How do you solve abs(2x-3)<=4?

Feb 4, 2015

First of all, you have to determine the absolute value. Since $| a | = a$ if $a > 0$ and $- a$ if $a < 0$, we need to determine for which $x$ $2 x - 3$ is greater or lesser than zero.

This is easily found:
$2 x - 3 \setminus \ge 0 \setminus \iff 2 x \setminus \ge 3 \setminus \iff x \setminus \ge \frac{3}{2}$

Thus, we need to study two different disequality, and then put them back together:

Case 1: $x \setminus \ge \frac{3}{2}$

In this case, $| 2 x - 3 | = 2 x - 3$, and so we have
$2 x - 3 \setminus \le 4 \setminus \iff 2 x \setminus \le 7 \setminus \iff x \setminus \le \frac{7}{2}$

We must be very careful: our answer is accepted only if $x \setminus \ge \frac{3}{2}$, and since we found that the answer is $x \setminus \le \frac{7}{2}$, putting the two requests together, we have $x \setminus \in \setminus \left[\frac{3}{2} , \frac{7}{2} \setminus\right]$

Case 2: $x \setminus \le \frac{3}{2}$

In this case, $| 2 x - 3 | = - 2 x + 3$, and so we have
$- 2 x + 3 \setminus \le 4 \setminus \iff - 2 x \setminus \le 1 \setminus \iff x \setminus \ge - \frac{1}{2}$

As before, we must accept the request $x \setminus \ge - \frac{1}{2}$ only for $x$ values smaller than $\frac{3}{2}$, which means $x \setminus \in \setminus \left[- \frac{1}{2} , \frac{3}{2} \setminus\right]$

Our final answer is the sum of the two cases, so $\setminus \left[- \frac{1}{2} , \frac{3}{2} \setminus\right] \setminus \cup \setminus \left[\frac{3}{2} , \frac{7}{2} \setminus\right] = \setminus \left[- \frac{1}{2} , \frac{7}{2} \setminus\right]$

Here's WolframAlpha for a visual representation