Prove that the mean deviation for $y$ is equal to the mean deviation for $x$, i.e., $$ \mu.D_y = \mu.D_x $$ if $$ y_{i} = x_{i} \pm 3 ,\mbox{ where } i=1,2,\ldots, n, $$ and $x_i$ are ungrouped data.
$\textbf{Attempt}$: \begin{align*} \mu.D_y &= \mu.D_x \\ \dfrac{\displaystyle{\sum_{i=1}^n} |y_i-\overline{y}|}{n} &= \dfrac{\displaystyle{\sum_{i=1}^n} |x_i-\overline{x}|}{n} \\ \sum_{i=1}^{n}|(x_i\pm 3)-\overline{y}| &= \sum_{i=1}^{n}|(y_i\pm 3)-\overline{x}| \\ \end{align*}
I'm trying to prove it like this, but I don't know how to complete the solution.
Let $\overline{x}$ be the mean of ungrouped data $x_1,\ldots, x_n$. Then the mean deviation for $x_i$ is $$ \mu.D_x = \frac{1}{n}\sum_{i=1}^{n}|x_i-\overline{x}|. $$ Suppose $y_i=x_i+3$ for each $1\leq i\leq n$. Then the mean for $y_i$ is $$ \color{blue}{\overline{y}} =\frac{1}{n} \sum_{i=1}^n y_i =\frac{1}{n}\sum_{i=1}^n (x_i+3) = \frac{1}{n} \sum_{i=1}^n x_i + \frac{3n}{n} \color{blue}{= \overline{x} + 3}. $$ Thus the mean deviation for $y_i$ simplifies as \begin{align*} \mu.D_y &= \frac{1}{n}\sum_{i=1}^n |y_i-\overline{y}| \\ &= \frac{1}{n}\sum_{i=1}^n |(x_i+3)-(\overline{x}+3)| \\ &= \frac{1}{n}\sum_{i=1}^n |x_i-\overline{x}| \\ &= \mu.D_x. \end{align*}
We obtain a similar result if $y_i=x_i-3$ for each $1\leq i\leq n$.