1. **State the problem:** Write the expression $3(2 - i) + i(5 + 2i)$ in the form $x + yi$, where $x, y \in \mathbb{R}$. 2. **Apply distributive property:** $$3(2 - i) + i(5 + 2i) = 3 \cdot 2 - 3 \cdot i + i \cdot 5 + i \cdot 2i = 6 - 3i + 5i + 2i^2$$ 3. **Recall that $i^2 = -1$:** $$6 - 3i + 5i + 2(-1) = 6 - 3i + 5i - 2$$ 4. **Combine like terms:** $$6 - 2 + (-3i + 5i) = 4 + 2i$$ 5. **Final answer:** The expression in the form $x + yi$ is $$\boxed{4 + 2i}$$ --- 1. **State the problem:** Given real numbers $p$ and $k$ such that $$2 - 3pi - 9 - ki = 4k + 1 - i,$$ find the values of $p$ and $k$. 2. **Group real and imaginary parts:** Real parts: $2 - 9 = 4k + 1$ Imaginary parts: $-3p - k = -1$ 3. **Simplify real parts equation:** $$2 - 9 = -7$$ So, $$-7 = 4k + 1$$ Subtract 1 from both sides: $$-7 - 1 = 4k$$ $$-8 = 4k$$ Divide both sides by 4: $$\cancel{\frac{-8}{4}} = \cancel{\frac{4k}{4}}$$ $$-2 = k$$ 4. **Simplify imaginary parts equation:** $$-3p - k = -1$$ Substitute $k = -2$: $$-3p - (-2) = -1$$ $$-3p + 2 = -1$$ Subtract 2 from both sides: $$-3p = -1 - 2$$ $$-3p = -3$$ Divide both sides by -3: $$\cancel{\frac{-3p}{-3}} = \cancel{\frac{-3}{-3}}$$ $$p = 1$$ 5. **Final answer:** $$\boxed{p = 1, \quad k = -2}$$