SOLUTION: given the three side lengths of triangle ABC which are 13-x, 6+x, and 15 what are all the possible vaules of X?

If the three sides of a triangle are a,b, and c, then
the triangular inequality states:

a + b > c
a + c > b
b + c > a

Here we have

Side a = 13-x, 
Side b = 6+x,
Side c = 15


The sum of each pair of sides is greater than the third side, so there
are three possibilities

a + b > c

(13-x)+(6+x) > 15
13-x+6+x > 15
19 > 15

That would be true regardless of what x is, so any
restrictions on x will comes from the other two
-----------------------------------
a + c > b

(13-x)+15 > 6+x
13-x+15 > 6+x
28-x > 6+x
-2x > -22
x < 11    (the inequality reverses when dividing by a negative)

So that's one restriction on x  
-----------------------------------
b + c > a

(6+x)+15 > 13-x
6+x+15 > 13-x
21+x > 13-x
2x > -8
x > -4

That's another restriction on x
----------------------------------

Put those restrictions toget0her

x < 11 and x > -4

Turn x > -4 around as -4 < x since it is equivalent
and we can then write that as a continued inequality
with x in the middle:

-4 < x < 11

Edwin