The length of a rectangle is 2 in more than it's w i d t h the area of a rectangle is equal to 1 inch less than three times a perimeter find the length and width of the rectangle

If we let x and y represent length and width, respectively, then we can write equations according to the problem statement.
.. x = y +2
.. xy = 3(2(x +y)) -1

This can be solved a variety of ways. I find a graphing calculator provides an easy solution: (x, y) = (13, 11).

The length of the rectangle is 13 inches.
The width of the rectangle is 11 inches.

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Just so you're aware, the problem statement is nonsensical. You cannot compare perimeter (inches) to area (square inches). You can compare their numerical values, but the units are different, so there is no direct comparison.

3x+9+4x+x=8x+9that's the answer for your question.