if $\frac{p}{q}, \frac{r}{s}$ are positive simplified fractions such that $qr - ps=1$, prove that $\frac{p+r}{q+s}$ is also a simplified fraction

Question

if $\frac{p}{q}, \frac{r}{s}$ are positive simplified fractions such that $qr - ps=1$, prove that $\frac{p+r}{q+s}$ is also a simplified fraction

It's not hard to prove through Pick's theorem actually. Let $(p,q), (r,s)$ be two points on the plane, then $(0,0), (p,q), (r,s), (p+r, q+s)$ formed a parallelogon whose area is $1$. But there is no lattice point within $(0,0), (p,q), (r,s), (p+r, q+s)$, Thus $\frac{p+r}{q+s}$ cannot be further reduced. Otherwise it'd produce a lattice point on the diagnal.

However Pick's theorem is kind of too deep for this. Is there an elementary proof?

Answer 1

$$ r(q+s)-s(r+p) = rq +rs -rs - ps = qr - ps=1 $$

Answer 2

Hint: Show that $$\frac pq <\frac{p+r}{q+s}<\frac rs.$$ Then prove by contradiction.

Edit: Alternatively $$\left|\begin{array}{cc}p+r&p\\ q+s&q\end{array}\right|=1,$$ hence $\gcd(p+r,q+s)=1$.