Find the dimensions of a rectangle with a perimeter of 54 meters if the length is 3 meters less than..?

Question

Find the dimensions of a rectangle with a perimeter of 54 meters if the length is 3 meters less than twice the width..Can someone help how to solve it not just the answer

Answer 1

You have the width - W

the length is 2W-3  (3meters less than twice the width)

So...these  (2W-3) and(W) added together equal 27 (1/2 the perimeter)

2W-3+W=27

2W+W=30 (after adding 3 to each side)

3W=30

W=10

L=17 (2W-3)

Answer 2

2L + 2W = 54

L = 2W - 3

Now plug in for L.

2(2W - 3) + 2W = 54   <- Distribute

4W - 6 + 2W = 54  <- combine W

6W - 6 = 54  <- add 6 to both sides

6W = 60  <- divide both sides by 6

W = 10

Now plug W back in and solve for L (remember L = 2W - 3)

L = 17

Now plug everything back into the original equation and check it.

2(17) + 2(10) = 54

34 + 20 = 54   DING DING DING! YAY.

Answer 3

Let L = length

      W = width

Perimeter = 2L + 2 W

               = 54 meters

L = 2W -3. ---> substitute this to the equation of perimeter

Perimeter = 2(2W - 3) + 2W

2(2W - 3) + 2W = 54

4W - 6 + 2W = 54

6W = 60

  W = 10 meter

  L = 2W - 3

    =  2(10) -3

    = 17 meters

so length is 17 meters and width is 10 meters.