Algebra-2

Consider different cases based on the signs of expressions within absolute values:

Case 1: When both 2x + 1 and x - 3 are non-negative.
This means 2x + 1 ≥ 0 and x - 3 ≥ 0.
Solving 2x + 1 ≥ 0 gives x ≥ -1/2.
Solving x - 3 ≥ 0 gives x ≥ 3.
In this case, the equation becomes 2x + 1 = x - 3.
Rearranging this equation, we get x + 1 = -3.
Simplifying, x = -4, but this value doesn't satisfy the condition x ≥ 3, so this case has no solution.

Case 2: When 2x + 1 is non-negative and x - 3 is negative.
2x + 1 ≥ 0 gives x ≥ -1/2.
x - 3 < 0 gives x < 3.
The equation now is 2x + 1 = -(x - 3).
Expanding the right si...