A rectangle has a length that is 4 metres more than three times its width. The perimeter is 96 metres. Write an equation for the rectangle and find its dimensions.

Step-by-step solution

1. 1

   Let the width be the variable. Everything in the problem is measured against the width, so set

   $w=\text{width in metres}$

2. 2

   Build the length expression from the words. Three times the width is $3w$, and 4 more than that adds $4$:

   $\ell=3w+4$

3. 3

   Write the perimeter equation. Substituting into $P=2(\ell+w)$ gives the required equation:

   $2\big((3w+4)+w\big)=96$

   or, after combining like terms,

   $2(4w+4)=96$

4. 4

   Solve for the width. Dividing both sides by $2$ keeps the arithmetic simple:

   $4w+4=48$

   $4w=44$

   $w=11$

5. 5

   Find the length and verify.

   $\ell=3(11)+4=37$

   $P=2(37+11)=2(48)=96\ \text{m}\quad\checkmark$

   So the rectangle is $11$ m by $37$ m. Notice that the whole method never changed — only the multiplier and the constant did.