Additional Mathematics – Equations, inequalities and graphs | e-Consult

First require the right‑hand side to be non‑negative:

2x − 1 ≥ 0 ⟹ x ≥ ½

Now remove the modulus by writing two simultaneous inequalities:

1. |x² − 4x + 3| ≤ 2x − 1 ⟹ 

   • x² − 4x + 3 ≤ 2x − 1 ⟹ x² − 6x + 4 ≤ 0
   • x² − 4x + 3 ≥ −(2x − 1) ⟹ x² − 2x + 2 ≥ 0

Second inequality x² − 2x + 2 ≥ 0 holds for all real x (discriminant

Solve the first quadratic inequality:

Δ = (−6)² − 4·1·4 = 20, √Δ = 2√5

Roots: x = [6 ± 2√5]/2 = 3 ± √5

Since the coefficient of x² is positive, the expression ≤ 0 between the roots:

3 − √5 ≤ x ≤ 3 + √5

Combine with the domain x ≥ ½. Because 3 − √5 ≈ 0.764 > ½, the domain condition is already satisfied.

Therefore the solution set is