1. Let's consider a challenging algebra problem: Solve for $x$ in the equation $$\frac{2x^2 - 3x + 1}{x - 1} = 4x - 5.$$\n\n2. The problem requires solving a rational equation where the numerator is a quadratic expression and the denominator is linear. We will multiply both sides by the denominator to eliminate the fraction, but we must remember that $x \neq 1$ because it would make the denominator zero.\n\n3. Multiply both sides by $x - 1$: $$2x^2 - 3x + 1 = (4x - 5)(x - 1).$$\n\n4. Expand the right side: $$2x^2 - 3x + 1 = 4x^2 - 4x - 5x + 5 = 4x^2 - 9x + 5.$$\n\n5. Bring all terms to one side to set the equation to zero: $$0 = 4x^2 - 9x + 5 - (2x^2 - 3x + 1) = 4x^2 - 9x + 5 - 2x^2 + 3x - 1 = 2x^2 - 6x + 4.$$\n\n6. Simplify the quadratic equation: $$2x^2 - 6x + 4 = 0.$$\n\n7. Divide the entire equation by 2 to simplify: $$\cancel{2}x^2 - \cancel{2} \cdot 3x + \cancel{2} \cdot 2 = 0 \Rightarrow x^2 - 3x + 2 = 0.$$\n\n8. Factor the quadratic: $$(x - 1)(x - 2) = 0.$$\n\n9. Solve for $x$: $$x = 1 \quad \text{or} \quad x = 2.$$\n\n10. Recall the restriction $x \neq 1$ because it makes the denominator zero in the original equation. So, discard $x=1$.\n\n11. The only valid solution is $$\boxed{x = 2}.$$