Question 1039974
If grouping symbols are not missing or not needed, then you have equation {{{a/b+c + d/b-c=2/b+c}}}.  The common denominator would be {{{b}}}.


{{{(a/b+c + d/b-c)b=(2/b+c)b}}}


{{{a+bc+d-bc=2+bc}}}


{{{a+d=2+bc}}}


{{{a+d-2=bc}}}


{{{b=(a+d-2)/c}}}



In case that is not what you expect, then express your equation precisely.