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If y = 2x + 1 and y = x2 - 2x + 1 then x2 - 2x + 1 = 2x + 1 x2 - 4x = 0 x(x - 4) = 0 x = 0 or 4

4x - 3 = 3x + y = x - y = 3.......(i) 3x + y = 2y + 5x - 12.........(ii) eqn(ii) + eqn(i) 3x = 15 x = 5 substitute for x in equation (i) 5 - y = 3 y = 2

\(\frac{av}{1 - v}\) = \(\sqrt[3]{\frac{2v + T}{a + 2T}}\) \(\frac{(av)^3}{(1 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) \(\frac{a^3v^3}{(1^3 - v)^3}\) = \(\frac{2v + T}{a + 2T}\) = \(\frac{2v(1 - v)^3 - a^4 v^3}{2a^3v^ 3 - (1 - v)^3}\)

x2 + y - 8 = 0, y + 5x - 2 = 0 Rearranging, x2 + y = 8.....(i) 5x + y = 2.......(ii) Subtract eqn(ii) from eqn(i) x2 - 5x - 6 = 0 (x - 6)(x + 1) = 0 x = 6, -1

(\(\frac{2}{3}\))m (\(\frac{3}{4}\))n = \(\frac{256}{729}\) \(\frac{2^m}{4^n}\) x \(\frac{3^n}{3^m}\) = \(\frac{2^{8}}{3^{6}}\) \(2^{m} \div 2^{2n} = 2^{8}; 3^{n} \div 3^{m} = 3^{-6}\) m - 2n = 8........(i) -m + n = -6........(ii) Solving the equations simultaneously m = 4, n = -2

The roots of the function are -1 and \(\frac{1}{3}\) sum of roots = -1 + \(\frac{1}{3}\) = -\(\frac{2}{3}\) product of roots = -1 x \(\frac{1}{3}\) = -\(\frac{1}{3}\) \(x^2\) - (sum of roots)x + (product of roots) = 0 \(x^2\) - (-\(\frac{2}{3}\))x + (-\(\frac{1}{3}\)) = 0 \(x^2\) + \(\frac{2x}{3}\) - \(\frac{1}{3}\) = 0 3\(x^2\) + 2x - 1 = 0( multiple through by -1 since the graph of the quadratic equation is maximum) 1 - 2x - 3\(x^2\).

7 years ago, Father(x - 7) years old, Son (y - 7) years x - 7 = 3(y - 7) x - 7 = 3y - 21 3y - x = -7 + 21 = 14 3y - x = 14 ... (1) In six years time, x + 6 = 2(y + 6) x + 6 = 2y + 12 2y + 12 = x + 6 12 - 6 = x - 2y 6 = x - 2y ... (2)

bx\(^2\) + cx + a = 0 a = b; b = c; c = a x = -b \(\pm\) \(\frac{\sqrt{b^2 - 4ac}}{2a}\) x = -c \(\pm\) \(\frac{\sqrt{c^2 - 4ab}}{2b}\)

The graphical method of solving the equation x3 + 3x2 + 4x - 28 = 0 is by drawing the graphs of the curves y = x2 + 3x + 4 and y = \(\frac{28}{x}\)`.

Two lines are said to be parallel if the slope of the two lines are equal. The equation : \(3y + 2x + 7 = 0\) \(3y = -2x - 7\) \(y = \frac{-2}{3} x - \frac{7}{3}\) \(\frac{\mathrm d y}{\mathrm d x} = - \frac{2}{3}\) All the options have the same slope except 15y + 10x - 13 = 0.

5 + 3x - 2x2 = 0, (x + 1)(5 - 2x) (expand to check) (x + 1)(5 - 2x) = 0 when x = 1 = 0, x = -1 when 5 - 2x = 0, x = \(\frac{5}{2}\) - 1

3x + 5y = 4, 4x + 3y = 5 3x + 5y = 4 x 4 4x + 3y = 5 x 3 12x + 20y = 16.....(i) 12x + 9y = 15.......(ii) subtract eqn.(ii) from eqn.(i) 11y = 1 y = \(\frac{1}{11}\) 12x + 20 x \(\frac{1}{11}\) = 16 12x = \(\frac{156}{11}\) x = \(\frac{13}{11}\) = \(\frac{13}{11}, \frac{1}{11}\)

The parabola is best described by the equation x = y2 - k because all other equations do not give the equation of the parabola in this position. C for example is the equation of a hyperbola facing downwards. D is the equation of a hyperbola A and E are equations of parabola facing upwards

\(4t^{2} + 7t - 1 = 0\) Using a = 4, b = 7, c = -1. \(x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\) = \(\frac{-7 \pm \sqrt{7^{2} - 4(4)(-1)}}{2(4)}\) = \(\frac{-7 \pm \sqrt{49 + 16}}{8}\) = \(\frac{-7 \pm \sqrt{65}}{8}\) = \(\frac{-7 \pm 8.0623}{8}\) The positive answer = \(\frac{-7 + 8.0623}{8} = \frac{1.0623}{8}\) \(\approxeq 0.1328\) (4 decimal place)

1st term a = 3, 5th term = 9, sum of n = 81 nth term = a + (n - 1)d, 5th term a + (5 - 1)d = 9 3 + 4d = 9 4d = 9 - 3 d = \(\frac{6}{4}\) = \(\frac{3}{2}\) = 6 Sn = \(\frac{n}{2}\)(6 + \(\frac{3}{4}\)n - \(\frac{3}{2}\)) 81 = \(\frac{12n + 3n^2}{4}\) - 3n = \(\frac{3n^2 + 9n}{4}\) 3n2 + 9n = 324 3n2 + 9n - 324 = 0 By almighty formula positive no. n = 9 = 3