# How do you solve absolute value inequalities with absolute value variables on both sides abs(2x)<=abs(x-3)?

Apr 7, 2015

All values of x in the interval [-3,2]

To remove the absolute value sign, square both sides, so that both would be positive only. Accordingly,
4${x}^{2}$ $\le$ x-3)^2
4${x}^{2}$ $\le$ ${x}^{2}$ -6x+9
3${x}^{2}$ +6x - 9 $\le 0$
${x}^{2}$ +2x -3 $\le$0
(x+3)(x-2) $\le 0$.

Now, there can be two options for this inequality to hold good.
Case 1 x+3 is positive, that is x $\ge$-3. and x-2 is negative, that is x$\le$2.

Case 2 x+3 is negative, that is x$\le$ -3 and x-2 is positive, that is x$\ge$2

To understand the solution mark the above inequalities on the number line. The solution to the inequality would be x$\ge$-3 and x$\le$2. In the interval notation it would be [-3,2]