assign charges to the ions, and see what works together, and take out the spectator ions
something like this example:
CaCl2(aq) + 2AgNO3(aq) → Ca(NO3)2(aq) + 2AgCl(s)
Ca2+ + 2Cl- + 2Ag+ + 2NO3- → Ca2+ + 2NO3- + 2AgCl(s) fully ionic equation
and from this you can see whatever remains in solutions are spectator ions, so this equation can be reduced even more, try harder.