1. **Problem statement:** Given points on triangle $ABC$ with $D$ on $AB$ such that $\overrightarrow{AD} : \overrightarrow{DB} = 2 : 1$ and $E$ on $AC$ such that $\overrightarrow{AC} : \overrightarrow{CE} = 3 : 2$, express $\overrightarrow{DE}$ as $x \overrightarrow{AB} + y \overrightarrow{AC}$ and find $x$, $y$. 2. **Set up vectors:** Let $\overrightarrow{AB} = \mathbf{b}$ and $\overrightarrow{AC} = \mathbf{c}$. 3. **Find position vectors of $D$ and $E$ relative to $A$:** - Since $D$ divides $AB$ in ratio $2:1$, $$\overrightarrow{AD} = \frac{2}{2+1} \overrightarrow{AB} = \frac{2}{3} \mathbf{b}$$ - Since $E$ divides $AC$ in ratio $3:2$, note $\overrightarrow{AC} : \overrightarrow{CE} = 3 : 2$ means $E$ lies beyond $C$ or between $A$ and $C$? Actually, $\overrightarrow{CE}$ is from $C$ to $E$, so $E$ lies on $AC$ extended beyond $C$ or between $A$ and $C$? Since $\overrightarrow{AC} : \overrightarrow{CE} = 3 : 2$, $CE = \frac{2}{3} AC$, so $E$ lies beyond $C$ along $AC$. Express $\overrightarrow{AE}$: $$\overrightarrow{AE} = \overrightarrow{AC} + \overrightarrow{CE} = \mathbf{c} + \frac{2}{3} \mathbf{c} = \frac{5}{3} \mathbf{c}$$ 4. **Express $\overrightarrow{DE}$:** $$\overrightarrow{DE} = \overrightarrow{AE} - \overrightarrow{AD} = \frac{5}{3} \mathbf{c} - \frac{2}{3} \mathbf{b}$$ 5. **Rewrite $\overrightarrow{DE}$ in terms of $\mathbf{b}$ and $\mathbf{c}$:** $$\overrightarrow{DE} = -\frac{2}{3} \mathbf{b} + \frac{5}{3} \mathbf{c}$$ 6. **Compare with $x \overrightarrow{AB} + y \overrightarrow{AC}$:** $$x = -\frac{2}{3}, \quad y = \frac{5}{3}$$ **Final answer:** $$x = -\frac{2}{3}, \quad y = \frac{5}{3}$$