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y = \(\frac {3x - 9c}{4x + 5d}\) \(\frac {y}{1} = \frac {3x - 9c}{4x + 5d}\) = y(4x + 5d) = 3x - 9c = 4xy + 5dy = 3x - 9c = 4xy - 3x = -9c - 5dy = (4y - 3)x = -9c - 5dy = x = \(\frac {-9c - 5dy}{4y - 3}\) x = \(\frac {-(9c + 5dy)}{4y - 3}\)

\(y = \frac{ax^3 - b}{3z}\) cross multiply \(ax^3 - b\) = 3yz \(ax^3\) = 3yz + b divide both sides by a \(x^3 = \frac{3yz + b}{a}\) take cube root of both sides therefore, x = \(\sqrt[3] \frac{3yz + b}{a}\)

Find the roots of the equations: \(3m^2 - 2m - 65 = 0\) = \( m^2 - 15m + 13m - 65 = 0\) = 3m(m - 5) + 13( m - 5) = 0 ( m - 5)(3m + 13) = 0 m-5 = 0 or 3m + 13 = 0 therefore, m = 5 or \(\frac{-13}{3}\) therefore the roots of the quadratic equation = ( 5, \(\frac{-13}{3})\)

\(x^2 - (y + z)^2\) Using difference of two squares (\(a^2 - b^2 = (a + b)(a - b)|) ⇒ (x + (y + z))(x - (y + z)) ⇒ (x + y + z)(x - y - z)

Let p = \(\frac{2}{3}\) and q = \(\frac{- 3}{4}\) using (y - p)(y - q) = 0 = ( y - \(\frac{2}{3})\)( y - (\(\frac{- 3}{4})) = 0\) = (\( y - \frac{2}{3})( y + \frac{3}{4})\) = 0 \( y^2 + \frac{3}{4}y - \frac{2}{3}y - \frac{6}{12} = 0 \) \( y^2 + \frac{1}{12}y - \frac{1}{2}\) = 0 = multiply through by the l. c. m of 3 and 4 = 12 ∴ the quadratic equation is \(12y^2 + y - 6 = 0\)

a. x -3 -2 -1 0 1 2 3 \(2x^2\) 18 8 2 0 2 8 -18 - x 3 2 1 0 - 1 - 2 - 3 - 4 - 4 - 4 - 4 -4 -4 -4 -4 y 17 6 -1 -4 -3 2 11 b. the graph above, solution to question 8b. ci. The roots of the equation from the graph are -1.2 and 1.7. ii. Let y = \(2x^2 - x - 4\) \(\frac{dy}{dx}\) = 4x - 1 At critical point \(\frac{dy}{dx}\) = 0 so that 4x - 1 = 0; x = \(\frac{1}{4}\) Using test points: 1. we use a test point less than \(\frac{1}{4}\), e.g x = 0 \(\frac{dy}{dx}\)(0) = 4(0) - 1 = -1 at this point, y = \(2x^2 - x - 4\) is decreasing 2. we use a test point greater than \(\frac{1}{4}\) e.g x = 1 \(\frac{dy}{dx}\)(1) = 4(1) - 1 = 4 - 1 = 3. at this point, y = \(2x^2 - x - 4\) is increasing. so, the function y = \(2x^2 - x - 4\) increases as x increases when x is greater than \(\frac{1}{4}\) Therefore, the value of x for which y increases are x is greater than \(\frac{1}{4}\) iii. The minimum point of y = -4.1.

3(x – 1) \le 2 (x – 3) = 3x – 3 \le 2x – 6 Add 3 to both sides = 3x – 3+ 3 \le 2x – 6+ 3 = 3x \le 2x – 3 Subtract 2x from both the sides = 3x – 2x \le 2x – 3 – 2x ∴ x \le -3

\(x^2 - x - 4 \le 2\) Subtract two from both sides to rewrite it in the quadratic standard form: = \(x^2 - x - 4 - 2 \le 2 - 2\) = \(x^2 - x - 6 \le 0\) Now set it = 0 and factor and solve like normal. = \(x^2 - x\) - 6=0 = \((x - 3)(x + 2)\)=0 \(x\) + 2 = 0 or \(x\) - 3 = 0 \(x\) = -2 or \(x\) = 3 So the two zeros are -2 and 3, and will mark the boundaries of our answer interval. To find out if the interval is between -2 and 3, or on either side, we simply take a test point between -2 and 3 (for instance, \(x\) = 0) and evaluate the original inequality. = \(x2 - x - 4 \le 2\) = \((0)^2 - (0) - 4 \le 2\) = \(0 - 0 - 4 \le 2\) \(−4 \le 2\) Since the above is a true statement, we know that the solution interval is between -2 and 3, the same region where we picked our test point. Since the original inequality was less than or equal, we include the endpoints. ∴ \(-2 \le x \le 3.\)

