1. Set up the initial dimensions and the reduction
- Initial Length $(L)$ : 11 meters
- Initial Width $(W)$ : 5 meters
- Reduction: $x$ meters from each dimension
2. Set up the new dimensions
- New Length $\(\left(L^{\prime}\right): 11-x\)$
- New Width ( $\(W^{\prime}\)$ ): $\(5-x\)$
3. Set up the equation based on the new ratio

The new ratio of length to width ( $\(L^{\prime}\)$ to $\(W^{\prime}\)$ ) is 8 to 3 :

$$
\(\frac{L^{\prime}}{W^{\prime}}=\frac{8}{3}\)
$$


Substitute the new dimensions:

$$
\(\frac{11-x}{5-x}=\frac{8}{3}\)
$$

4. Solve for $x$

Cross-multiply the equation:

$$
\(3(11-x)=8(5-x)\)
$$


Distribute the numbers on both sides:

$$
\(33-3 x=40-8 x\)
$$

Gather the $x$ terms on one side (add $8 x$ to both sides):

$$
\(\begin{gathered}
33-3 x+8 x=40 \\
33+5 x=40
\end{gathered}\)
$$


Gather the constant terms on the other side (subtract 33 from both sides):

$$
\(\begin{gathered}
5 x=40-33 \\
5 x=7
\end{gathered}\)
$$


Solve for $x$ :

$$
\(\begin{aligned}
& x=\frac{7}{5} \\
& x=1.4
\end{aligned}\)
$$

5. Check the options

The calculated value of $x$ is 1.4 .
A. 1.4
B. 1.6
C. 1.8
D. 2.0
E. 2.2

The correct value of $x$ is $\(\mathbf{1 . 4}\)$.