What is the exact time ?

Let the 12-hour mark be $0^\circ$ position. At 10:00, the minute-hand position is $\theta_m (10:00) = 0^\circ$ . Since the minute-hand moves $360^\circ$ in 60 minutes or $6^\circ$ per minute, we have $\theta_m (10:x+6) = 6x+36$ in degrees.

At 10:00, the hour-hand positive is $\theta_h(10:00) = 300^\circ$ . Since hour-hand moves $30^\circ$ in 60 minutes or $0.5^\circ$ per minute, we have $\theta_h (10:x-3) = 300 + 0.5(x-3)$ and the opposite position of $\theta_h (10:x-3)$ is $\theta_h (10:x-3) - 180$ .

Then we have:

$\begin{aligned} \theta_m (10:x+6) & = \theta_h(10:x-3) - 180 \\ 6x+36 & = 300 + 0.5(x-3) - 180 \\ 5.5x & = 82.5 \\ \implies x & = \boxed{11} \end{aligned}$

We can solve this problem in $2$ ways.

Method 1 (Angles) Each minute, the hour hand moves $\frac{1}{2}$ more of a degree, and starts at degree $300.$ Then, $3$ minutes ago, the hour hand was at degree $300+\frac{x-3}{2}.$ The minute hand moves $6$ degrees every minute, and starts at degree $0.$ Then, in $6$ minutes, the minute hand will be at $6(x+6).$ We have $6(x+6)+180=300+\frac{x-3}{2}.$ Solving this we get, $\boxed{x=15}$ .

Method 2 (Time) In $60$ minutes, the hour hand actually only moves $5$ minutes on the face of the clock. So, by the time minute hand moves $1$ minute, the hour hand only moves $\frac{5}{60}=\frac{1}{12}$ minutes. The hour hand starts at minute $50$ and the minute hand starts at minute $x.$ Now, we can easily create the equation $50+\frac{x}{12}-\frac{3}{12}=6+x+30$ . Solving this we get, $\boxed{x=15}$ .