Help there,
Let's solve the problem:

Let, L and B be the length and breadth of the rectangle, respectively.
Now, given that,
L= 3B-2, and,
Area= L*B= 16 sq.unit

According to the given problem,
B(3B-2) = 16
or, 3B - 2B = 16
or, 3B - 2B - 16 = 0
or, 3B + 6B - 8B - 16 = 0
or, 3B (B+2) - 8 (B+2) = 0
or, (B+2) (3B-8) = 0

Now, B cannot be -2, since breadth cannot be negative.
Hence, B is 8/3, which is 2.6 unit.
L= 3B -2 = 6 unit.
Perimeter= 2 (L+B) = 17.2 unit.

Answer:
L= 6 unit.
B= 2.6 unit.
Perimeter= 17.2 unit.

Hope this answer helps you.
Thank you!!

Hello!

Let the width(b) of the rectangle be 'x'.

Length, l = 3x - 2

Area of the rect. = l × b

16 = (3x - 2)*x

16 = 3x^2 - 2x

the equation is 3x^2 - 2x - 16 = 0

Now we will factorise this equation:-

3x^2 - 2x - 16

3x^2 - 8x + 6x - 16

3x^2 + 6x - 8x - 16

3x(x + 2) -8(x + 2)

(3x - 8)(x + 2)

Hence possible values of x can be 8/3 or -2.

Since the breadth can't be negative,

x = 8/3

Now, 3x - 2 = 3*(8/3) - 2

= 8 - 2 = 6 cm

Now, perimeter, P = 2(l + b)

P = 2(6 + 8/3)

P = 2(18 + 8/3)

P = 26/3 × 2

P = 52/3 cm

Hence the perimeter of the rectangle is 52/3 cm.

Thankyou!

Hope it helps!