1. **State the problem:** We need to find the dimensions of a rectangular room where the length is twice the width and the perimeter is 48 meters. 2. **Define variables:** Let the width be $w$ meters. 3. Since the length is twice the width, the length is $l = 2w$ meters. 4. **Formula for perimeter of a rectangle:** $$P = 2(l + w)$$ 5. Substitute the known values into the formula: $$48 = 2(2w + w)$$ 6. Simplify inside the parentheses: $$48 = 2(3w)$$ 7. Multiply: $$48 = 6w$$ 8. Solve for $w$ by dividing both sides by 6: $$\frac{48}{\cancel{6}} = \frac{6w}{\cancel{6}}$$ $$8 = w$$ 9. Find the length $l$: $$l = 2w = 2 \times 8 = 16$$ 10. **Answer:** The width is 8 meters and the length is 16 meters.