1. Simplify the following expressions. (a) Simplify $28a - 3[5a - (-7b)] + 18b$. Step 1: Remove the double negative inside the bracket: $5a - (-7b) = 5a + 7b$. Step 2: Multiply inside the bracket by $-3$: $-3(5a + 7b) = -15a - 21b$. Step 3: Combine like terms: $28a - 15a - 21b + 18b = (28a - 15a) + (-21b + 18b) = 13a - 3b$. Answer: $13a - 3b$. (b) Simplify $8y - 3 \left[ \frac{1}{2}(4x - 2y) - \frac{2}{3}(6x - 5y) \right]$. Step 1: Distribute inside the brackets: $\frac{1}{2}(4x - 2y) = 2x - y$. $\frac{2}{3}(6x - 5y) = 4x - \frac{10}{3}y$. Step 2: Subtract inside the bracket: $(2x - y) - (4x - \frac{10}{3}y) = 2x - y - 4x + \frac{10}{3}y = -2x + \left(-1 + \frac{10}{3}\right)y = -2x + \frac{7}{3}y$. Step 3: Multiply by $-3$: $-3(-2x + \frac{7}{3}y) = 6x - 7y$. Step 4: Add $8y$: $8y + 6x - 7y = 6x + y$. Answer: $6x + y$. (c) Simplify $\frac{32x}{y^3} \div \frac{8x^3 z}{3xy}$. Step 1: Rewrite division as multiplication by reciprocal: $\frac{32x}{y^3} \times \frac{3xy}{8x^3 z}$. Step 2: Multiply numerators and denominators: Numerator: $32x \times 3xy = 96x^2 y$. Denominator: $y^3 \times 8x^3 z = 8x^3 y^3 z$. Step 3: Simplify: $\frac{96x^2 y}{8x^3 y^3 z} = \frac{96}{8} \times \frac{x^2}{x^3} \times \frac{y}{y^3} \times \frac{1}{z} = 12 \times \frac{1}{x} \times \frac{1}{y^2} \times \frac{1}{z} = \frac{12}{x y^2 z}$. Answer: $\frac{12}{x y^2 z}$. (d) Simplify $\sqrt[3]{\frac{125 p^{12} q^9}{8}} \div \left( \frac{q^4}{2p} \right)^2$. Step 1: Simplify cube root: $\sqrt[3]{\frac{125 p^{12} q^9}{8}} = \frac{\sqrt[3]{125} \sqrt[3]{p^{12}} \sqrt[3]{q^9}}{\sqrt[3]{8}} = \frac{5 p^4 q^3}{2}$. Step 2: Simplify denominator: $\left( \frac{q^4}{2p} \right)^2 = \frac{q^8}{4 p^2}$. Step 3: Division becomes multiplication by reciprocal: $\frac{5 p^4 q^3}{2} \times \frac{4 p^2}{q^8} = \frac{5 \times 4 \times p^{4+2} \times q^{3-8}}{2} = \frac{20 p^6 q^{-5}}{2} = 10 p^6 q^{-5} = \frac{10 p^6}{q^5}$. Answer: $\frac{10 p^6}{q^5}$. 2. Express each as a single fraction. (a) Simplify $\frac{5x - 6}{3} - \frac{3(5x - 1)}{4} + 2$. Step 1: Write 2 as $\frac{6}{3}$ to have common denominator 12. Step 2: Find common denominator 12: $\frac{5x - 6}{3} = \frac{4(5x - 6)}{12} = \frac{20x - 24}{12}$. $\frac{3(5x - 1)}{4} = \frac{3(5x - 1) \times 3}{12} = \frac{9(5x - 1)}{12} = \frac{45x - 9}{12}$. $2 = \frac{24}{12}$. Step 3: Combine: $\frac{20x - 24}{12} - \frac{45x - 9}{12} + \frac{24}{12} = \frac{20x - 24 - 45x + 9 + 24}{12} = \frac{-25x + 9}{12}$. Answer: $\frac{-25x + 9}{12}$. (b) Simplify $\frac{3}{x - 1} + \frac{5 + x}{2(1 - x)}$. Step 1: Note $1 - x = -(x - 1)$, so rewrite second fraction: $\frac{5 + x}{2(1 - x)} = \frac{5 + x}{2(- (x - 1))} = -\frac{5 + x}{2(x - 1)}$. Step 2: Find common denominator $2(x - 1)$: $\frac{3}{x - 1} = \frac{6}{2(x - 1)}$. Step 3: Combine: $\frac{6}{2(x - 1)} - \frac{5 + x}{2(x - 1)} = \frac{6 - (5 + x)}{2(x - 1)} = \frac{6 - 5 - x}{2(x - 1)} = \frac{1 - x}{2(x - 1)}$. Step 4: Simplify numerator and denominator: $\frac{1 - x}{2(x - 1)} = \frac{-(x - 1)}{2(x - 1)} = -\frac{1}{2}$. Answer: $-\frac{1}{2}$. 