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x2 + y - 5 = 0.....(i) y - 7x + 3 = 0.........(ii) y = 7x - 3, substituting the value of y in equation (i) x2 + (7x - 3) - 5 = 0 x2 + 7x + 3 = 0 (x + 8)(x - 1) = 0 x = -8 or 1

(3x - 2)(5x - 4) = (3x - 2)2 = 5x2 - 22x + 6 = 9x2 = 12x + 4 6x2 - 10x + 4 = 0 6x2 - 6x - 4x + 4 = 0 6x(x - 1) -4(x - 1) = (6x - 4)(x -1) = 0 x = 1 or \(\frac{2}{3}\)

f(x + 2) = f(-1) f(-1) is gotten when x = -3. \(2x^{2} + 7x - 5\) at x = -3. f(-1) = \(2(-3)^{2} + 7(-3) - 5\) = \(18 - 21 - 5 = -8\)

No expanded working has been attached for this question yet. The verified answer is Option D.

\(\frac{1}{p}\) - \(\frac{1}{q}\) \(\div\) \(\frac{p}{q}\) - \(\frac{q}{p}\) = \(\frac{q - p}{pq}\) \div \(\frac{p^2 - q^2}{pq}\) \(\frac{q - p}{pq}\) x \(\frac{pq}{p^2q^2}\) = \(\frac{q - p}{p^2 - q^2}\) \(\frac{-(p - q)}{(p + q)(p - q)}\) = \(\frac{-1}{p + q}\)

y\(^2\) - 3y > 18 = y\(^2\) - 3y - 18 > 0 y\(^2\) - 6y + 3y - 18 > 0 = y(y - 6) + 3 (y - 6) > 0 = (y + 3) (y - 6) > 0 Case 1 (+, +) \(\to\) (y + 3) > 0, (y - 6) > 0 = y > -3, y > 6 By simply substituting the value of y into the inequality, y > -3 is invalid, and y < 6 is also invalid. Therefore, y < - 3 or y > 6

\(\frac{x + 1}{3}\) > \(\frac{1}{X + 3}\) = \(\frac{x + 1}{3}\) > \(\frac{x + 3}{X + 3}\) = (x + 1)(x + 3)2 > 3(x + 3) = (x + 1)[x2 + 6x + 9] > 3(x + 3) x3 + 7x2 + 15x + 9 > 3x + 9 = x3 + 7x2 + 12x > 0 = x(x + 3)9x + 4) > 0 Case 1 (+, +, +) = x > 0 , x + 3 > 0, x + 4 > 0 = x > -4 (solution only) Case 2 (+, -, -) = x > 0, x + 4 0, x -3, x 0 = x -4 = x < -0, -4 < x < -3(solution) combining the solutions -4 < x < -3

x \(\ast\) y = xy x \(\ast\) 2 = x2 x \(\ast\) 2 = x ∴ x2 - x = 0 x(x - 1) = 0 x = 0 or 1

\(\frac{2k - 1}{k + 1} = \frac{3k + 1}{2k - 1}\) \((k + 1)(3k + 1) = (2k - 1)(2k - 1)\) \(3k^{2} + 4k + 1 = 4k^{2} - 4k + 1\) \(4k^{2} - 3k^{2} - 4k - 4k + 1 - 1 = 0\) \(k^{2} - 8k = 0\) \(k(k - 8) = 0\) \(\therefore \text{k = 0 or 8}\) The terms of the sequence given k = 0: (1, -1, 1) \(\implies \text{The common ratio r = -1}\) The terms of the sequence given k = 8: (9, 15, 25) \(\implies \text{The common ratio r = } \frac{5}{3}\) The possible values of the common ratio are -1 and \(\frac{5}{3}\).

Initial salary = N540 increment = N36 (every 6 months) Period of increment = 2 years and 6 months amount(increment) = N36 x 5 = N180 The man's new salary = N540 + N180 = N720.00

A rectangular polygon has each interior angle to be 150o let the polygon has n-sides therefore, Total interior angle 150 x n = 150n hence 150n = (2n - 4)90 150n = 180n - 360 360 = (180 - 150)n 30n = 360 n = 12

Diameter = 8cm ∴ Radius = 4cm Length of arc = \(\frac{\theta}{360}\) x 2 \(\pi\)r but Q = 22\(\frac{1}{2}\) ∴ Length \(\frac{22\frac{1}{2}}{360}\) x 2 x \(\pi\) x 4 = \(\frac{22\frac{1}{2} \times 8\pi}{360}\) = \(\frac{180}{360}\) = \(\frac{\pi}{2}\)

2x + 3 \(\neq\) 2x - 3 for any value of x ∴ for the \(\bigtriangleup\) to be isosceles, either 2x - 3 = x + 3 or 2x + 3 = x + 3 solve the two equations we arrive at x = 6 or x = 0 When x = 6, the sides are 9, 15, 9 When x = 0, the sides are 3, 4, -3 since lengths of a \(\bigtriangleup\)can never be negative then the value of x = 6

Surface area = 154cm2 (area of sphere) 4\(\pi\)r2 = 154 r\(\sqrt{\frac{154}{4\pi}}\) = 3.50cm

Area of sector \(\frac{\theta}{360}\) x \(\pi\)r2, \(\theta\) = 60o, r = 3m = \(\frac{60}{360}\) x \(\frac{12}{7}\) x 3 x 3 \(\frac{1}{6}\) x \(\frac{22}{7}\) x 9 = \(\frac{33}{7}\) = 4.7m2