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# #2x-3# is greater or lesser than zero.

This is easily found:

#2x-3\ge 0 \iff 2x \ge 3 \iff x \ge 3/2#

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

**Case 1: #x\ge 3/2#**

In this case, #|2x-3|=2x-3#, and so we have

#2x-3\le 4 \iff 2x \le 7 \iff x \le 7/2#

We must be very careful: our answer is accepted only if #x\ge 3/2#, and since we found that the answer is #x \le 7/2#, putting the two requests together, we have #x \in \[3/2, 7/2\]#

**Case 2: #x\le 3/2#**

In this case, #|2x-3|=-2x+3#, and so we have

#-2x+3\le 4 \iff -2x \le 1 \iff x \ge -1/2#

As before, we must accept the request #x \ge -1/2# only for #x# values smaller than #3/2#, which means #x \in \[-1/2, 3/2\]#

Our final answer is the sum of the two cases, so #\[-1/2, 3/2\] \cup \[3/2, 7/2\]=\[-1/2, 7/2\]#

Here's WolframAlpha for a visual representation