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.
