S3 E Maths Textbook Solutions >> New Discovering Mathematics 3A Chapters 1 & 2 Solutions >>

Ex 1.4

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Solutions

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(a)

\begin{align} {x \over x - 2} + {x + 7 \over x + 2} & = {5 \over 1} \\ {x(x + 2) \over (x - 2)(x + 2)} + {(x + 7)(x - 2) \over (x - 2)(x + 2)} & = {5(x - 2)(x + 2) \over (x - 2)(x + 2)} \\ {x(x + 2) + (x + 7)(x - 2) \over (x - 2)(x + 2)} & = {5(x - 2)(x + 2) \over (x - 2)(x + 2)} \\ \\ x (x + 2) + (x + 7)(x - 2) & = 5(x - 2)(x + 2) \\ x^2 + 2x + x^2 - 2x + 7x - 14 & = 5( \underbrace{x^2 - 2^2}_{(a - b)(a + b) = a^2 - b^2}) \\ 2x^2 + 7x - 14 & = 5(x^2 - 4) \\ 2x^2 + 7x - 14 & = 5x^2 - 20 \\ 0 & = 5x^2 - 2x^2 - 7x - 20 + 14 \\ 0 & = 3x^2 - 7x - 6 \\ 0 & = (3x + 2)(x - 3) \end{align} \begin{align} 3x + 2 & = 0 && \text{ or } & x - 3 & = 0 \\ 3x & = -2 &&& x & = 3 \\ x & = -{2 \over 3} \end{align}

(b)

\begin{align} {3 \over x - 1} + {2x + 10 \over x^2 + 2x - 3} & = {1 \over 3} \\ {3 \over x - 1} + {2x + 10 \over (x - 1)(x + 3) } & = {1 \over 3} \\ {3(3)(x + 3) \over 3(x - 1)(x + 3)} + {3(2x + 10) \over 3(x - 1)(x + 3)} & = {(x - 1)(x + 3) \over 3(x - 1)(x + 3)} \\ {9(x + 3) + 3(2x + 10) \over 3(x - 1)(x + 3)} & = {(x - 1)(x + 3) \over 3(x - 1)(x + 3)} \\ \\ 9(x + 3) + 3(2x + 10) & = (x - 1)(x + 3) \\ 9x + 27 + 6x + 30 & = x^2 + 3x - x - 3 \\ 15x + 57 & = x^2 + 2x - 3 \\ 0 & = x^2 + 2x - 15x - 3 - 57 \\ 0 & = x^2 - 13x - 60 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-13) \pm \sqrt{(-13)^2 - 4(1)(-60)} \over 2(1)} \\ & = {13 \pm \sqrt{409} \over 2} \\ & = 16.611 \text{ or } -3.6118 \\ & \approx 16.6 \text{ or } -3.61 \end{align}

(c)

\begin{align} {4x - 3 \over x - 5} & = {2 \over 1} - {x - 5 \over x - 7} \\ {(4x - 3)(x - 7) \over (x - 5)(x - 7)} & = {2(x - 5)(x - 7) \over (x - 5)(x - 7)} - {(x - 5)^2 \over (x - 5)(x - 7)} \\ {(4x - 3)(x - 7) \over (x - 5)(x - 7)} & = {2(x - 5)(x - 7) - (x - 5)^2 \over (x - 5)(x - 7)} \\ \\ (4x - 3)(x - 7) & = 2(x - 5)(x - 7) - (x - 5)^2 \\ 4x^2 - 28x - 3x + 21 & = 2(x^2 - 7x - 5x + 35) - [ \underbrace{(x)^2 - 2(x)(5) + (5)^2}_{(a - b)^2 = a^2 - 2ab + b^2} ] \\ 4x^2 - 31x + 21 & = 2(x^2 - 12x + 35) - (x^2 - 10x + 25) \\ 4x^2 - 31x + 21 & = 2x^2 - 24x + 70 - x^2 + 10x - 25 \\ 4x^2 - 31x + 21 & = x^2 - 14x + 45 \\ 4x^2 - x^2 - 31x + 14x + 21 - 45 & = 0 \\ 3x^2 - 17x - 24 & = 0 \\ \\ x & = { -b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-17) \pm \sqrt{(-17)^2 - 4(3)(-24)} \over 2(3)} \\ & = {17 \pm \sqrt{577} \over 6} \\ & = 6.8368 \text{ or } -1.1701 \\ & \approx 6.84 \text{ or } -1.17 \end{align}

(d)

\begin{align} {7 \over x + 3} - {4 \over (x - 3)} & = {5 \over 1} \\ {7(x - 3) \over (x + 3)(x - 3)} - {4(x + 3) \over (x + 3)(x - 3)} & = {5(x + 3)(x - 3) \over (x + 3)(x - 3)} \\ {7(x - 3) - 4(x + 3) \over (x + 3)(x - 3)} & = {5(x + 3)(x - 3) \over (x + 3)(x - 3)} \\ \\ 7(x - 3) - 4(x + 3) & = 5(x + 3)(x - 3) \\ 7x - 21 - 4x - 12 & = 5 ( \underbrace{x^2 - 3^2}_{(a + b)(a - b) = a^2 - b^2} ) \\ 3x - 33 & = 5 (x^2 - 9) \\ 3x - 33 & = 5x^2 - 45 \\ 0 & = 5x^2 - 3x - 45 + 33 \\ 0 & = 5x^2 - 3x - 12 \\ \\ x & = { -b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(5)(-12)} \over 2(5)} \\ & = {3 \pm \sqrt{249} \over 10} \\ & = 1.8779 \text{ or } -1.2779 \\ & \approx 1.88 \text{ or } - 1.28 \end{align}

