(i)

$$ x = 7 $$

(ii)

$$ x = - 2 $$

(iii)

$$ x = 7 $$

(a)

\begin{align*} 9x - 7 & \le 12 \\ 9x & \le 12 + 7 \\ 9x & \le 19 \\ x & \le {19 \over 9} \\ x & \le 2{1 \over 9} \end{align*}

(b)

\begin{align*} 7 - 2x & > 2 \\ -2x & > 2 - 7 \\ -2x & > -5 \\ x & < {-5 \over -2} \\ x & < 2.5 \end{align*}

(c)

\begin{align*} 3 + 5x & \ge 32 \\ 5x & \ge 32 - 3 \\ 5x & \ge 29 \\ x & \ge {29 \over 5} \\ x & \ge 5.8 \end{align*}

(d)

\begin{align*} 3x - 4 & \ge {1 \over 3}x - 2 \\ 3x - {1 \over 3}x & \ge -2 + 4 \\ 2{2 \over 3}x & \ge 2 \\ x & \ge 2 \div 2{2 \over 3} \\ x & \ge 0.75 \end{align*}

(e)

\begin{align*} 12 & < 3x - 1 &&& 3x - 1 & < 27 \\ -3x & < -1 - 12 &&& 3x & < 27 + 1 \\ -3x & < -13 &&& 3x & < 28 \\ x & > {-13 \over -3} &&& x & < {28 \over 3} \\ x & > 4{1 \over 3} &&& x & < 9{1 \over 3} \end{align*}
$$ 4{1 \over 3} < x < 9{1 \over 3} $$

(a)

\begin{align*} x + 2y & = 8 \phantom{0} \text{--- (1)} \\ \\ 3x + 2y & = 12 \phantom{0} \text{--- (2)} \\ \\ (1) & - (2), \\ x + 2y - (3x + 2y) & = 8 - 12 \\ x + 2y - 3x - 2y & = -4 \\ -2x & = -4 \\ x & = {-4 \over -2} \\ x & = 2 \\ \\ \text{Substitute } & x = 2 \text{ into (1),} \\ 2 + 2y & = 8 \\ 2y & = 8 - 2 \\ 2y & = 6 \\ y & = {6 \over 2} \\ y & = 3 \\ \\ \therefore x & = 2, y = 3 \end{align*}

(b)

\begin{align*} x + y & = {5 \over 6} \phantom{0} \text{--- (1)} \\ \\ x - y & = {1 \over 6} \phantom{0} \text{--- (2)} \\ \\ (1) & - (2), \\ x + y - (x - y) & = {5 \over 6} - {1 \over 6} \\ x + y - x + y & = {2 \over 3} \\ 2y & = {2 \over 3} \\ y & = {2 \over 3} \div 2 \\ y & = {1 \over 3} \\ \\ \text{Substitute } & y = {1 \over 3} \text{ into (1),} \\ x + {1 \over 3} & = {5 \over 6} \\ x & = {5 \over 6} - {1 \over 3} \\ x & = {1 \over 2} \\ \\ \therefore x & = {1 \over 2}, y = {1 \over 3} \end{align*}

(c)

\begin{align*} 3a - 2b & = 1 \phantom{0} \text{--- (1)} \\ \\ 5a + 3b & = -11 \\ 5a & = -3b - 11 \\ a & = {1 \over 5}(-3b - 11) \\ a & = -{3 \over 5}b - {11 \over 5} \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3 \left(-{3 \over 5}b - {11 \over 5}\right) - 2b & = 1 \\ -{9 \over 5}b - {33 \over 5} - 2b & = 1 \\ -{9 \over 5}b - 2b & = 1 + {33 \over 5} \\ -3.8b & = 7.6 \\ b & = {7.6 \over -3.8} \\ b & = -2 \\ \\ \text{Substitute } & b = -2 \text{ into (2),} \\ a & = -{3 \over 5}(-2) - {11 \over 5} \\ a & = -1 \\ \\ \therefore a & = -1, b = -2 \end{align*}

