$p + r = 6 \implies p = 6 - r.$ I'm with you so far.

$qs = 15 \implies s = \dfrac{15}{q}.$ Still humming along.

$pr + q + s = 16 \implies r(6 - r) + q + \dfrac{15}{q} = 16 \implies 6r - r^2 + \dfrac{q^2 + 15}{q} = 16 \implies$

$r^2 - 6r + 9 + 7 = \dfrac{q^2 + 15}{q} \implies (r - 3)^2 = \dfrac{q^2 - 7q + 15}{q} \implies r = 3 \pm \sqrt{\dfrac{q^2 - 7q + 15}{q}}.$

Best check my algebra, but I think you slipped up on step 3. I agree that step 4 will be tedious and prone to error in the best case, but if step 3 is messed up, you are doomed. Nothing wrong with your thought process.