Henry knows that the area of a rectangle is 30 square inches the perimeter is 12 inches if the length is 1 inch longer than the width what are the length and width henry's rectangle explain how you know

What you typed is impossible, the greatest possible area for a four sided polygon is a square.  If the perimeter is 12, the sides of the square would be 3 inches.  The area of that square would be 9 in^2.
 
Work through this and see that it is impossible.

P=2(W+L), we are told that P=12 so

12=2(W+L)

6=(W+L) and we are told that L=W+1 so

6=2W+1

5=2W

W=2.5, and then L=3.5 so the Area would be:

A=2.5(3.5)= 8.75in^2 NOT 30 in^2

Okay now that that crashed and burned, let's ignore the perimeter statement and just go with L=W+1

Then A=LW and using L=W+1 in the area equation we get:

A=W^2+W and since A=30 we have:

30=W^2+W

W^2+W-30=0

(W+6)(W-5)=0  since W>0 to have any meaning...

W=5 and L=6

Okay that's fine, the area fits the length and width parameters...

Now notice that P=2(5+6)=22in NOT 12 in

:D  So you mistyped???