(d)

\begin{align*} 3p - 4q - 24 & = 0 \phantom{0} \text{--- (1)} \\ \\ 5p - 6q - 38 & = 0 \\ 5p & = 6q + 38 \\ p & = {1 \over 5}(6q + 38) \\ p & = 1.2q + 7.6 \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ 3(1.2q + 7.6) - 4q - 24 & = 0 \\ 3.6q + 22.8 - 4q - 24 & = 0 \\ 3.6q - 4q & = -22.8 + 24 \\ -0.4q & = 1.2 \\ q & = {1.2 \over -0.4} \\ q & = -3 \\ \\ \text{Substitute } & q = -3 \text{ into (2),} \\ p & = 1.2(-3) + 7.6 \\ p & = 4 \\ \\ \therefore p & = 4, q = -3 \end{align*}

(e)

\begin{align*} {1 \over 4}x + {3 \over 5}y & = -4 \phantom{0} \text{--- (1)} \\ \\ {1 \over 5}x + {1 \over 4}y & = -{9 \over 10} \\ 5 \left({1 \over 5}x + {1 \over 4}y\right) & = 5 \left(-{9 \over 10}\right) \\ x + 1.25y & = -4.5 \\ x & = -1.25y - 4.5 \phantom{0} \text{--- (2)} \\ \\ \text{Substitute } & \text{(2) into (1),} \\ {1 \over 4}(-1.25y - 4.5) + {3 \over 5}y & = -4 \\ -0.3125y - 1.125 + 0.6y & = -4 \\ -0.3125y + 0.6y & = -4 + 1.125 \\ 0.2875y & = -2.875 \\ y & = {-2.875 \over 0.2875} \\ y & = -10 \\ \\ \text{Substitute } & y = -10 \text{ into (2),} \\ x & = -1.25 (-10) - 4.5 \\ x & = 8 \\ \\ \therefore x & = 8, y = -10 \end{align*}

(a)

\begin{align*} 3x - 2 & = 0 && \text{ or } & 2x + 7 & = 0 \\ 3x & = 2 &&& 2x & = -7 \\ x & = {2 \over 3} &&& x & = -{7 \over 2} \end{align*}

(b)

\begin{align*} 2x^2 + 5x - 12 & = 0 \\ (2x - 3)(x + 4) & = 0 \end{align*} \begin{align*} 2x - 3 & = 0 && \text{ or } & x + 4 & = 0 \\ 2x & = 3 &&& x & = -4 \\ x & = {3 \over 2} \end{align*}

(c)

\begin{align*} 4x^2 - 3x & = 7 \\ 4x^2 - 3x - 7 & = 0 \\ (4x - 7)(x + 1) & = 0 \end{align*} \begin{align*} 4x - 7 & = 0 && \text{ or } & x + 1 & = 0 \\ 4x & = 7 &&& x & = -1 \\ x & = {7 \over 4} \end{align*}

(d)

\begin{align*} 7x^2 & + 15x - 3 = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-15 \pm \sqrt{(15)^2 - 4(7)(-3)} \over 2(7)} \\ & = {-15 \pm \sqrt{309} \over 14} \\ & = 0.1842 \text{ or } -2.327 \\ & \approx 0.184 \text{ or } -2.33 \end{align*}

(e)

\begin{align*} (x - 5)(x + 3) & = 7 \\ x^2 + 3x - 5x - 15 & = 7 \\ x^2 - 2x - 15 & = 7 \\ x^2 - 2x - 15 - 7 & = 0 \\ x^2 - 2x - 22 & = 0 \end{align*}
\begin{align*} x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-2) \pm \sqrt{(-2)^2 - 4(1)(-22)} \over 2(1)} \\ & = {2 \pm \sqrt{92} \over 2} \\ & = 5.796 \text{ or } -3.796 \\ & \approx 5.80 \text{ or } -3.80 \end{align*}

(a)

\begin{align*} 3x(x + 4) + 28(x + 2) + 21 & = 0 \\ 3x^2 + 12x + 28x + 56 + 21 & = 0 \\ 3x^2 + 40x + 77 & = 0 \\ (3x + 7)(x + 11) & = 0 \end{align*} \begin{align*} 3x + 7 & = 0 && \text{ or } & x + 11 & = 0 \\ 3x & = -7 &&& x & = -11 \\ x & = -{7 \over 3} \end{align*}

(b)

\begin{align*} 2(4x^2 + 23x) & = 105 \\ 8x^2 + 46x & = 105 \\ 8x^2 + 46x - 105 & = 0 \\ (2x + 15)(4x - 7) & = 0 \end{align*} \begin{align*} 2x + 15 & = 0 && \text{ or } & 4x - 7 & = 0 \\ 2x & = - 15 &&& 4x & = 7 \\ x & = -{15 \over 2} &&& x & = {7 \over 4} \end{align*}

