Both L7 and L8 of my previous post will result to exactly the same code stress values if the combination method is "ALGEBRAIC" following the code rule. Just to explain:

on METHOD-1: USE ALGEBRAIC
L7=L5-L6; substitute L5, L6
L7=(L1-L3)-(L2-L4); substitute L1,L2,L3,L4
L7=[(W+T1+P1)-(W+P1)]-[(WNC+T2)-WNC]
= W+T1+P1-W-P1-WNC-T2+WNC
L7= T1-T2 (THERMAL STRESS RANGE)

on METHOD-2: USE ALGEBRAIC
L7=L5-L6; substitute L5, L6
L7=(L1-L3)-(L2-L3); substitute L1,L2,L3
L7=[(W+T1+P1)-(W+P1)]-[(W+T2+P1)-(W+P1)]
= W+T1+P1-W-P1-W-T2-P1+W+P1
L7= T1-T2 (THERMAL STRESS RANGE)

hence, if L5= +130F, and L6= -90F
L7= +130F - (-90F); ALGEBRAIC COMBINATION
L7= +220F (THERMAL STRESS RANGE)

also,
L8=L1-L2; (USE ALGEBRAIC) substitute L1, L2
=(W+T1+P1)-(W+T2+P1)
=W+T1+P1-W-T2-P1
L8=T1-T2 (THERMAL STRESS RANGE)
then, if T1= +200F, and T2= -20F
L8= +200F - (-20F); ALGEBRAIC COMBINATION
L8= +220F (THERMAL STRESS RANGE)

***FOR L7=L5+L6, the combination method to use shall be "ABSOLUTE". This means;
L7=|L5| + |L6| ; for L5= +130F, and L6= -90F
=|130F| + |-90F| ; ABSOLUTE COMBINATION
L7= +220F (THERMAL STRESS RANGE)
Note that using the ABSOLUTE combination if both T1,T2 are above zero will result into unacceptable stress range value because it will force to get the SUM (not the ALGEBRAIC difference). So, say T1=+200F, T2=+20F will have L7=|200F|+|20F|=220F (Not ok). However, for L7=L5-L6 using ALGEBRAIC combination, L7=(200F)-(20F)=180F (correct stress range between T1 & T2). This means that the latter is better to use than the former method for any +/- values of T2.