The length of a rectangle is 8 in more than its width the area of a rectangle is equal to 3 in less than three times the perimeter find the length and width of the rectangle

This is the sort of question that is easily solved graphically.

The length is 17 inches; the width is 9 inches.

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Let x and y represent the width and length, respectively.
  y = x +8 . . . . . the length is 8 more than the width
  xy = 3(2(x +y)) -3 . . . . . the area is 3 less than 3 times the perimeter*

You can use the first expression for y to substitute into the second equation to get a quadratic in x. Only the positive solution is of interest.
  x(x +8) = 6(2x +8) -3
  x^2 -4x -45 = 0
  (x -9)(x +5) = 0

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* This part of the problem statement is nonsensical. Area is in square inches; perimeter is in inches. You cannot compare these quantities; you can only compare their numerical values. Subtracting 3 inches from some number of square inches cannot be done.