First, we find the equation of the boundary line using the two intercepts.The slope is m = \(\frac{4 - 0}{0 - 5} = {4}{5}\) The y-intercept is 4 The slope-intercept form of the equation is therefore y = -\(\frac{4}{5} x + 4\) \(\implies y + \frac{4}{5} x = 4\) Multiply both sides by \(\frac{5}{4}\) \(\implies\frac{5}{4}(y +\frac{4}{5} x) = 4\times\frac{5}{4}\) \(\implies\frac{5}{4} y + x = 5\) Multiply both sides by 4 ⇒ \(5y + 4x = 20\) The inequality is therefore either \(5y + 4x \le 20\) or \(5y + 4x \ge 20.\) Using the test point (0, 0) -The origin ⇒ 5(0) + 4(0) \le 20 ⇒ 0 \le 20 (True) ∴ The inequality is \(5y + 4x \le 20\)

The fraction \(\frac{ x^2 + 2 }{ 10x^2 - 13x - 3}\) is undefined when the denominator is equal to zero \(then 10x^2 - 13x - 3 = 0\) by factorisation, \(10x^2 - 13x - 3\) = 0 becomes \( 10x^2 - 15x +2x -3\) = 0 \(5x(2x - 3) + 1(2x - 3) = 0\) \((5x + 1)(2x - 3) = 0\) \(then, x = \frac{-1}{5}\) or \(\frac{3}{2}\)

16\(x^4 - y^4\) = 2\(^4x^4 - y^4\) = \((2x)^4 - y^4\) = \(((2x)^2)^2 - (y^2)^2\) Using a\(^2 - b^2\) = (a - b)(a + b) identity = ((2x)\(^2 - y^2)((2x)^2 + y^2)\) Using the identity one more time = \((2x - y)(2x + y)((2x)^2 + y^2)\) ∴ \((2x - y)(2x + y)(4x^2 + y^2)\)

Let \(p(x) = -2x^3 + 6x^2 + 17x - 21\) Using the remainder theorem Let \(x + 1 = 0\) ∴ \(x = -1\) Since, \((x + 1)\) divides \(p(x)\), then, remainder will be p(-1) ⇒ p(-1) = -2(-1)\(^3 + 6(-1)^2\) + 17(-1) - 21 ∴ p(-1) = -30

Given the sequence 3, 8, 13, 18, ... which is an arithmetic sequence a = 3 d = T\(_2\) - T\(_1\) = 8 - 3 = 5 The general term of an A.P is: T\(_n\) = a + (n - 1)d ⇒ T\(_n\) = 3 + (n - 1)5 = T\(_n\) = 3 + 5n - 5 ∴ T\(_n\) = 5n - 2

T\(_3\) = 6 T\(_5\) = 12 S\(_{12}\) = ? T\(_n\) = a + (n - 1)d ⇒ T\(_3\) = a + 2d = 6 ----- (i) ⇒ T\(_5\) = a + 4d = 12 ----- (ii) Subtract equation (ii) from (i) ⇒ -2d = -6 ⇒ d\(\frac{-6}{-2}\) = 3 Substitute 3 for d in equation (i) ⇒ a + 2(3) = 6 ⇒ a + 6 = 6 ⇒ a = 6 - 6 = 0 S\(_n\) = \(\frac{n(2a + (n - 1)d)}{2}\) ⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\) ⇒ S\(_{12}\) = 6(0 + 11 x 3) ⇒ S\(_{12}\) = 6(33) ∴ S\(_{12}\) = 198

\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\) \(T_n = ar^n - 1\) ∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i) \(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii) = 2a = 3(1 - r) = 2a = 3 - 3r ∴ a = \(\frac{3 - 3r}{2}\) Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i) = \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\) = \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\) = 3(3r - 3r\(^2\)) = -4 = 9r - 9r\(^2\) = -4 = 9r\(^2\) - 9r - 4 = 0 = 9r\(^2\) - 12r + 3r - 4 = 0 = 3r(3r - 4) + 1(3r - 4) = 0 = (3r - 4)(3r + 1) = 0 ∴ r = \(\frac{4}{3} or - \frac{1}{3}\) For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1. ∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)