That was presumptuous of me. It's a little different when you are the tutor...
Ok, so did you understand the diagram, and where the labelled bits came from?
If you're up to the equations...
"The length of the picture is 1 inch greater than the width."
Remember that we said L is the length and W is the width of the picture.
This sentence therefore translates to L = W + 1.
"the frame's area is 8inches square"
We need to form an equation for the area of the frame.
Looking at the diagram, we can see the area of teh frame can be given by the two vertical rectangles on the left and right side of the picture, with height W + 0.5 +0.5 = W + 1, and width 0.5.
The area of each of these rectangles is base x height = \(\displaystyle 0.5 \times (W+1)\)
The remaining area will be given by the two horizontal rectangles above and below the picture with width L and height 0.5.
The area of each of these rectangles is base x height = \(\displaystyle L \times 0.5\).
The total area of the frame is therefore given by
\(\displaystyle A = 2 \times 0.5(W + 1) + 2 \times 0.5L\) [equ 1]
Now, we said earlier that L = W + 1 and we can substitute this into [equ 1]:
\(\displaystyle A = 2 \times 0.5L + 2 \times 0.5L\)
\(\displaystyle A = L + L\)
\(\displaystyle A = 2L\)
We know the area of the frame is 8 inches square:
\(\displaystyle 8 = 2L\)
\(\displaystyle L = 4\)
That is, the length of the picture is 4 inches.
The width of the picture is 4-1=3 inches.
Let's check:
With these dimensions of the picture, the area of the frame is given by
A = 2( base x height of left and right rectangles) + 2(base x height of top and bottem rectangles)
\(\displaystyle A = 2\left(0.5\times(0.5+0.5+3))\right) + 2\left(4\times0.5\right) = 8\) ... excellent!