1. **Problem 9:** Find $x$ if $KD = x + 7$ and $KN = 2x - 4$. Since $K$ is the centroid, it divides the median into a ratio of 2:1, so $KD = 2 imes KN$. 2. Write the equation: $$x + 7 = 2(2x - 4)$$ 3. Expand the right side: $$x + 7 = 4x - 8$$ 4. Subtract $x$ from both sides: $$\cancel{x} + 7 = 4x - 8 - \cancel{x}$$ $$7 = 3x - 8$$ 5. Add 8 to both sides: $$7 + 8 = 3x - 8 + 8$$ $$15 = 3x$$ 6. Divide both sides by 3: $$\frac{15}{\cancel{3}} = \frac{3x}{\cancel{3}}$$ $$5 = x$$ --- 7. **Problem 10:** Find $x$ if $AK = 2x - 8$ and $AG = 2x - 5$. Since $K$ is the centroid, $AK$ is two-thirds of $AG$: $$AK = \frac{2}{3} AG$$ 8. Write the equation: $$2x - 8 = \frac{2}{3}(2x - 5)$$ 9. Multiply both sides by 3 to clear the fraction: $$3(2x - 8) = 2(2x - 5)$$ 10. Expand both sides: $$6x - 24 = 4x - 10$$ 11. Subtract $4x$ from both sides: $$6x - 4x - 24 = 4x - 4x - 10$$ $$2x - 24 = -10$$ 12. Add 24 to both sides: $$2x - 24 + 24 = -10 + 24$$ $$2x = 14$$ 13. Divide both sides by 2: $$\frac{2x}{\cancel{2}} = \frac{14}{\cancel{2}}$$ $$x = 7$$ **Final answers:** Problem 9: $x = 5$ Problem 10: $x = 7$