrt(3x-3)=1

My work: 
sqrt(x)-sqrt(3x-3)=1
(sqrt(x))^2=(1+sqrt(3x-3))^2
x=(1+sqrt(3x-3))(1+sqrt(3x-3))
x=1+sqrt(3x-3)+sqrt(3x-3)+(sqrt(3x-3))^2
x=3x-2+2(sqrt(3x-3))
(-2x+2)/2=2(sqrt(3x-3))/2
(-x)^2=(sqrt(3x-3))^2
x^2=3x-3
0=x^2+3x-3
After this step, I get lost. I know the answer is supposed to be 1 but I am not sure where I went wrong or where to go from the last step I took.

Question: sqrt(x)-sqrt(3x-3)=1
          
                 <===== ERROR IS HERE!!
              <===== s/b THIS, instead!
     
     (x - 1)  (x - 4) = 0
x - 1 = 0   |   x - 4 = 0 ---- Equating each factor to 0
    x = 1   |       x = 4

                        C H E C K
      x = 1                               x = 4
                  
             1 = 1 (TRUE)                - 1 = 1 (FALSE)
          x = 1 is the ONLY SOLUTION     x = 4 is an EXTRANEOUS SOLUTION