1. Solve the simultaneous linear equations with complex coefficients: Given: $$2z + (3 + i)w = 9 - i$$ $$(3 - i)z - iw = -1 + i$$ 2. Use substitution or elimination. Let's use elimination: Multiply the first equation by $i$: $$i(2z) + i(3 + i)w = i(9 - i)$$ $$2iz + (3i - 1)w = 9i + 1$$ 3. Now add this to the second equation: $$(3 - i)z - iw + 2iz + (3i - 1)w = -1 + i + 9i + 1$$ Simplify: $$(3 - i)z + 2iz + (-i + 3i - 1)w = 10i$$ $$(3 - i + 2i)z + (2i - 1)w = 10i$$ $$(3 + i)z + (2i - 1)w = 10i$$ 4. From the first original equation: $$2z + (3 + i)w = 9 - i$$ Express $z$: $$z = \frac{9 - i - (3 + i)w}{2}$$ 5. Substitute $z$ into the combined equation: $$ (3 + i)\frac{9 - i - (3 + i)w}{2} + (2i - 1)w = 10i$$ Multiply both sides by 2: $$(3 + i)(9 - i - (3 + i)w) + 2(2i - 1)w = 20i$$ Expand: $$(3 + i)(9 - i) - (3 + i)^2 w + 2(2i - 1)w = 20i$$ Calculate $(3 + i)(9 - i)$: $$3 \times 9 + 3 \times (-i) + i \times 9 + i \times (-i) = 27 - 3i + 9i - i^2 = 27 + 6i + 1 = 28 + 6i$$ Calculate $(3 + i)^2$: $$(3 + i)^2 = 9 + 6i + i^2 = 9 + 6i - 1 = 8 + 6i$$ So: $$(28 + 6i) - (8 + 6i)w + 2(2i - 1)w = 20i$$ Simplify terms with $w$: $$-(8 + 6i)w + (4i - 2)w = [-(8 + 6i) + 4i - 2]w = (-8 - 6i + 4i - 2)w = (-10 - 2i)w$$ Equation becomes: $$28 + 6i - (10 + 2i)w = 20i$$ 6. Isolate $w$: $$-(10 + 2i)w = 20i - 28 - 6i = -28 + 14i$$ $$w = \frac{-28 + 14i}{-(10 + 2i)} = \frac{-28 + 14i}{-10 - 2i}$$ Multiply numerator and denominator by conjugate of denominator $-10 + 2i$: $$w = \frac{(-28 + 14i)(-10 + 2i)}{(-10 - 2i)(-10 + 2i)}$$ Calculate denominator: $$(-10)^2 - (2i)^2 = 100 - (-4) = 104$$ Calculate numerator: $$(-28)(-10) + (-28)(2i) + 14i(-10) + 14i(2i) = 280 - 56i - 140i + 28i^2 = 280 - 196i + 28(-1) = 280 - 196i - 28 = 252 - 196i$$ So: $$w = \frac{252 - 196i}{104} = \frac{63}{26} - \frac{49}{26}i$$ 7. Substitute $w$ back to find $z$: $$z = \frac{9 - i - (3 + i)w}{2}$$ Calculate $(3 + i)w$: $$(3 + i)\left(\frac{63}{26} - \frac{49}{26}i\right) = 3 \times \frac{63}{26} - 3 \times \frac{49}{26}i + i \times \frac{63}{26} - i \times \frac{49}{26}i = \frac{189}{26} - \frac{147}{26}i + \frac{63}{26}i - \frac{49}{26}i^2 = \frac{189}{26} - \frac{84}{26}i + \frac{49}{26} = \frac{238}{26} - \frac{84}{26}i = \frac{119}{13} - \frac{42}{13}i$$ Calculate numerator: $$9 - i - \left(\frac{119}{13} - \frac{42}{13}i\right) = 9 - i - \frac{119}{13} + \frac{42}{13}i = \left(9 - \frac{119}{13}\right) + \left(-1 + \frac{42}{13}\right)i = \frac{117}{13} - \frac{119}{13} + \frac{-13}{13} + \frac{42}{13}i = -\frac{2}{13} + \frac{29}{13}i$$ Divide by 2: $$z = \frac{-\frac{2}{13} + \frac{29}{13}i}{2} = -\frac{1}{13} + \frac{29}{26}i$$ --- 8. Simplify the algebraic expression: $$\frac{a}{(a+b)^2 - 2ab} \div \frac{(a+b)^3 - 3ab(a+b)}{a^4 - b^4} \div \frac{(a+b)^2 - 4ab}{(a+b)^2 - 3ab}$$ 9. Simplify denominators and numerators: $$(a+b)^2 - 2ab = a^2 + 2ab + b^2 - 2ab = a^2 + b^2$$ $$(a+b)^3 - 3ab(a+b) = (a+b)((a+b)^2 - 3ab) = (a+b)(a^2 + 2ab + b^2 - 3ab) = (a+b)(a^2 - ab + b^2)$$ $$a^4 - b^4 = (a^2 - b^2)(a^2 + b^2) = (a - b)(a + b)(a^2 + b^2)$$ $$(a+b)^2 - 4ab = a^2 + 2ab + b^2 - 4ab = a^2 - 2ab + b^2 = (a - b)^2$$ $$(a+b)^2 - 3ab = a^2 + 2ab + b^2 - 3ab = a^2 - ab + b^2$$ 10. Rewrite expression: $$\frac{a}{a^2 + b^2} \div \frac{(a+b)(a^2 - ab + b^2)}{(a - b)(a + b)(a^2 + b^2)} \div \frac{(a - b)^2}{a^2 - ab + b^2}$$ 11. Division becomes multiplication by reciprocal: $$= \frac{a}{a^2 + b^2} \times \frac{(a - b)(a + b)(a^2 + b^2)}{(a+b)(a^2 - ab + b^2)} \times \frac{a^2 - ab + b^2}{(a - b)^2}$$ 12. Cancel common factors: - $(a+b)$ cancels - $(a^2 + b^2)$ cancels - $(a^2 - ab + b^2)$ cancels - $(a - b)$ in numerator cancels with one $(a - b)$ in denominator Remaining: $$a \times \frac{a - b}{1} = a(a - b) = a^2 - ab$$ --- 13. Solve the system: $$x - y = 2$$ $$x^2 + 2xy + y^2 = 3$$ 14. Note that: $$x^2 + 2xy + y^2 = (x + y)^2$$ 15. From first equation: $$x = y + 2$$ 16. Substitute into second: $$(y + 2 + y)^2 = 3$$ $$(2y + 2)^2 = 3$$ $$4(y + 1)^2 = 3$$ $$(y + 1)^2 = \frac{3}{4}$$ 17. Solve for $y$: $$y + 1 = \pm \frac{\sqrt{3}}{2}$$ $$y = -1 \pm \frac{\sqrt{3}}{2}$$ 18. Find $x$: $$x = y + 2 = 1 \pm \frac{\sqrt{3}}{2}$$ Solutions: $$(x, y) = \left(1 + \frac{\sqrt{3}}{2}, -1 + \frac{\sqrt{3}}{2}\right), \left(1 - \frac{\sqrt{3}}{2}, -1 - \frac{\sqrt{3}}{2}\right)$$ --- 19. Find solution region for inequalities: $$2x + y \geq 4$$ $$x - y \leq 4$$ 20. Rewrite inequalities: $$y \geq 4 - 2x$$ $$y \geq x - 4$$ 21. The solution region is the intersection where: $$y \geq 4 - 2x$$ $$y \leq x - 4$$ 22. Graphically, this is the region above or on line $2x + y = 4$ and below or on line $x - y = 4$. Final answers: - Complex system: $z = -\frac{1}{13} + \frac{29}{26}i$, $w = \frac{63}{26} - \frac{49}{26}i$ - Simplified expression: $a^2 - ab$ - Quadratic system solutions: $(1 \pm \frac{\sqrt{3}}{2}, -1 \pm \frac{\sqrt{3}}{2})$ - Inequality solution region: intersection of $y \geq 4 - 2x$ and $y \leq x - 4$