In general, thermodynamic systems prefer to minimize their potential energy (in much the same way a ball rolls downhill to minimize its gravitational potential energy) and maximize their entropy (in much the same way a child's room will gradually become more disorganized over time without expending effort to clean it up).
Often, these two goals will come into conflict as in the case of water freezing. When water freezes, more stable bonds are able to form between water molecules as they crystallize into a solid. Forming these stronger bonds reduces their chemical potential energy and releases heat to the surroundings. Because the chemical potential energy is lowered during freezing, it is an exothermic process that lowers the enthalpy of the system (ΔH < 0). However, because the water molecules can no longer move around freely as is the case in the liquid state, freezing is associated with a loss in entropy (ΔS < 0).
In cases like these, what sets the balance between the propensity of the system to minimize enthalpy versus maximize entropy? Scientists have devised a measure, called free energy, for just such a purpose. In the case of Gibbs free energy (G), this quantity is defined as G = H - TS, and the change in free energy of a system will be ΔG = ΔH - TΔS. As you should be able to tell from the equation, a process will be thermodynamically favorable if the process lowers the free energy of the system (ΔG < 0). The equation also tells us that temperature is the key factor determining whether the system will prefer to minimize enthalpy or maximize entropy. At low temperatures, minimizing enthalpy becomes more important than maximizing entropy and at high temperatures, maximizing entropy becomes more important than minimizing enthalpy.
The freezing point of water occurs when the liquid and solid states of water are at equilibrium, and for a process at equlibrium, ΔG = 0. Therefore, 0 = ΔH - TΔS, which gives an equation for the melting point of water: T = ΔH/ΔS.
What happens when we add salt to the water? Salt will not appreciably affect the strength of the bonds between water molecules in the liquid and solid states, so ΔH is unchanged. Salt also does not get incorporated into the ice, so the entropy of the ice is not changed. However, salt does make the liquid phase more disordered, and raises the entropy of the liquid phase. Since salt water has an even higher entropy than pure water, there is an even greater loss in entropy associated with freezing salt water than pure water.
So, if the magnitude of ΔH is unchanged and the magnitude of ΔS is larger in the presence of salt, the equation T = ΔH/ΔS tells us that ice should melt at a lower temperature in the presence of salt (which is what actually happens).