1. **Make x the subject of the formula** **a)** Given: $$3p = 9r - \frac{3x^2}{q^2}$$ Step 1: Isolate the term with $x^2$: $$3p - 9r = -\frac{3x^2}{q^2}$$ Step 2: Multiply both sides by $-q^2$ to clear the denominator: $$-(3p - 9r)q^2 = 3x^2$$ Step 3: Simplify the left side: $$-(3p - 9r)q^2 = 3x^2 \implies (9r - 3p)q^2 = 3x^2$$ Step 4: Divide both sides by 3: $$\frac{(9r - 3p)q^2}{3} = x^2 \implies (3r - p)q^2 = x^2$$ Step 5: Take the square root of both sides: $$x = \pm q \sqrt{3r - p}$$ **b)** Given: $$x(2 + a) = b(x + 3)$$ Step 1: Expand the right side: $$x(2 + a) = bx + 3b$$ Step 2: Bring all $x$ terms to one side: $$x(2 + a) - bx = 3b$$ Step 3: Factor out $x$: $$x[(2 + a) - b] = 3b$$ Step 4: Divide both sides by $[(2 + a) - b]$: $$x = \frac{3b}{2 + a - b}$$ --- 2. **Factorize, simplify and make x the subject:** Given: $$\frac{4xy + 8y^2}{2y^2 - 16xy} = -2c$$ Step 1: Factor numerator and denominator: $$\frac{4y(x + 2y)}{2y^2 - 16xy}$$ Step 2: Factor denominator: $$2y^2 - 16xy = 2y(y - 8x)$$ Step 3: Rewrite: $$\frac{4y(x + 2y)}{2y(y - 8x)} = -2c$$ Step 4: Simplify common factor $2y$: $$\frac{2(x + 2y)}{y - 8x} = -2c$$ Step 5: Multiply both sides by denominator: $$2(x + 2y) = -2c(y - 8x)$$ Step 6: Expand both sides: $$2x + 4y = -2cy + 16cx$$ Step 7: Bring $x$ terms to one side and $y$ terms to the other: $$2x - 16cx = -2cy - 4y$$ Step 8: Factor $x$ and $y$: $$x(2 - 16c) = -y(2c + 4)$$ Step 9: Solve for $x$: $$x = \frac{-y(2c + 4)}{2 - 16c}$$ --- 3. **Make b the subject:** Given: $$\frac{1}{a} = \frac{1}{b} + \frac{1}{c}$$ Step 1: Combine right side over common denominator $bc$: $$\frac{1}{a} = \frac{c + b}{bc}$$ Step 2: Cross multiply: $$bc = a(b + c)$$ Step 3: Expand right side: $$bc = ab + ac$$ Step 4: Bring all terms involving $b$ to one side: $$bc - ab = ac$$ Step 5: Factor $b$: $$b(c - a) = ac$$ Step 6: Solve for $b$: $$b = \frac{ac}{c - a}$$ --- 4. **Expand and simplify:** Given: $$(2\sqrt{5} - \sqrt{10})^2 - (2\sqrt{5} + 1)(1 - 2\sqrt{5})$$ Step 1: Expand first square: $$(2\sqrt{5})^2 - 2 \times 2\sqrt{5} \times \sqrt{10} + (\sqrt{10})^2 = 4 \times 5 - 4 \sqrt{50} + 10 = 20 - 4 \times 5 \sqrt{2} + 10 = 30 - 20\sqrt{2}$$ Step 2: Expand second product: $$2\sqrt{5} \times 1 - 2\sqrt{5} \times 2\sqrt{5} + 1 \times 1 - 1 \times 2\sqrt{5} = 2\sqrt{5} - 4 \times 5 + 1 - 2\sqrt{5} = 2\sqrt{5} - 20 + 1 - 2\sqrt{5} = -19$$ Step 3: Subtract second from first: $$(30 - 20\sqrt{2}) - (-19) = 30 - 20\sqrt{2} + 19 = 49 - 20\sqrt{2}$$ **Final answer:** $$49 - 20\sqrt{2}$$