 Eliminating "p"3 = "up"2 first, and then using the linear constraint to eliminate "p"1 = 1 − "p"2 − "up"2 = 1 − (1 + "u")"p"2, the quadratic constraint reduces to a [[quadratic equation]] in "p"2
with [[root of an equation|roots]] "p"2 = 0 (obvious) and "p"2 = (1 + "u") / (1 + "u" + "u"2), from which "p"1 and "p"3 are then obtained by [[back substitution]].One can define six such parameters "uab" = "pa" / "pb", for which "pc" ≤ "pb" ≤ "pa" when ("c", "b", "a") is a [[cyclic permutation]] of (1, 2, 3).All different values of "p"1, "p"2, "p"3 ordered as above are obtained with "u" running in the range "u" ≥ 1.