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rs + tr - pt - ps = rs - ps - tr - pt = (r - p)s + (r - p)t = (r - p)(s + t), hence r - p is a factor

\(3x + y = 8 ... (i)\) \(x^{2} + xy = 6 ... (ii)\) From (i), \(y = 8 - 3x\) From (ii), \(xy = 6 - x^{2} \implies y = \frac{6 - x^{2}}{x}\) Equating the two values of y, we have \(8 - 3x = \frac{6 - x^{2}}{x} \implies x(8 - 3x) = 6 - x^{2}\) \(8x - 3x^{2} = 6 - x^{2} \implies 6 - x^{2} - 8x + 3x^{2} = 0\) \(2x^{2} - 8x + 6 = 0\) \(x^{2} - 4x + 3 = 0\) \(x^{2} - 3x - x + 3 = 0 \implies x(x - 3) - 1(x - 3) = 0\) \((x - 1)(x - 3) = 0 \therefore \text{x = 1 or 3}\) \(y = 8 - 3x \) When x = 1, \(y = 8 - 3(1) = 5\) When x = 3, \(y = 8 - 3(3) = -1\) \(\therefore \text{y = -1 or 5}\)

\(\frac{x}{2}\) + \(\frac{x}{3}\) + \(\frac{x}{4}\)< 1 = \(\frac{6x + 4x + 3x < 12}{12}\) i.e. 13 x < 12 = x < \(\frac{12}{13}\)

3n2 = 12n = 3n2 - 12n = 0 = 3n(n - 4) = 0 ∴ n = 4

Given 27, 9, 3,......this is a G.P r = \(\frac{9}{27}\) = \(\frac{1}{3}\) T = arn - 1 = 27\(\frac{1}{3}\) n - 1

(x - 2) (x - 3) = 12 x\(^2\) - 3x - 2x + 6 = 12 x\(^2\) - 5x - 6 = 0 (x +1)(x - 6) = 0 x = -1 or 6

\(\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}\) = \((\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} - \sqrt{x}}) (\frac{\sqrt{1 + x} + \sqrt{x}}{\sqrt{1 + x} + \sqrt{x}})\) = \(\frac{(1 + x) + \sqrt{x(1 + x)} + \sqrt{x(1 + x)} + x}{(1 + x) - x}\) = \(\frac{1 + 2x + 2\sqrt{x(1 + x)}}{1}\) = \(1 + 2x + 2\sqrt{x(1 + x)}\)

x2(x2 - 1)-\(\frac{1}{2}\) - (x2 - 1)\(\frac{1}{2}\) = \(\frac{x^2}{(x^2 - 1)^\frac{1}{2}}\) - \(\frac{(x^2 - 1)^\frac{1}{2}}{1}\) = \(\frac{x^2 - (x^2 - 1)}{(x^2 - 1) ^\frac{1}{2}}\) = \(\frac{x^2 - x^2 + 1}{(x^2 - 1)^\frac{1}{2}}\) = (x2 - 1)-\(\frac{1}{2}\)

Given (-2, 0) and (0, -4) Gradient = \(\frac{-4 - 0}{0 - (-2)}\) = \(\frac{-4}{2}\) = -2

For y = ax2 - bx + c for minimum y \(\frac{dy}{dx}\) = 2x - 2 = \(\frac{dy}{dx}\) = 0 ∴ 2x - 2 = 0 x = 1

Given the first and last term of an A.P, the sum of the terms is given by \(S_{n} = \frac{n}{2} [a + l]\) where a = first term; l = last term and n = number of terms. \(\therefore S_{20} = \frac{20}{2} [7 + 117]\) = \(10 (124)\) = 1240

BD = \(\sqrt{x^2 + x^2}\) = \(\sqrt{12^2 + 12^2}\) = \(\sqrt{144 + 144}\) = 2(144) = 12\(\sqrt{2cm}\)

\(\frac{x}{4}\) = \(\frac{\tan 60}{1}\) x = 4 tan60 = 4\(\sqrt{3}\) BD = 2x = 8\(\sqrt{3}\)

The sum of the exterior angles of a regular polygon = 360^\circ. \(\therefore x + (x + 5) + (x + 10) + (x + 15) + (x + 20) = 360^\circ\) \(5x + 50 = 360^\circ \implies 5x = 360^\circ - 50^\circ = 310^\circ\) \(x = \frac{310^\circ}{5} = 62^\circ\)

\(\frac{BC}{60}\) = \(\frac{tan 62}{1}\) BC = 60 tan 62 \(\frac{AC}{60}\) = \(\frac{tan 62}{1}\) AC = 60 tan 64 AB = AC - BC = 60(tan 64o - tan 62o)