Current calculation in circuit
Just another method. The key point is to judge if the diode is on.

Assume the LED is on, then the voltage drop on the LED and R2 should be 2V or so, then

$$
V_{D1} = V_{R2} = 2V
$$ 

Then the current on \$R_{2}\$ should be 

$$
I_{R2}= \frac{2V}{200\Omega}=10mA
$$

Then the voltage drop on \$R_{1}\$ should at least

$$
V_{R1} = (I_{R2} + I_{D1}) \times R_{1} &gt;10mA \times 1k\Omega=10V
$$

Apparently, this voltage is greater than the source voltage, so the diode can NOT be on. So we think it's open circuit. Then it's a voltage source in series with two resistor now. And the math will be easy.