Posted by James Caulfield on .
abbc+addc/ab+bc+ad+dc
Simplify and state the domain

Trig/algebra2 
Reiny,
you probably meant ...
(abbc+addc)/(ab+bc+ad+dc)
= (b(ac)  d(ac))/(b(a+c) + d(a+c))
= (ac)(bd)/((a+c)(b+d))
a ≠ c , b≠  d