#### Answer

$x=4$

#### Work Step by Step

Squaring both sides and then isolating the radical expression, the given equation is equivalent to
\begin{array}{l}\require{cancel}
1+\sqrt{x}=\sqrt{3x-3}
\\\\
\left( 1+\sqrt{x} \right)^2=\left(\sqrt{3x-3}\right)^2
\\\\
(1)^2+2(1)(\sqrt{x})+(\sqrt{x})^2=3x-3
\\\\
1+2\sqrt{x}+x=3x-3
\\\\
2\sqrt{x}=3x-x-3-1
\\\\
2\sqrt{x}=2x-4
.\end{array}
Squaring both sides again, the given equation is equivalent to
\begin{array}{l}\require{cancel}
2\sqrt{x}=2x-4
\\\\
\left( 2\sqrt{x} \right)^2=(2x-4)^2
\\\\
4x=(2x)^2-2(2x)(4)+(-4)^2
\\\\
4x=4x^2-16x+16
\\\\
0=4x^2-16x-4x+16
\\\\
0=4x^2-12x+16
\\\\
\dfrac{0}{4}=\dfrac{4x^2-12x+16}{4}
\\\\
x^2-3x+4=0
.\end{array}
Using the FOIL Method which is given by $(a+b)(c+d)=ac+ad+bc+bd,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
(x-4)(x+1)=0
.\end{array}
Equating each factor to zero and then isolating the variable, the solutions are $
x=\{-1,4 \}$
If $x=-1,$ then
\begin{array}{l}\require{cancel}
1+\sqrt{x}=\sqrt{3x-3}
\\\\
1+\sqrt{-1}=\sqrt{3(-1)-3}
\text{ (not a real number)}
.\end{array}
Hence, only $
x=4
$ satisfies the original equation.