#### Answer

$x=6$ satisfies the given conditions

#### Work Step by Step

$y=x-\sqrt{x-2}$ and $y=4$
Substitute $y$ by $4$:
$x-\sqrt{x-2}=4$
Take $\sqrt{x-2}$ to the right side and $4$ to the left side:
$x-4=\sqrt{x-2}$
Square both sides of the equation:
$(x-4)^{2}=(\sqrt{x-2})^{2}$
$x^{2}-8x+16=x-2$
Take all terms to the left side of the equation:
$x^{2}-8x+16-x+2=0$
Simplify:
$x^{2}-9x+18=0$
Solve by factoring:
$(x-6)(x-3)=0$
Set both factors equal to $0$ and solve each individual equation for $x$:
$x-6=0$
$x=6$
$x-3=0$
$x=3$
Check the solutions found by plugging them into the original equation:
$x=3$
$3-\sqrt{3-2}=4$ False
$x=6$
$6-\sqrt{6-2}=4$ True
$x=6$ satisfies the given conditions