What is the solution set for #abs(x – 2) + 4 = 3#?
As it is written, your absolute value equation has no real solutions.
That happens because the absolute value of any real number always returns the positive value of said number, irrespective of its sign.
In your case, you know that if you isolate the modulus on one side of the equation by adding
#|x-2| + color(red)(cancel(color(black)(4))) - color(red)(cancel(color(black)(4))) = 3-4#
#|x-2| = -1#
In this case, you need the absolute value of a real number, which is what
This contradicts the definition of the absolute value, which implies that