(c)

\begin{align*} (x - 1)^2 - 16 & = 0 \\ (x - 1)^2 & = 16 \\ x - 1 & = \pm \sqrt{16} \\ x - 1 & = \pm 4 \end{align*} \begin{align*} x - 1 & = 4 && \text{ or } & x - 1 & =- 4 \\ x & = 4 + 1 &&& x & = -4 + 1 \\ x & = 5 &&& x & = -3 \end{align*}

(d)

\begin{align*} {5 \over x - 1} & = {3 \over 1} + {4 \over x} \\ {5 \over x - 1} & = {3x \over x} + {4 \over x} \\ {5 \over x - 1} & = {3x + 4 \over x} \\ {5x \over x(x - 1)} & = { (3x + 4)(x - 1) \over x(x - 1)} \\ 5x & = (3x + 4)(x - 1) \\ 5x & = 3x^2 - 3x + 4x - 4 \\ 5x & = 3x^2 + x - 4 \\ 0 & = 3x^2 + x - 5x - 4 \\ 0 & = 3x^2 - 4x - 4 \\ 0 & = (3x + 2)(x - 2) \end{align*} \begin{align*} 3x + 2 & = 0 && \text{ or } & x - 2 & = 0 \\ 3x & = -2 &&& x & = 2 \\ x & = -{2 \over 3} \end{align*}

(e)

\begin{align*} {x \over 1 + x} + {x + 1 \over 1 - 3x} & = {1 \over 4} \\ {4x(1 - 3x) \over 4(1 + x)(1 - 3x)} + { 4(x + 1)(1 + x) \over 4(1 + x)(1 - 3x)} & = {(1 + x)(1 - 3x) \over 4(1 + x)(1 - 3x)} \\ 4x(1 - 3x) + 4(x + 1)^2 & = (1 + x)(1 - 3x) \\ 4x - 12x^2 + 4 \underbrace{ [ (x)^2 + 2(x)(1) + (1)^2 ] }_{ (a + b)^2 = a^2 + 2ab + b^2 } & = 1 - 3x + x - 3x^2 \\ 4x - 12x^2 + 4(x^2 + 2x + 1) & = 1 - 2x - 3x^2 \\ 4x - 12x^2 + 4x^2 + 8x + 4 & = 1 - 2x - 3x^2 \\ - 8x^2 + 12x + 4 & = 1 - 2x - 3x^2 \\ -8x^2 + 3x^2 + 12x + 2x + 4 - 1 & = 0 \\ -5x^2 + 14x + 3 & = 0 \\ 5x^2 - 14x - 3 & = 0 \\ (5x + 1)(x - 3) & = 0 \end{align*} \begin{align*} 5x + 1 & = 0 && \text{ or } & x - 3 & = 0 \\ 5x & = -1 &&& x & = 3 \\ x & = -{1 \over 5} \end{align*}

(f)

\begin{align*} {2 \over 3x} & = {x - 2 \over 4(x + 3)} \\ { 2[4(x + 3)] \over 12x(x + 3)} & = { 3x(x - 2) \over 12x(x + 3)} \\ 2[4(x + 3)] & = 3x(x - 2) \\ 8(x + 3) & = 3x^2 - 6x \\ 8x + 24 & = 3x^2 - 6x \\ 0 & = 3x^2 - 6x - 8x - 24 \\ 0 & = 3x^2 - 14x - 24 \\ 0 & = (3x + 4)(x - 6) \end{align*} \begin{align*} 3x + 4 & = 0 && \text{ or } & x - 6 & = 0 \\ 3x & = -4 &&& x & = 6 \\ x & = -{4 \over 3} \end{align*}

(g)

\begin{align*} {5 \over x + 2} - {5 \over x^2 -4} & = 0 \\ {5 \over x + 2} & = {5 \over x^2 - 2^2} \\ {5 \over x + 2} & = {5 \over (x + 2)(x - 2)} \phantom{000000} [a^2 - b^2 = (a + b)(a - b)] \\ {5(x - 2) \over (x + 2)(x - 2)} & = {5 \over (x + 2)(x - 2)} \\ 5(x - 2) & = 5 \\ 5x - 10 & = 5 \\ 5x & = 5 + 10 \\ 5x & = 15 \\ x & = {15 \over 5} \\ x & = 3 \end{align*}

