This question is related to the function topic. Here we are faced with two functional parametric equations that are added together and their value is also written in terms of the parameter $x$. Pay attention that $f$, $g$ and $y$ are different functions:
$$ f(2x+1) + g(x-1) = 3x+2 $$ $$ f\left(1+\frac{1}{x}\right) + 3y\left(\frac{1-2x}{2x}\right) = \frac{1}{2x} + 4 $$
Now the basic question is how to get the result of $f(2015) + g(2015)$? I tried to get their main rule by changing the expression variables inside the functions but it was very long and involved confusing calculations. What is the best way to answer this question?
Suppose $$g(t) = \begin{cases}1&t = 2015\\0&t \ne 2015\end{cases}$$ It follows that $$f(t) = \begin{cases}6049&t = 4033\\\frac{3t+1}2&t \ne 4033\end{cases}$$ and $$y(t) = \begin{cases}\frac{2020}3&t = 2015\\-\frac{2t}3&t \ne 2015\end{cases}$$ And $f(2015) + g(2015) = 3023 + 1 = 3024$.
Now suppose that $$g(t) = 3t + 5$$ Then $$f(t) = 0$$ and $$y(t) = \frac{t+5}3$$ Now $f(2015) + g(2015) = 0 + 6050 = 6050$.
There is nothing in the problem that eliminates either of these possibilities. The second equation puts no limits on the functions $f$ and $g$. It just defines a third function $y$ which is irrelevant to the question. In fact, by choosing an appropriate function $g$, you can obtain any value at all for $f(2015) + g(2015)$. The problem as stated has no solution.