Current Year and Algebra

we can make 2 unique equations as follows,

first Let $x=2019$ ,

$3f(2019) -5\times 2019 f(\dfrac{1}{2019}) =2012$

now let $x=\dfrac{1}{2019}$

$3f(\dfrac{1}{2019}) -\dfrac{5}{2019} f(2019) =\dfrac{1}{2019} -7$

We have two linear unique equations,

Let $f(2019) =a$ and $f(\dfrac{1}{2019}) =b$

$3a- 10095b=2012$

$3b -\dfrac{5}{2019} a=\dfrac{-14132}{2019}$

we can solve this system , which gives $a=4039$ and $b=\dfrac{2021}{2019}$

$a=f(2019)=4039$

$\begin{cases} \text{Given}: & 3f(x) - 5xf\left(\dfrac 1x\right) = x -7 & ...(1) \\ \text{Replace }x \text{ with }\dfrac 1x: & 3f\left(\dfrac 1x\right) - \dfrac 5x f(x) = \dfrac 1x -7 & \small \color{#3D99F6} \text{Multiply both sides by }x \\ & 3xf\left(\dfrac 1x\right) - 5f(x) = 1 -7x & ...(2) \end{cases}$

$\begin{aligned} 3(1)+5(2): \quad -16f(x) & = -32x - 16 \\ \implies f(x) & = 2x + 1 \\ f(2019) & = 2(2019)+ 1 = \boxed{4039} \end{aligned}$