3. Expand and simplify $(5x + y)(x - 2y) - 2(3x - 2y)^2$. Step 1: Expand first product: $(5x + y)(x - 2y) = 5x \times x + 5x \times (-2y) + y \times x + y \times (-2y) = 5x^2 - 10xy + xy - 2y^2 = 5x^2 - 9xy - 2y^2$. Step 2: Expand square: $(3x - 2y)^2 = (3x)^2 - 2 \times 3x \times 2y + (2y)^2 = 9x^2 - 12xy + 4y^2$. Step 3: Multiply by $-2$: $-2(9x^2 - 12xy + 4y^2) = -18x^2 + 24xy - 8y^2$. Step 4: Add results: $5x^2 - 9xy - 2y^2 - 18x^2 + 24xy - 8y^2 = (5x^2 - 18x^2) + (-9xy + 24xy) + (-2y^2 - 8y^2) = -13x^2 + 15xy - 10y^2$. Answer: $-13x^2 + 15xy - 10y^2$. 4. Factorise completely. (a) Factorise $6pq - 10p + 5r - 3qr$. Step 1: Group terms: $(6pq - 10p) + (5r - 3qr)$. Step 2: Factor each group: $2p(3q - 5) + r(5 - 3q)$. Step 3: Note $5 - 3q = -(3q - 5)$, so rewrite: $2p(3q - 5) - r(3q - 5) = (3q - 5)(2p - r)$. Answer: $(3q - 5)(2p - r)$. (b) Factorise $49(x - y)^2 - 4y^2$. Step 1: Recognize difference of squares: $49(x - y)^2 - (2y)^2 = [7(x - y)]^2 - (2y)^2$. Step 2: Apply difference of squares formula: $[7(x - y) - 2y][7(x - y) + 2y]$. Step 3: Simplify each bracket: $7x - 7y - 2y = 7x - 9y$, $7x - 7y + 2y = 7x - 5y$. Answer: $(7x - 9y)(7x - 5y)$. (c) Factorise $(5x + 4y)^2 - 20x - 16y$. Step 1: Expand square: $(5x + 4y)^2 = 25x^2 + 40xy + 16y^2$. Step 2: Rewrite expression: $25x^2 + 40xy + 16y^2 - 20x - 16y$. Step 3: Group terms: $25x^2 + 40xy - 20x + 16y^2 - 16y$. Step 4: Factor by grouping: $5x(5x + 8y - 4) + 8y(2y - 2)$. Step 5: Note $2y - 2 = 2(y - 1)$, but this does not match the first group. Step 6: Try to factor as quadratic in $x$: $25x^2 + (40y - 20)x + 16y^2 - 16y$. Step 7: Use quadratic formula or factorization: Try $(5x + a)(5x + b) = 25x^2 + 5x(a + b) + ab$. We want $5x(a + b) = 40y - 20$ and $ab = 16y^2 - 16y$. Try $a = 4y - 4$, $b = 4y - 4$: $5x(a + b) = 5x(8y - 8) = 40xy - 40x$ (close but not exact). Try $a = 8y - 4$, $b = 2y - 4$: Sum: $8y - 4 + 2y - 4 = 10y - 8$, times 5x is $50xy - 40x$ (too large). Try $a = 8y - 2$, $b = 2y - 8$: Sum: $10y - 10$, times 5x is $50xy - 50x$ (too large). Step 8: Try factoring by grouping differently: Rewrite expression as: $25x^2 + 40xy - 20x + 16y^2 - 16y = (25x^2 + 40xy) - 20x + (16y^2 - 16y)$. Factor: $5x(5x + 8y) - 4(5x) + 16y(y - 1)$. No common factor in all terms. Step 9: Try completing the square or accept no simple factorization. Answer: Expression does not factor nicely over integers; leave as is or factor partially. (d) Factorise $2a^5 b - 32ab^5$. Step 1: Factor out common terms: $2ab(a^4 - 16b^4)$. Step 2: Recognize difference of squares: $a^4 - 16b^4 = (a^2)^2 - (4b^2)^2 = (a^2 - 4b^2)(a^2 + 4b^2)$. Step 3: Further factor $a^2 - 4b^2$: $a^2 - 4b^2 = (a - 2b)(a + 2b)$. Answer: $2ab(a - 2b)(a + 2b)(a^2 + 4b^2)$.