1. **State the problem:** We want to understand how the terms 27 and -10 appeared in the equation after multiplying both sides by the common denominator $ (x-3)(x+3) $. 2. **Recall the original step:** $$ (x - 3)(x + 3) \left( \frac{4x}{x-3} - 3 \right) = (x - 3)(x + 3) \cdot \frac{3x - 1}{x + 3} $$ 3. **Distribute the multiplication on the left side:** - Multiply $ \frac{4x}{x-3} $ by $ (x-3)(x+3) $: $$ \frac{4x}{x-3} \cdot (x-3)(x+3) = 4x \cancel{(x-3)} (x+3) / \cancel{(x-3)} = 4x(x+3) $$ - Multiply $ -3 $ by $ (x-3)(x+3) $: $$ -3 \cdot (x-3)(x+3) = -3(x^2 - 9) $$ (since $ (x-3)(x+3) = x^2 - 9 $) 4. **Simplify $ -3(x^2 - 9) $:** $$ -3x^2 + 27 $$ 5. **So the left side becomes:** $$ 4x(x+3) - 3x^2 + 27 $$ 6. **Expand $ 4x(x+3) $:** $$ 4x^2 + 12x $$ 7. **Therefore, the left side is:** $$ 4x^2 + 12x - 3x^2 + 27 $$ 8. **Now, for the right side:** $$ (3x - 1)(x - 3) $$ 9. **Expand the right side:** $$ 3x \cdot x - 3x \cdot 3 - 1 \cdot x + 1 \cdot 3 = 3x^2 - 9x - x + 3 $$ 10. **Combine like terms:** $$ 3x^2 - 10x + 3 $$ **Summary:** - The 27 comes from $ -3 \times (-9) $ when expanding $ -3(x^2 - 9) $. - The $ -10x $ comes from combining $ -9x $ and $ -x $ in the expansion of $ (3x - 1)(x - 3) $. Thus, the terms 27 and -10 appear naturally from expanding and simplifying the expressions after multiplying by the common denominator.