A rectangle has a length that is 2 metres more than twice its width. The perimeter is 64 metres. Write an equation for the rectangle and find its dimensions.

Step-by-step solution

1. 1

   Name the quantity that everything else is described from. The length is described in terms of the width, so the width is the natural variable. Let

   $w=\text{width in metres}$

2. 2

   Translate the phrase one piece at a time. Twice the width is $2w$; 2 more than that adds $2$. So the length is

   $\ell=2w+2$

   The order matters: $2w+2$ is not the same as $2(w+2)$, which would be twice the quantity two more than the width.

3. 3

   Substitute into the perimeter formula. For any rectangle $P=2(\ell+w)$, so

   $2\big((2w+2)+w\big)=64$

   This is the requested equation. Combining like terms inside gives the tidier equivalent form

   $2(3w+2)=64$

4. 4

   Solve for the width. Divide by $2$ first so the numbers stay small:

   $3w+2=32$

   $3w=30$

   $w=10$

5. 5

   Recover the length and check the perimeter. With $w=10$:

   $\ell=2(10)+2=22$

   $P=2(22+10)=2(32)=64\ \text{m}\quad\checkmark$

   The width is $10$ m and the length is $22$ m.