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

Accepted Solution

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

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

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

* 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.