A rectangle has a width of 2 cm and a length of 9 cm. The width is increased by 1 cm and the length is increased by $x$ cm. The area of the resulting rectangle is twice the area of the original rectangle. Find the value of $x$.

1. Problem Description

A rectangle has a width of 2 cm and a length of 9 cm. The width is increased by 1 cm and the length is increased by $x$ cm. The area of the resulting rectangle is twice the area of the original rectangle. Find the value of $x$.

2. Solution Steps

Let $w$ be the width and $l$ be the length of the original rectangle. We are given $w = 2$ cm and $l = 9$ cm.

The area of the original rectangle is $A_1 = w \times l = 2 \times 9 = 18$ cm$^2$.

The new width is $w' = w + 1 = 2 + 1 = 3$ cm.

The new length is $l' = l + x = 9 + x$ cm.

The area of the new rectangle is $A_2 = w' \times l' = 3(9+x)$ cm$^2$.

We are given that the area of the new rectangle is twice the area of the original rectangle, so $A_2 = 2A_1$.

Substituting the values, we get

$3(9+x) = 2(18)$

$27 + 3x = 36$

$3x = 36 - 27$

$3x = 9$

$x = \frac{9}{3}$

$x = 3$

3. Final Answer

The value of $x$ is 3.