(i)

\begin{align*} 2x^2 + qx - 2 & = 0 \\ \\ \text{Let } & x = -2, \\ 2(-2)^2 + q(-2) - 2 & = 0 \\ 8 - 2q - 2 & = 0 \\ -2q & = -8 + 2 \\ -2q & = -6 \\ q & = {-6 \over -2} \\ q & = 3 \end{align*}

(ii)

\begin{align*} 2x^2 + qx - 2 & = 0 \\ 2x^2 + 3x - 2 & = 0 \\ (2x - 1)(x + 2) & = 0 \end{align*} \begin{align*} 2x - 1 & = 0 && \text{ or } & x + 2 & = 0 \\ 2x & = 1 &&& x & =- 2 \\ x & = {1 \over 2} \end{align*}
$$ \text{Other solution is } x = {1 \over 2} $$

(a)

\begin{align*} \text{Smallest possible length} & = 8.5 \text{ cm} \phantom{000000} [8.5 \approx 9] \\ \\ \text{Smallest possible breadth} & = 5.5 \text {cm} \phantom{000000} [5.5 \approx 6] \\ \\ \text{Smallest perimeter} & = 2(8.5) + 2(5.5) \\ & = 28 \text{ cm} \end{align*}

(b)

\begin{align*} \text{Largest possible length} & = 9.499... \text{ cm} \phantom{000000} [9.499... \approx 9] \\ \\ \text{Largest possible breadth} & = 6.499... \text {cm} \phantom{000000} [6.499... \approx 6] \\ \\ \text{Largest possible area} & = 9.499 \times 6.499 \\ & = 61.75 \text{ cm}^2 \end{align*}

(a)

\begin{align*} -5 & < 12 - 3a &&& 12 - 3a & < -1 \\ 3a & < 12 + 5 &&& -3a & < -1 -12 \\ 3a & < 17 &&& -3a & < -13 \\ a & < {17 \over 3} &&& a & > {-13 \over -3} \\ a & < 5{2 \over 3} &&& a & > 4{1 \over 3} \end{align*} \begin{align*} 4{1 \over 3} & < a < 5{2 \over 3} \\ \\ a & = 5 \end{align*}

(b)

\begin{align*} b - 5 & \le 7 &&& 3b - 2 & \ge 11 \\ b & \le 7 + 5 &&& 3b & \ge 11 + 2 \\ b & \le 12 &&& 3b & \ge 13 \\ & &&& b & \ge {13 \over 3} \\ & &&& b & \ge 4{1 \over 3} \end{align*} \begin{align*} 4{1 \over 3} & \le b \le 12 \\ \\ \text{Odd integers: } & 5, 7, 9, 11 \end{align*}

(c)

\begin{align*} {2 \over 7} & < 2x - 1 &&& {2x - 1 \over 1} & < {9 + 2x \over 12} \\ -2x & < -1 - {2 \over 7} &&& {12(2x - 1) \over 12} & < {9 + 2x \over 12} \\ -2x & < -{9 \over 7} &&& 12(2x - 1) & < 9 + 2x \\ x & > -{9 \over 7} \div -2 &&& 24x - 12 & < 9 + 2x \\ x & > {9 \over 14} &&& 24x - 2x & <9 + 12 \\ & &&& 22x & < 21 \\ & &&& x & < {21 \over 22} \end{align*}
$$ {9 \over 14} < x < {21 \over 22} $$

(a)

\begin{align*} -5 & \le 4x + 1 &&& 4x + 1 & \le 2x + 9 \\ -4x & \le 1 + 5 &&& 4x - 2x & \le 9 - 1 \\ -4x & \le 6 &&& 2x & \le 8 \\ x & \ge {6 \over -4} &&& x & \le 4 \\ x & \ge -1.5 \end{align*} $$ -1.5 \le x \le 4 $$
\begin{align*} -6 & \le 2y - 2 &&& 2y - 2 & \le 8 \\ -2y & \le -2 + 6 &&& 2y & \le 8 + 2 \\ -2y & \le 4 &&& 2y & \le 10 \\ y & \ge {4 \over -2} &&& y & \le 5 \\ y & \ge -2 \end{align*} $$ -2 \le y \le 5 $$
\begin{align*} \text{Greatest value of } x - y & = 4 - (-2) \\ & = 6 \end{align*}

