How to Prove that |x-y|≤3k Given |x-3|≤k and |y-3|≤2k in Algebra?

[Albert1]

given :

$\left | x-3 \right |\leq k$, and $\left | y-3 \right |\leq 2k$

prove :

$\left | x-y \right |\leq 3k$

 

[kaliprasad]

given :

$\left | x-3 \right |\leq k$, and $\left | y-3 \right |\leq 2k$

prove :

$\left | x-y \right |\leq 3k$

Spoiler

in the straight line maximum distance from 3 to x is k and maximum from 3 to y is 2k. the distance from x to y can be maximum if they are on opposite sides and the sum of distance = k + 2k = 3k
hence the result

algebraically

Spoiler

$\left | x-3 \right |\leq k$
$\left | y-3 \right |\leq 2k$ or $\left | 3-y \right |\leq 2k$
hence
$\left | x-y \right | = \left | x-3 + 3 -y \right |$
$ \le \left | x-3 \right | + \left | 3 -y \right |$
$ \le k + 2k$
$ \le 3k $