#0 < (4x-5)/(x+3) = ((4x+12) - 17)/(x+3) = (4(x+3) -17)/(x+3) = 4 - 17/(x+3).#
Add #17/(x+3)# to both sides to get:
#17/(x+3) < 4#
There are now two permissible cases:
Case 1: When #x < -3#, #x + 3 < 0#, so if we multiply both sides by #(x+3)# we have to reverse the inequality to get:
#17>4(x+3)=4x+12#
Subtract 12 from both sides to get:
#5>4x#.
Divide both sides by 4 to get:
#5/4>x#, i.e. #x < 5/4#.
Since in this case we already know #x<-3#, this condition is already fulfilled.
Case 2: When #x > -3#, #x + 3 > 0#, so we can multiply both sides by #(x+3)# without reversing the inequality to get:
#17<4(x+3)=4x+12#
Subtract 12 from both sides to get:
#5<4x#.
Divide both sides by 4 to get:
#5/4<x#, i.e. #x > 5/4#.
If #x > 5/4# then it satisfies #x > -3#
So in Case 2, we just require #x > 5/4#.
Combining the 2 cases, we find that #x < -3# or #x > 5/4#.
Note that #x = -3# is not allowed due to the resulting division by 0, which is undefined.