(b)

\begin{align*} (x + y)(x - y) & = x^2 - y^2 \\ \\ \text{Smallest value of } x^2 - y^2 & = (0)^2 - (5)^2 \\ & = -25 \end{align*}

(i)

\begin{align*} \text{Let } & d = 40, \\ 40 & = 35 + 7t - 2t^2 \\ 2t^2 - 7t + 40 - 35 & = 0 \\ 2t^2 - 7t + 5 & = 0 \\ (2t - 5)(t - 1) & = 0 \end{align*} \begin{align*} 2t - 5 & = 0 && \text{ or } & t - 1 & =0 \\ 2t & = 5 &&& t & = 1 \text{ second} \\ t & = {5 \over 2} \\ t & = 2.5 \text{ seconds} \end{align*}

(ii)

\begin{align*} \text{Let } & d = 0, \\ 0 & = 35 + 7t - 2t^2 \\ 2t^2 - 7t - 35 & = 0 \\ \\ t & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-7) \pm \sqrt{ (-7)^2 - 4(2)(-35)} \over 2(2)} \\ & = {7 \pm \sqrt{ 329 } \over 4 } \\ & = 6.2845 \text{ or } -2.7845 \text{ (NA, since } t \ge 0) \\ & \approx 6.28 \text{ seconds} \end{align*}

(i)

\begin{align*} \text{Greatest value of } 2x - y & = 2(5) - (-1) \\ & = 11 \end{align*}

(ii)

\begin{align*} \text{Least possible value of } 2xy & = 2(5)(-1) \\ & = -10 \end{align*}

(iii)

\begin{align*} \text{Greatest possible value of } {y \over x} & = { 7 \over 1 } \\ & = 7 \end{align*}

\begin{align*} {x \over x - y} - {y \over x - y} & = {x - y \over x - y} \\ & = 1 \\ \\ \therefore {x \over x - y } & \text{ has a greater value} \end{align*}

\begin{align*} \text{Let no. of correct answers} & = x \\ \\ \text{No. of incorrect answers} & = 21 - x \\ \\ \text{Total marks} & = (2)(x) - (1)(21 - x) \\ & = 2x - 21 + x \\ & = 3x - 21 \\ \\ \text{Total marks} & \le 33 \phantom{000000} [\text{Not more than 33}] \\ 3x - 21 & \le 33 \\ 3x & \le 33 + 21 \\ 3x & \le 54 \\ x & \le 18 \\ \\ \text{Max. no. of correct questions} & = 18 \end{align*}

(i)

\begin{align*} \text{Length} & = (x + 3.4) \text{ cm} \\ \\ \text{Area} & = (x)(x + 3.4) \\ & = ( x^2 + 3.4x ) \text{ cm}^2 \\ \\ 125 & = x^2 + 3.4x \\ 0 & = x^2 + 3.4x - 125 \end{align*}

(ii)

\begin{align*} 0 & = x^2 + 3.4x - 125 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-3.4 \pm \sqrt{(3.4)^2 - 4(1)(-125)} \over 2(1)} \\ & = {-3.4 \pm \sqrt{511.56} \over 2} \\ & = 9.6088 \text{ or } -13.008 \text{ (NA, since } x > 0) \\ \\ \text{Length} & = 9.6088 + 3.4 \\ & = 13.008 \\ & \approx 13.01 \text{ cm (2 d.p.)} \end{align*}

\begin{align*} 9 \text{ minutes 10 seconds} & = (9 \times 60) + 10 \\ & = 550 \text{ seconds} \\ \\ 2.4 \text{ km} & = 2.4 \times 1000 \\ & = 2400 \text{ m} \\ \\ \text{Let time taken for } A & = x \text{ seconds} \\ \\ \text{Distance for } A & = \text{Speed} \times \text{Time} \\ & = 4 \times x \\ & = 4x \text{ m} \\ \\ \text{Time taken for } B & = (550 - x) \text{ seconds} \\ \\ \text{Distance for } B & = 5 \times (550 - x) \\ & = ( 2750 - 5x ) \text{ m} \\ \\ \text{Total distance} & = 4x + 2750 - 5x \\ 2400 & = 2750 - x \\ x & = 2750 - 2400 \\ x & = 350 \end{align*} \begin{align*} \text{Distance for } A & = 4(350) &&& \text{Distance for } B & = 2750 - 5(350) \\ & = 1400 \text{ m} &&& & = 1000 \text{ m} \\ & = 1.4 \text{ km} &&& & = 1 \text{ km} \end{align*}

