An identity holds if and only if the identity between the inverses hold (as long as you don't have something like #0=0# of course). So, in you case, you have that

#3/x=12/{x+7} \iff x/3={x+7}/12#

The only thing we have to take care about is to make sure that the original denominators aren't zero, i.e. #x \ne 0# and #x+7 \ne 0#. This means that, when we'll eventually find solutions, we have to make sure that these solutions are nor #0# nor #-7#.

Now let's solve the equality between inverses:

#x/3={x+7}/12 \iff 12x=3(x+7)#.

Expanding, we have #12x=3x+21#. Isolating the #x#-terms and the costants, we get #9x=21#, and thus we can solve for #x#: #x=21/9#, which you can simplify (dividing by #3# both numerator and denominator) into #7/3#.