I will do my best to explain what is done, but feel free to ask if you have questions on what I did:
\(cy+3=6d-2y\)
bring all the "y"s to one side
\(cy+3-3=6d-2y-3\)
\(cy+2y=6d-3-2y+2y\)
this should be your result...
\(cy+2y=6d-3\)
now divide y out to isolate it:
\(y(c+2)=6d-3\)
then divide our c+2 so your y is alone
\(\frac{y(c+2)}{c+2}= \frac{6d-3}{c+2}\)
you should get this as your answer:
\(y= \frac{6d-3}{c+2}\)
