Do you mean when can (a/b) + (c/d) = (a+c)/(bd) happen? You have to be very careful with parentheses when writing expressions like this. Technically, you wrote
(a/b) + (c/d) = a + (c/bd), which is very different.
I was in further maths today, and came across an interesting situation within adding fractions. I wondered if there was any way you could have 2 different fractions a/b + c/d =a+c/bd where a,b,c and d are all integers. I would appreciate a response soon.
Many thanks
My post # 4 was intended as a sort of prod to get you to see how to do the next step in your problem. The situation you describe can indeed happen. However, it is not the philosophy of this forum (nor me) to simply hand out solutions to problems. We're here to help people get unstuck with problems and then solve the rest of the problem themselves. It's much better for the student that way, because then you own the solution. You understand the solution. If I just tell a student the answer, much data shows that the likelihood of the student remembering is quite small. Whereas, if the student solves the problem by himself/herself, they will own the solution and remember the thought process by which they arrived at the solution much better. So take another look at post # 4, and see if you can see where to go next.
Your arithmetic is incorrect. Check out your added fractions again. This approach, incidentally, may eventually give you a particular case that works. But you would not have found all situations in which it can happen. In fact, I think there's only one equation with four unknowns that characterizes the situations in which this can occur. So you'd get a 3-parameter family of solutions. That is, there are infinitely many possible ways to make this work.