The absolute value ∣A∣ is the distance of A from 0, so it is never negative. ∣A∣=4 means A is 4 units from 0: A=4 or A=−4.
Isolate the absolute value first. In 2∣x−3∣+1=9, subtract 1 and divide by 2 before touching the bars: ∣x−3∣=4.
Then split into two cases. x−3=4 gives x=7, and x−3=−4 gives x=−1.
Count before you solve. Once ∣A∣=c is isolated, with c a number: if c>0, there are two cases (and, when A is linear in x, two solutions). If c=0, there is one case, A=0 (one solution when A is linear). If c<0, there are none, since a distance can't be negative.
When x also appears outside the bars, as in ∣x−4∣=2x+1, solve both cases and check each answer in the original. A case can produce a value that fails. Here x−4=2x+1 gives x=−5, but then the right side is −9, and an absolute value can't equal a negative, so x=−5 fails. The other case, x−4=−(2x+1), gives 3x=3 and x=1, which checks: ∣−3∣=3=2(1)+1. The only solution is x=1.
On a graph, y=∣x−3∣ is a V with its corner at (3,0). The solutions of ∣x−3∣=4 are where the V meets the horizontal line y=4.
Desmos When Desmos is faster
Graph each side as its own y=. Type abs( or the | key for the bars: y=3∣2x−1∣+4 and y=19. Click the points where the graphs cross; their x-coordinates are the solutions, here −2 and 3.
This is especially useful when x is also outside the bars, as in ∣x−4∣=2x+1: graph y=∣x−4∣ and y=2x+1, and they cross only at (1,3). Every crossing on the graph is a real solution, so there is nothing to check afterward.
For a simple equation such as ∣x+2∣=6, the two cases take seconds by hand.