4. 122 Solve the system of equations $\left\{\begin{array}{l}\sqrt{x-1}+\sqrt{y-3}=\sqrt{x+y}, \\ \lg (x-10)+\lg (y-6)=1 .\end{array}\right.$
[Solution] The original system of equations can be transformed into the system of equations
$$\left\{\begin{array}{l}
x y-3 x-y-1=0 \\
x y-6 x-10 y+50=0
\end{array}\right.$$
Subtracting (4) from (3) and simplifying, we get
$$x=17-3 y .$$

Substituting (5) into (3) and rearranging, we get
$$3 y^{2}-25 y+52=0$$

Thus $\square$
$$y_{1}=4, y_{2}=\frac{13}{3}$$

Substituting the values of $y$ into (5), we get $x_{1}=5, x_{2}=4$, so the solutions to the system of equations are
$$\left\{\begin{array} { l } 
{ x _ { 1 } = 5 , } \\
{ y _ { 1 } = 4 ; }
\end{array} \quad \left\{\begin{array}{l}
x_{2}=4 \\
y_{2}=\frac{13}{3}
\end{array}\right.\right.$$

Upon verification, neither of these pairs satisfies equation (2), so the original system of equations has no solution.