0 < (4x-5)/(x+3) = ((4x+12) - 17)/(x+3) = (4(x+3) -17)/(x+3) = 4 - 17/(x+3).0<4x−5x+3=(4x+12)−17x+3=4(x+3)−17x+3=4−17x+3.
Add 17/(x+3)17x+3 to both sides to get:
17/(x+3) < 417x+3<4
There are now two permissible cases:
Case 1: When x < -3x<−3, x + 3 < 0x+3<0, so if we multiply both sides by (x+3)(x+3) we have to reverse the inequality to get:
17>4(x+3)=4x+1217>4(x+3)=4x+12
Subtract 12 from both sides to get:
5>4x5>4x.
Divide both sides by 4 to get:
5/4>x54>x, i.e. x < 5/4x<54.
Since in this case we already know x<-3x<−3, this condition is already fulfilled.
Case 2: When x > -3x>−3, x + 3 > 0x+3>0, so we can multiply both sides by (x+3)(x+3) without reversing the inequality to get:
17<4(x+3)=4x+1217<4(x+3)=4x+12
Subtract 12 from both sides to get:
5<4x5<4x.
Divide both sides by 4 to get:
5/4<x54<x, i.e. x > 5/4x>54.
If x > 5/4x>54 then it satisfies x > -3x>−3
So in Case 2, we just require x > 5/4x>54.
Combining the 2 cases, we find that x < -3x<−3 or x > 5/4x>54.
Note that x = -3x=−3 is not allowed due to the resulting division by 0, which is undefined.
