NP-hardness given some reducible language While reading a passage in an older textbook I came upon this problem which I am having difficulty in justifying whether its true or false. Is this possible? 

If some problem $A$ is NP-hard, and if that problem $A$ can be reduced to problem $B$, and there exists some polynomial-time algorithm which solves $B$, then P=NP?

While I understand that the view is $P≠NP$, if the above case were true would $P=NP$?
Hint: if $A$ is NP-hard, and it can be reduced in polynomial time to $B$, what can you say about $B$? Then if you have a polynomial time algorithm for $B$, what does that imply? (Check out the definition of NP-hardness and NP-completeness).