1. **State the problem:** We want to find constants $A$ and $B$ such that $$\frac{x}{(x-4)(x-1)} = \frac{A}{x-4} + \frac{B}{x-1}$$ 2. **Multiply both sides by the denominator $(x-4)(x-1)$ to clear fractions:** $$x = A(x-1) + B(x-4)$$ 3. **Expand the right side:** $$x = A x - A + B x - 4 B$$ 4. **Group like terms:** $$x = (A + B) x - (A + 4 B)$$ 5. **Equate coefficients of powers of $x$ on both sides:** - Coefficient of $x$: $1 = A + B$ - Constant term: $0 = - (A + 4 B)$ 6. **Solve the system of equations:** From the constant term: $$0 = - (A + 4 B) \implies A + 4 B = 0 \implies A = -4 B$$ Substitute into the first equation: $$1 = A + B = -4 B + B = -3 B \implies B = -\frac{1}{3}$$ Then, $$A = -4 \times \left(-\frac{1}{3}\right) = \frac{4}{3}$$ 7. **Final result:** $$A = \frac{4}{3}, \quad B = -\frac{1}{3}$$ This shows how the values $A=\frac{4}{3}$ and $B=-\frac{1}{3}$ are derived by equating coefficients after clearing denominators.