Combine like terms to get #12g-5t=15g+7#. Now subtract the #g# from the LHS: #-5t=3g+7#, and divide by -5 to get #t=\frac{3g+7}{-5}#. If we had a second equation to substitute this into than we would be able to get our answer, but there is not, so this is as far as we can go. If there is a definition for #g,t# that I can't see than by all means PM me and I'll update my answer.