Question 1190187
<pre>

If you are required to go strictly by the definition 
of absolute value, you would do it this way:

{{{abs(x-3)}}}{{{""=""}}}{{{2}}}

{{{sqrt((x-3)^2)}}}{{{""=""}}}{{{2}}}

Square both sides:

{{{(x-3)^2}}}{{{""=""}}}{{{2^2}}}

{{{x^2-6x+9}}}{{{""=""}}}{{{4}}}

Subtract 4 from both sides:

{{{x^2-6x+5}}}{{{""=""}}}{{{0}}}

Factor the left sides:

{{{(x-5)(x-1)}}}{{{""=""}}}{{{0}}}

x - 5 = 0;  x - 1 = 0
    x = 5;      x = 1 

Edwin</pre>