Your first equation assumes that the full voltage is across the resistor. Then your second equation finds out that the full voltage is across the resistor.

What's really going on is that within a pretty broad range of currents, the voltage drop across your LED is roughly 3V. So you assume that 3V drop. Then you find out that the voltage across the resistor is \$V_r = 5\mathrm{V} - 3\mathrm{V} = 2\mathrm{V}\$. Then you solve for current: \$I = \frac{V_r}{100\Omega} = 20\mathrm{mA}\$.

Your first equation assumes that the full voltage is across the resistor. Then your second equation finds out that the full voltage is across the resistor.

What's really going on is that within a pretty broad range of currents, the voltage drop across your LED is roughly 3V. So you assume that 3V drop. Then you find out that the voltage across the resistor is \$V_r = 5\mathrm{V} - 3\mathrm{V} = 2\mathrm{V}\$. Then you solve for current: \$I = \frac{V_r}{100\Omega} = 20\mathrm{mA}\$.

Edit: then you make sure that the resulting current is within that broad range where you can expect a 3V drop -- there's some (really) lower limit where more current will flow through parasitic parallel resistances than through the junction, and a high limit where the diode will burn up.