1. **State the problem:** Simplify and solve the equation $2i (a + ib) - 7 - 5i + 1 = 3i - (a - ib)$ for $a$ and $b$. 2. **Expand and simplify both sides:** Left side: $$2i(a + ib) - 7 - 5i + 1 = 2i a + 2i (ib) - 7 - 5i + 1 = 2ia + 2i^2 b - 7 - 5i + 1$$ Recall $i^2 = -1$, so: $$2ia + 2(-1) b - 7 - 5i + 1 = 2ia - 2b - 6 - 5i$$ Right side: $$3i - (a - ib) = 3i - a + ib = -a + ib + 3i$$ Combine like terms: $$-a + i(b + 3)$$ 3. **Equate real and imaginary parts:** Real parts: $$-2b - 6 = -a$$ Imaginary parts: $$2a - 5 = b + 3$$ 4. **Rewrite the system:** $$-2b - 6 = -a \implies a = 2b + 6$$ $$2a - 5 = b + 3$$ 5. **Substitute $a$ into the second equation:** $$2(2b + 6) - 5 = b + 3$$ $$4b + 12 - 5 = b + 3$$ $$4b + 7 = b + 3$$ 6. **Solve for $b$:** $$4b - b = 3 - 7$$ $$3b = -4$$ $$b = \frac{-4}{3}$$ 7. **Find $a$ using $a = 2b + 6$:** $$a = 2 \times \frac{-4}{3} + 6 = \frac{-8}{3} + 6 = \frac{-8}{3} + \frac{18}{3} = \frac{10}{3}$$ **Final answer:** $$a = \frac{10}{3}, \quad b = \frac{-4}{3}$$