A rectangle has a length that is 2 less than 3 times the width. If the area of this rectangle is 16, find the dimensions and the perimeter.

w(3w-2) = 16

w = 8/3

p = 2(8/3 + 6) = ?

To solve this problem, we need to set up equations based on the given information.

Let's represent the width of the rectangle as "w" and the length as "l."

From the first statement, we know that the length is 2 less than 3 times the width: l = 3w - 2.

We are also given that the area of the rectangle is 16: lw = 16.

Now we have two equations:
1) l = 3w - 2
2) lw = 16

We can substitute the value of "l" from equation 1 into equation 2, and solve for "w":

(3w - 2)w = 16

Expanding and simplifying the equation:
3w^2 - 2w - 16 = 0

Now we have a quadratic equation. We can either factor it or use the quadratic formula to find the value of "w."

Factoring this equation, we get:
(3w + 4)(w - 4) = 0

Setting each factor to zero and solving for "w":
3w + 4 = 0 -> w = -4/3
w - 4 = 0 -> w = 4

Since we can't have a negative width, we discard w = -4/3 and consider w = 4 as our solution.

Now that we have the value of "w," we can substitute it back into equation 1 to find the value of "l":
l = 3w - 2 -> l = 3(4) - 2 -> l = 10

The dimensions of the rectangle are:
Width = 4
Length = 10

Finally, to find the perimeter, we use the formula:
Perimeter = 2(length + width)

Substituting the values we found:
Perimeter = 2(10 + 4) = 2(14) = 28

Therefore, the dimensions of the rectangle are width = 4 and length = 10, and the perimeter is 28 units.