Difficult coin weighing puzzle: 14 coins, 1 fake (heavier or lighter), 3 pre-determined weighings

Suppose a triple of weighing results determines a coin. If a weighing result is "equal" then the coin did not appear in that weighing. Otherwise, the coin appeared on either the "less" side of each weighing or the "greater" side of each weighing depending on whether the coin was lighter or heavier.

For each coin, then, choose a distinct weighing result pattern that will determine that coin. (Weighing result patterns that are completely flipped must identify the same coin with the opposite weight, so we won't use these.)

A < = =
B = < =
C = = <
D < < =
E < = <
F = < <
G < > =
H < = >
I = < >
J < < <
K < < >
L < > <
M > < <
N = = =

Then we know exactly how to assemble each weighing (ie A appears in the first weighing only; G appears on opposite sides of the first two weighings; J appears on the same side of all weighings; etc) except that we don't know which side to put the coins on, but deciding the sides turns out to be easy, as we merely need to balance the number of coins in each weighing. Coin X (the known good coin) is needed because there are otherwise nine coins involved in each weighing. We will not be able to distinguish between coin N being lighter or heavier.

One solution is

AGJKL-DEHMX
BIJKM-DFGLX
CHJLM-EFIKX