1. The problem is to express the equation $$(2x + 1)(x - 3) = \frac{\square}{2x + 1} - \frac{\square}{x - 3}$$ by finding the numerators in the fractions on the right side. 2. Start by expanding the left side: $$ (2x + 1)(x - 3) = 2x \cdot x + 2x \cdot (-3) + 1 \cdot x + 1 \cdot (-3) = 2x^2 - 6x + x - 3 = 2x^2 - 5x - 3 $$ 3. To write the right side as a single fraction, find a common denominator: $$ \frac{A}{2x + 1} - \frac{B}{x - 3} = \frac{A(x - 3) - B(2x + 1)}{(2x + 1)(x - 3)} $$ 4. Since the left side equals the right side, their numerators must be equal: $$ 2x^2 - 5x - 3 = A(x - 3) - B(2x + 1) $$ 5. Expand the right side: $$ A x - 3A - 2Bx - B = (A - 2B)x - (3A + B) $$ 6. Equate coefficients of like terms: - For $x^2$: Left side has $2x^2$, right side has none, so this suggests the right side must be adjusted to include $x^2$ terms. This means the initial assumption that the right side is a difference of two fractions with numerators $A$ and $B$ constants is incorrect. 7. Instead, let the numerators be linear expressions: $$ \frac{Ax + C}{2x + 1} - \frac{Bx + D}{x - 3} $$ 8. Then the combined numerator is: $$ (Ax + C)(x - 3) - (Bx + D)(2x + 1) $$ 9. Expand: $$ A x^2 - 3 A x + C x - 3 C - 2 B x^2 - B x - 2 D x - D $$ 10. Group like terms: $$ (A - 2B) x^2 + (-3A + C - B - 2D) x + (-3C - D) $$ 11. Set equal to left side numerator: $$ 2 x^2 - 5 x - 3 $$ 12. Equate coefficients: $$ A - 2B = 2 $$ $$ -3A + C - B - 2D = -5 $$ $$ -3C - D = -3 $$ 13. Solve the system: From the third equation: $$ -3C - D = -3 \implies D = -3C + 3 $$ Substitute $D$ into the second equation: $$ -3A + C - B - 2(-3C + 3) = -5 $$ $$ -3A + C - B + 6C - 6 = -5 $$ $$ -3A - B + 7C = 1 $$ From the first equation: $$ A = 2 + 2B $$ Substitute $A$ into the above: $$ -3(2 + 2B) - B + 7C = 1 $$ $$ -6 - 6B - B + 7C = 1 $$ $$ -7B + 7C = 7 $$ $$ -B + C = 1 $$ $$ C = 1 + B $$ Substitute $C$ into $D$: $$ D = -3(1 + B) + 3 = -3 - 3B + 3 = -3B $$ 14. Choose $B = 0$ for simplicity: $$ C = 1, D = 0, A = 2 $$ 15. Therefore, numerators are: $$ \frac{2x + 1}{2x + 1} - \frac{0x + 0}{x - 3} = \frac{2x + 1}{2x + 1} - 0 $$ But this does not match the original expression. Try $B = 1$: $$ C = 2, D = -3, A = 4 $$ Check: $$ (4x + 2)(x - 3) - (1x - 3)(2x + 1) = 4x^2 - 12x + 2x - 6 - (2x^2 + x - 6x - 3) = 4x^2 - 10x - 6 - 2x^2 - x + 6x + 3 = 2x^2 - 5x - 3 $$ This matches perfectly. 16. Final answer: $$ \frac{4x + 2}{2x + 1} - \frac{x - 3}{x - 3} $$ Simplify the second fraction: $$ \frac{x - 3}{x - 3} = 1 $$ So the right side is: $$ \frac{4x + 2}{2x + 1} - 1 $$ This equals the left side expression. **Answer:** Numerators are $4x + 2$ and $x - 3$ respectively.