the correct question is
What values make the inequality (x+6)/(6-x)<0 true?
we know that
the denominator cannot be zero;
so
(6-x)------------> will be 0 if x=6
the inequality is true
when
(x+6) > 0 and (6-x) < 0--------> case 1
or
(x+6) < 0 and (6-x) > 0--------> case 2
case 1
(x+6) > 0 and (6-x) < 0
(x+6) > 0 --------------> x> -6
(6-x) < 0 ----------------> x> 6
solution--------> (6,∞)
case 2
(x+6) < 0 and (6-x) > 0
(x+6) < 0 --------------> x< -6
(6-x) > 0 ----------------> x < 6
solution--------> (-∞,-6)
the answer is
the values that makes the inequality true are
(-∞,-6) ∩ (6,∞)
