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.
Since you haven't identified this figure, I'm going to assume that it's a rectangle.
The Perimeter of a rectangle of length L and width W is P = 2L + 2W.
Here you are given the Perimeter and the length, and are to find the width, W.
Solving the above equation for W, we get P - 2L = 2W.
Dividing by 2 (to isolate that W), we get
P
-- - L = W
2
Substitute P= 6 yds and L = 6 feet (or 2 yds), find W (in yards).
The first one = 5 and the second one = -6
hope this helps
x = 5 and y = 3. That is the answer