how do I solve 5/(b+1)>5

Question

I cannot seem to be able to understand how to move the b+1, I know I multiply both sides by it but I get

Answer 1

5/(b+1)>5 so 1/(b+1)>1.

If b+1>0, that is, if b>-1, 1>b+1, 0>b, so b<0. Therefore -1<b<0, b must be between 0 and -1 (exclusive).

If b+1<0, that is, b<-1, then the inequality must be reversed when we multiply both sides by b+1:

1<b+1, 0<b, so b>0, which contradicts the requirement b<-1.

Solution is -1<b<0. (Example: b=-½, so 5/(1-½)=10 which is greater than 5, so the inequality is satisfied. But if b=1, 5/2 is not greater than 5 so the inequality is not satisfied. The numerator is bound to be a fraction between 0 and 1 (exclusive) and this will always produce a quotient greater than 5.)