Let $f(x)=(3x-5):(x-2)$
a) Find the value of constants $p$ and $q$ such that $f(x)=p+ q:(x-2)$.
b) Hence describe a single transformation which transforms the graph of $y=1:x$ to the graph of $y= f(x)$.
I am in second grade IB student on SL. I have problems especially with the point a). I tried to put sign equal between $(3x-5):(x-2)$ and $p+q:(x-2)$ and do it by algebraic method but I found myself having 3 unknowns. Then I tried to think about it as transformations to the functions $1:(x-2)$ but it did not work either so I have no idea what to do :(
Thank you
\begin{align*} (3x-5):(x-2)&=p+\frac{q}{x-2} & \text{multiply all by $x-2$}\\ 3x-5 &= p\cdot (x-2) +q\\ 3(x-2)+1&= p\cdot (x-2)+q \end{align*} Hence $p=3$ and $q=-1$.