How do you solve absolute value inequalities with absolute value variables on both sides $ \left| 2x \right|\le \left| x-3 \right| $ ?

Hint: We start solving the problem by making use of the fact that if $ a\le b $ , $ a\ge 0 $ , $ b\ge 0 $ , then $ {{a}^{2}}\le {{b}^{2}} $ . We then expand the squares and then make the necessary calculations to proceed through the problem. We then make use of the fact that if $ \left( x-a \right)\left( x-b \right)\le 0 $ and $ a < b $ , then the solution of the inequality is $ a\le x\le b $ to get the required answer for the given inequation.

Complete step by step answer:
According to the problem, we are asked to solve absolute value inequalities with absolute value variables on both sides $ \left| 2x \right|\le \left| x-3 \right| $ .
We have given the inequality $ \left| 2x \right|\le \left| x-3 \right| $ ---(1).
We know that if $ a\le b $ , $ a\ge 0 $ , $ b\ge 0 $ , then $ {{a}^{2}}\le {{b}^{2}} $ . Let us use this result in equation (1).
 $ \Rightarrow {{\left| 2x \right|}^{2}}\le {{\left| x-3 \right|}^{2}} $ .
 $ \Rightarrow 4{{x}^{2}}\le {{x}^{2}}-6x+9 $ .
 $ \Rightarrow 3{{x}^{2}}+6x-9\le 0 $ .
 $ \Rightarrow {{x}^{2}}+2x-3\le 0 $ .
 $ \Rightarrow {{x}^{2}}+3x-x-3\le 0 $ .
 $ \Rightarrow x\left( x+3 \right)-1\left( x+3 \right)\le 0 $ .
 $ \Rightarrow \left( x+3 \right)\left( x-1 \right)\le 0 $ .
 $ \Rightarrow \left( x-\left( -3 \right) \right)\left( x-1 \right)\le 0 $ ---(2).
We know that if $ \left( x-a \right)\left( x-b \right)\le 0 $ and $ a < b $ , then the solution of the inequality is defined as $ a\le x\le b $ . Let us use this result in equation (2).
So, we get the solution of the inequality in equation (2) as $ -3\le x\le 1 $.
$ \therefore $ The solution set (i.e., the values of x) for the given inequality $ \left| 2x \right|\le \left| x-3 \right| $ is $ -3\le x\le 1 $ .

Note:
 We should not use the property if $ a\le b $ , then $ {{a}^{2}}\le {{b}^{2}} $ if the values of a and b are negative in such case the direction of inequality will change. We should perform each step carefully in order to avoid confusion and calculation mistakes. We can also report the obtained answer as $ \left[ -3,1 \right] $ which represents similar values of x as $ -3\le x\le 1 $ . Similarly, we can expect problems to find the solution set for the inequality $ \left| 3x \right| > \left| 1-2x \right| $ .