\begin{align*} \text{Let first digit} & = x \\ \\ \text{Second digit} & = 11 - x \\ \\ \text{Value of original number} & = 10x + 11 - x \phantom{000000.} [\text{I.e. } 23 = 20 + 3] \\ & = 9x + 11 \\ \\ \text{Value of reversed number} & = 10(11 - x) + x \phantom{00000}[\text{I.e. } 32 = 30 + 2] \\ & = 110 - 10x + x \\ & = 110 - 9x \\ \\ \text{Original number} - \text{Reversed number} & = 27 \\ 9x + 11 - (110 - 9x) & = 27 \\ 9x + 11 - 110 + 9x & = 27 \\ 9x + 9x & = 27 - 11 + 110 \\ 18x & = 126 \\ x & = {126 \over 18} \\ x & = 7 \\ \\ \text{First digit} = x = 7, & \text{ Second digit} = 11 - 7 = 4 \\ \\ \text{Original number} & = 74 \end{align*}

\begin{align*} \text{Let no. of postcards Cheryl has} & = x \\ \\ \text{No. of postcards Bernard has} & = 4x \\ \\ \text{No. of postcards Albert has} & = 2.5(4x) \\ & = 10x \\ \\ x + 4x + 10x & = 180 \\ 15x & = 180 \\ x & = {180 \over 15} \\ x & = 12 \\ \\ C: 12, B: 4(12) = 48, & \phantom{.} A: 10(12) = 120 \end{align*}

\begin{align*} \text{Let Yihao's initial amount} & = 3x \\ \\ \text{Raju's initial amount} & = 5x \\ \\ { \text{Yihao's current amount} \over \text{Raju's current amount} } & = {3x + 30 \over 5x - 30} \\ {9 \over 11} & = {3x + 30 \over 5x - 30} \\ {9(5x - 30) \over 11(5x - 30)} & = {11(3x + 30) \over 11(5x - 30)} \\ 9(5x - 30) & = 11(3x + 30) \\ 45x - 270 & = 33x + 330 \\ 45x - 33x & = 330 + 270 \\ 12x & = 600 \\ x & = {600 \over 12} \\ x & = 50 \\ \\ \text{Yihao's current amount} & = 3(50) + 30 \\ & = \$ 180 \end{align*}

(i)

\begin{align*} \text{Distance} & = \text{Speed} \times \text{Time} \\ \\ \text{Distance between B and D} & = (x)(2) + (x - 3)(1.5) \\ & = 2x + 1.5x - 4.5 \\ & = ( 3.5x - 4.5 ) \text{ km} \end{align*}

(ii)

\begin{align*} \text{Distance between B and D} & = (x + 1)(3) \\ & = (3x + 3) \text{ km} \end{align*}

(iii)

\begin{align*} 3.5x - 4.5 & = 3x + 3 \\ 3.5x - 3x & = 3 + 4.5 \\ 0.5x & = 7.5 \\ x & = {7.5 \over 0.5} \\ x & = 15 \\ \\ \text{Distance} & = 3(15) + 3 \\ & = 48 \text{ km} \end{align*}

(iv)

\begin{align*} \text{Average speed} & = { \text{Total distance} \over \text{Total time} } \\ & = { 2(48) \over 2 + 1.5 + 0.5 + 3 } \phantom{000000} [\text{Include half an hour rest time}] \\ & = 13{5 \over 7} \text{ km/h} \end{align*}

(i)

\begin{align*} \text{Time taken} & = { \text{Distance} \over \text{Speed} } \\ & = {200 \over x} \text{ h} \end{align*}

