On a straight line, the consecutive points $P, Q, R$ and $S$ are considered which form a harmonic quatern. If $QR=\frac{47}{QS}$ and $PS =\frac{96}{PQ}$, calculate $PR$ (Answer:$7$)
My progress:
I used the property of the quatern in line segments and the theorem of Descartes
$\frac{PQ}{QR}=\frac{PS}{RS}(I)$ (property)$
R. Descartes:$ \frac{2}{PR} = \frac{1}{PQ} + \frac{1}{PS}\implies \frac{2}{PR} = \frac{PS}{96}+\frac{PQ}{96}$
$PR=\frac{192}{(PS+PQ)}$
From(I): $ \frac{PQ}{PR-PQ} =\frac{PS+RS}{RS}\implies PQRS = PRPS+PRRS-PQPS-PQRS$
$2PQRS+PQPS=PR(PS+RS)\implies PQ(2RS+PS) = PR(PS+RS) $
I can't find the algebra to finish;;;?