Solve Algebra with 3 variables

Question

If $p$, $q$ and $r$ are positive integers and $p + \displaystyle\frac{1}{q + \displaystyle\frac{1}{r}} = \frac{129}{31}$ then what is the value of $p + q + r$?

I tried getting a common denominator, but nothing seems to work as the correct answer is an actual number

Answer 1

The fraction $$\frac1{q+\dfrac1r}$$ is lesser than $1$, so $$p=\left\lfloor\frac{129}{31}\right\rfloor=4$$ (The brackets $\lfloor\quad\rfloor$ mean "integer part").

Now, $$\frac1{q+\dfrac1r}=\frac{129}{31}-4=\frac5{31}$$ Therefore $$q+\frac1r=\frac{31}5$$

Proceed similarly to find $q$ and $r$.

Answer 2

One Liner: $$\frac{129}{31} = 4+\frac{5}{31} = 4+\frac{1}{\frac{31}{5}} = 4+\frac{1}{6+\frac{1}{5}}$$ Fundamentally, this just comes down to writing fractions in simplest form, where denominator exceeds the numerator