(ii)

\begin{align*} \text{Time taken} & = { 200 \over x + 5 } \text{ h} \end{align*}

(iii)

\begin{align*} 1 \text{ hour 15 minutes} & = 1 + {15 \over 60} \text{ hour} \\ & = {5 \over 4} \text{ hour} \\ \\ {200 \over x} - {200 \over x + 5} & = {5 \over 4} \\ {200(x + 5) \over x(x + 5)} - {200x \over x(x + 5)} & = {5 \over 4} \\ {200(x + 5) - 200x \over x(x + 5)} & = {5 \over 4} \\ {200x + 1000 - 200x \over x(x + 5)} & = {5 \over 4} \\ {1000 \over x(x + 5)} & = {5 \over 4} \\ {4(1000) \over 4x(x + 5)} & = {5x(x + 5) \over 4x(x + 5)} \\ 4(1000) & = 5x(x + 5) \\ 4000 & = 5x^2 + 25x \\ 0 & = 5x^2 + 25x - 4000 \\ 0 & = x^2 + 5x - 800 \phantom{0} \text{ (Shown)} \end{align*}

(iv)

\begin{align*} 0 & = x^2 + 5x - 800 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-5 \pm \sqrt{(5)^2 - 4(1)(-800)} \over 2(1)} \\ & = {-5 \pm \sqrt{3225} \over 2} \\ & = 25.894 \text{ or } -30.894 \\ \\ \text{Speed from Segamat to Singapore} & = x + 5 \\ & = 25.894 + 5 \\ & = 30.894 \\ & \approx 30.9 \text{ km/h} \end{align*}

\begin{align*} \text{Let numerator} & = x \\ \\ \text{Denominator} & = x + 2 \\ \\ \text{Original fraction} & = {x \over x + 2} \\ \\ \text{New fraction} & = {x + 3 \over x + 2 + 3} \\ & = {x + 3 \over x + 5} \\ \\ {x + 3 \over x + 5} - {x \over x + 2} & = {3 \over 20} \\ {20(x + 2)(x + 3) \over 20(x + 5)(x + 2)} - {20x(x + 5) \over 20(x + 5)(x + 2)} & = {3(x + 5)(x + 2) \over 20(x + 5)(x + 2)} \\ 20(x + 2)(x + 3) - 20x(x + 5) & = 3(x + 5)(x + 2) \\ 20(x^2 + 3x + 2x + 6) - 20x^2 - 100x & = 3(x^2 + 2x + 5x + 10) \\ 20(x^2 + 5x + 6) - 20x^2 - 100x & = 3(x^2 + 7x + 10) \\ 20x^2 + 100x + 120 - 20x^2 - 100x & = 3x^2 + 21x + 30 \\ 120 & = 3x^2 + 21x + 30 \\ 0 & = 3x^2 + 21x + 30 - 120 \\ 0 & = 3x^2 + 21x - 90 \\ 0 & = x^2 + 7x - 30 \\ 0 & = (x - 3)(x + 10) \end{align*} \begin{align*} x - 3 & = 0 && \text{ or } & x + 10 & = 0 \\ x & = 3 &&& x & = -10 \text{ (NA)} \end{align*}
\begin{align*} \text{Original fraction} & = {3 \over 3 + 2} \\ & = {3 \over 5} \end{align*}

(a)(i)

\begin{align*} \text{Cost price} & = \$ \left( {32 \over x} \right) \end{align*}

(a)(ii)

\begin{align*} 5 \text{ cent} & = \$ 0.05 \\ & = \$ \left(1 \over 20\right) \\ \\ \text{Selling price} & = {32 \over x} + {1 \over 20} \\ & = {20(32) \over 20x} + {x \over 20x} \\ & = \$ \left( {640 + x \over 20x} \right) \end{align*}

(b)

\begin{align*} \text{Revenue} & = (x - 20) \left( 640 + x \over 20x\right) \\ & = \left( x - 20 \over 1\right)\left(640 + x \over 20x\right) \\ & = { (x - 20)(640 + x) \over 20x } \\ & = { 640x + x^2 - 12 \phantom{.} 800 - 20x \over 20 x }\\ & = \$ \left( { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \right) \\ \\ 35 & = { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \\ {35 \over 1} & = { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \\ {20x(35) \over 20x} & = { x^2 + 620x - 12 \phantom{.} 800 \over 20x } \\ 20x(35) & = x^2 + 620x - 12 \phantom{.} 800 \\ 700x & = x^2 + 620x - 12 \phantom{.} 800 \\ 0 & = x^2 + 620x - 700x - 12 \phantom{.} 800 \\ 0 & = x^2 - 80x - 12 \phantom{.} 800 \phantom{0} \text{ (Shown)} \end{align*}

