Hard function problem --- help!

Hi proyaop!

$f(x)=\dfrac{x}{x-1}$

So: $f(5x)=\dfrac{5x}{5x-1}$

Now equate the given with this:
That is, 

$\dfrac{5x}{5x-1}=\dfrac{pf(x)+q}{rf(x)+1}$

But, we know $f(x)$:

$\dfrac{5x}{5x-1}=\dfrac{p(\dfrac{x}{x-1})+q}{r(\dfrac{x}{x-1})+1}$

Multiply the right with (x-1)/(x-1):

$\dfrac{5x}{5x-1}=\dfrac{p*x+q(x-1)}{r*x+(x-1)}$

Expand and equate.

$\dfrac{5x}{5x-1}=\dfrac{(p+q)x-q}{(r+1)x-1}$

Well by comparing we get:

 $p+q=5 \\ q=0 \\ r+1=5 \\$

Hence, $p=5,q=0,r=4$

So, $p+q+r=5+0+4=9$

I hope this helps!

I got it! The answer is 9!

How I did it:

First I plugged in x as 0, f(0) = 0 / -1 = 0

(0p + q)/(0r + 1) = f(5 * 0) = 0 => q/1 = 0 => q = 0

Next I plugged in x as -1, f(-1) = 1/2

(p/2 + 0)/(r/2 + 1) = f(-5) = 5/6

p/(r + 2) = 5/6

6p = 5r + 10  (1)

Next I plugged in x as -2, f(-2) = 2/3

(2p/3)/(2r/3 + 1) = f(-10) = 10/11

2p/(2r + 3) = 10/11

22p = 20r + 30

11p = 10r + 15   (2)

Take equation (1) and multiply by 2 => 12p = 10r + 20 (1b)

Take equation 1b - equation 2 => p = 5

Plug it back into equation 1 => 30 = 5r + 10 => r = 4

p + q + r = 5 + 0 + 4 = 9