Algebra-2

To solve the equation |x + 2| = |3x - 4|, we need to consider different cases based on the signs of the expressions inside the absolute value symbols.

Case 1: When both x + 2 and 3x - 4 are non-negative.
This means x + 2 ≥ 0 and 3x - 4 ≥ 0.
Solving x + 2 ≥ 0 gives x ≥ -2.
Solving 3x - 4 ≥ 0 gives 3x ≥ 4, so x ≥ 4/3.
In this case, the equation becomes x + 2 = 3x - 4.
Rearranging this equation, we get 4 + 2 = 3x - x.
Simplifying, 6 = 2x, so x = 3.

Case 2: When x + 2 is non-negative and 3x - 4 is negative.
x + 2 ≥ 0 gives x ≥ -2.
3x - 4 < 0 gives 3x < 4, so x < 4/3.
The equation now is x + 2 = -(3x - 4).
Expanding the right side, x + 2 = -3x + 4.
Adding 3x to both sides, 4x + 2 = 4.
Subtracting 2 from both sides, 4x = 2, so...