(c)

\begin{align*} 0 & = x^2 - 80x - 12 \phantom{.} 800 \\ 0 & = (x - 160)(x + 80) \end{align*} \begin{align*} x - 160 & = 0 && \text{ or } & x + 80 & = 0 \\ x & = 160 &&& x & = -80 \text{ (NA, since } x > 0) \end{align*}
$$ \text{Fruitseller bought 160 apples} $$

(a)

\begin{align*} A: x \text{ minutes} & \rightarrow \text{1 tank} \\ 1 \text{ minute} & \rightarrow {1 \over x} \text{ tank} \\ \\ B: (x + 40) \text{ minutes} & \rightarrow \text{1 tank} \\ 1 \text{ minute} & \rightarrow {1 \over x + 40} \text{ tank} \\ \\ A \&B: 48 \text{ minutes} & \rightarrow \text{1 tank} \\ 1 \text{ minute} & \rightarrow {1 \over 48} \text{ tank} \\ \\ {1 \over x} + {1 \over x + 40} & = {1 \over 48} \\ {48(x + 40) \over 48x(x + 40)} + {48x \over 48x(x + 40)} & = {x(x + 40) \over 48x(x + 40)} \\ 48(x + 40) + 48x & = x(x + 40) \\ 48x + 1920 + 48x & = x^2 + 40x \\ 96x + 1920 & = x^2 + 40x \\ 0 & = x^2 + 40x - 96x - 1920 \\ 0 & = x^2 - 56x - 1920 \phantom{0} \text{ (Shown)} \end{align*}

(b)(i)

\begin{align*} 0 & = x^2 - 56x - 1920 \\ 0 & = (x - 80)(x + 24) \end{align*} \begin{align*} x - 80 & = 0 && \text{ or } & x + 24 & = 0 \\ x & = 80 &&& x & = -24 \end{align*}

(b)(ii)

\begin{align*} \text{Cannot accept } x = -24 \text{ since time taken cannot be negative} \end{align*}

(c)

\begin{align*} \text{Time taken by } B & = x + 40 \\ & = 80 + 40 \\ & = 120 \text{ minutes} \\ & = 2 \text{ hours} \end{align*}

(a)

\begin{align*} x + y & = 3 \\ y & = -x + 3 \end{align*}

$x$	$0$	$1$	$2$
$y$	$3$	$2$	$1$

\begin{align*} x + 3y & = 1 \\ 3y & = - x + 1 \\ y & = -{1 \over 3}x + {1 \over 3} \end{align*}

$x$	$-2$	$1$	$4$
$y$	$1$	$0$	$-1$

$$ x= 4, y = -1 $$

(b)

\begin{align*} 2x - y & = 4 \\ -y & = -2x + 4 \\ y & = 2x - 4 \end{align*}

$x$	$0$	$1$	$2$
$y$	$-4$	$-2$	$0$

\begin{align*} 5x + 3y & = -1 \\ 3y & = -5x - 1 \\ y & = -{5 \over 3}x - {1 \over 3} \end{align*}

$x$	$-2$	$1$	$2.5$
$y$	$3$	$-2$	$-4.5$

$$ x = 1, y = -2 $$

(c)

\begin{align*} x + 3y - 12 & = 0 \\ 3y & = -x + 12 \\ y & = -{1 \over 3}x + 4 \end{align*}

$x$	$-3$	$0$	$3$
$y$	$5$	$4$	$3$

\begin{align*} 2x - 2y & = 0 \\ -2y & = -2x \\ y & = x \end{align*}

$x$	$-1$	$0$	$1$
$y$	$-1$	$0$	$1$

$$ x = 3, y = 3 $$

(d)

\begin{align*} 2x + 5y + 2 & = 0 \\ 5y & = -2x - 2 \\ y & = -{2 \over 5}x - {2 \over 5} \end{align*}

$x$	$-3.5$	$-1$	$1.5$
$y$	$1$	$0$	$-1$

\begin{align*} x + 5y - 3 & = 0 \\ 5y & = -x + 3 \\ y & = -{1 \over 5}x + {3 \over 5} \end{align*}

$x$	$-2$	$0$	$3$
$y$	$1$	$0.6$	$0$

$$ x = -5, y = 1.6 $$