In cases where you are using the conditional check with a _boolean_ to chose between two _numbers_, you can do math equations instead:

 (x ? num1 : num2) conclusions:
 
 	1)if num1 equals num2, there ARE savings
 	2)if num1 is (+1) or (-1) than num2, there ARE savings
 	3)if either num1 or num2 equals to 0, there ARE savings
 	4)it is MORE LIKELY to find greater savings on num1>num2 instead of num1<num2
 	5)in method (*A) and (*B), savings are NOT GUARANTEED
 	
 	a)num1>num2
 		i)(num1==(num2+1))
 			ex1: (x?5:4) to (x+4)
 			ex2: (x?8:7) to (x+7)
 		ii)num2==0
 			ex1: (x?3:0) to (x*3)
 			ex2: (x?7:0) to (x*7)
 		iii)
 			(*A) or (*B) //one might be shorter
 			
 	b)num1<num2
 		i)((num1+1)==num2)
 			ex1: (x?4:5) to (5-x)
 			ex2: (x?7:8) to (8-x)
 		ii)num1==0
 			ex1: (x?0:3) to (!x*3)
 			ex2: (x?0:7) to (!x*7)
 		iii)
 			(*A) or (*B) //one might be shorter
 			
 	c)num1==num2
 		i)
 			ex1: (x?5:5) to (5)
 			ex2: (x?-3:-3) to (-3)
 	
 	(*A) use ((x*(num1-num2))+num2)
 		ex1: (x?8:4) to ((x*4)+4)
 		ex2: (x?4:8) to ((x*-4)+8)
 		
 		ex3: (x?6:-4) to ((x*10)-4)
 		ex4: (x?-4:6) to ((x*-10)+6)
 		
 		ex5: (x?4:-6) to ((x*10)-6)
 		ex6: (x?-6:4) to ((x*-10)+4)
 		
 		ex7: (x?-5:-9) to ((x*4)-9)
 		ex8: (x?-9:-5) to ((x*-4)-5)
 		
 	(*B) use ((!x*(num2-num1))+num1)
 		ex1: (x?8:4) to ((!x*-4)+8)
 		ex2: (x?4:8) to ((!x*4)+4)
 		
 		ex3: (x?6:-4) to ((!x*-10)+6)
 		ex4: (x?-4:6) to ((!x*10)-4))
 		
 		ex5: (x?4:-6) to ((!x*-10)+4)
 		ex6: (x?-6:4) to ((!x*10)-6)
 		
 		ex7: (x?-5:-9) to ((!x*-4)-5)
 		ex8: (x?-9:-5) to ((!x*4)-9)

**Note:** In addition to this, you will need to remove the unnecessary `0-`, `+0`, `+-` etc.

> In case you don't find savings after running it with a JS compressor, simply use the former (x?y:z).

Previously I thought method B couldn't ever beat A, however exceptions do exist:

 (x?97:100) //original
 
 (-3*x+100)
 (3*!x+97)