(e)

\begin{align} {1 \over x} + {2 \over x - 1} + {3 \over x + 1} & = 0 \\ {(x + 1)(x - 1) \over x(x + 1)(x - 1)} + {2x(x + 1) \over x(x + 1)(x - 1)} + {3x(x - 1) \over x(x + 1)(x - 1)} & = 0 \\ {(x + 1)(x - 1) + 2x(x + 1) + 3x(x - 1) \over x(x + 1)(x - 1)} & = {0 \over x(x + 1)(x - 1)} \\ \\ (x + 1)(x - 1) + 2x(x + 1) + 3x(x - 1) & = 0 \\ \underbrace{ (x)^2 - (1)^2 }_{ (a + b)(a - b) = a^2 - b^2 } + 2x^2 + 2x + 3x^2 - 3x & = 0 \\ x^2 - 1^2 + 5x^2 - x & = 0 \\ x^2 - 1 + 5x^2 - x & = 0 \\ 6x^2 - x - 1 & = 0 \\ (3x + 1)(2x - 1) & = 0 \end{align} \begin{align} 3x + 1 & = 0 && \text{ or } & 2x - 1 & = 0 \\ 3x & = -1 &&& 2x & = 1 \\ x & = -{1 \over 3} &&& x & = {1 \over 2} \end{align}

(f)

\begin{align} {5 \over x - 15} - {2 \over x + 6} & = {3 \over x - 9} \\ {5(x + 6)(x - 9) \over (x - 15)(x + 6)(x - 9)} - {2 (x - 15)(x - 9) \over (x - 15)(x + 6)(x - 9)} & = {3(x - 15)(x + 6) \over (x - 15)(x + 6)(x - 9)} \\ {5(x + 6)(x - 9) - 2(x - 15)(x - 9) \over (x - 15)(x + 6)(x - 9)} & = {3(x - 15)(x + 6) \over (x - 15)(x + 6)(x - 9)} \\ \\ 5(x + 6)(x - 9) - 2(x - 15)(x - 9) & = 3(x - 15)(x + 6) \\ 5(x^2 - 9x + 6x - 54) - 2(x^2 - 9x - 15x + 135) & = 3(x^2 + 6x - 15x - 90) \\ 5(x^2 - 3x - 54) - 2(x^2 - 24x + 135) & = 3(x^2 - 9x - 90) \\ 5x^2 - 15x - 270 - 2x^2 + 48x - 270 & = 3x^2 - 27x - 270 \\ 3x^2 + 33x - 540 & = 3x^2 - 27x - 270 \\ 3x^2 - 3x^2 + 33x + 27x & = -270 + 540 \\ 60x & = 270 \\ x & = {270 \over 60} \\ x & = {9 \over 2} \end{align}

Question 6

(a)

\begin{align} {x \over 1} + {1 \over x + 3} & = {4 \over 1} \\ {x(x + 3) \over x + 3} + {1 \over x + 3} & = {4(x + 3) \over x + 3} \\ {x(x + 3) + 1 \over x + 3} & = {4(x + 3) \over x + 3} \\ \\ x(x + 3) + 1 & = 4(x + 3) \\ x^2 + 3x + 1 & = 4x + 12 \\ x^2 + 3x - 4x + 1 - 12 & = 0 \\ x^2 - x - 11 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-1) \pm \sqrt{(-1)^2 - 4(1)(-11)} \over 2(1)} \\ & = {1 \pm \sqrt{45} \over 2} \\ & = 3.8541 \text{ or } -2.8541 \\ & \approx 3.85 \text{ or } -2.85 \end{align}

(b)

\begin{align} {3 \over x - 2} + {5 \over 1} & = {1 \over x + 1} \\ {3(x + 1) \over (x - 2)(x + 1)} + {5(x - 2)(x + 1) \over (x - 2)(x + 1)} & = {x - 2 \over (x - 2)(x + 1)} \\ {3(x + 1) + 5(x - 2)(x + 1) \over (x - 2)(x + 1)} & = {x - 2 \over (x - 2)(x + 1)} \\ \\ 3(x + 1) + 5(x - 2)(x + 1) & = x - 2 \\ 3x + 3 + 5(x^2 + x - 2x - 2) & = x - 2 \\ 3x + 3 + 5(x^2 - x - 2) & = x - 2 \\ 3x + 3 + 5x^2 - 5x - 10 & = x - 2 \\ 5x^2 - 2x - 7 & = x - 2 \\ 5x^2 - 2x - x - 7 + 2 & = 0 \\ 5x^2 - 3x - 5 & = 0 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(5)(-5)} \over 2(5)} \\ & = {3 \pm \sqrt{109} \over 10} \\ & = 1.344 \text{ or } -0.74403 \\ & \approx 1.34 \text{ or } -0.744 \end{align}

(c)

\begin{align} {4 \over x - 3} + {1 \over x + 3} & = {6 \over 1} \\ {4(x + 3) \over (x - 3)(x + 3)} + {x - 3 \over (x - 3)(x + 3)} & = {6(x - 3)(x + 3) \over (x - 3)(x + 3)} \\ {4(x + 3) + x - 3 \over (x - 3)(x + 3)} & = {6(x - 3)(x + 3) \over (x - 3)(x + 3)} \\ \\ 4(x + 3) + x - 3 & = 6(x - 3)(x + 3) \\ 4x + 12 + x - 3 & = 6(x^2 - 3^2) \phantom{000000} [(a - b)(a + b) = a^2 - b^2] \\ 5x + 9 & = 6(x^2 - 9) \\ 5x + 9 & = 6x^2 - 54 \\ 0 & = 6x^2 - 5x - 54 - 9 \\ 0 & = 6x^2 - 5x - 63 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-5) \pm \sqrt{(-5)^2 - 4(6)(-63)} \over 2(6)} \\ & = {5 \pm \sqrt{1537} \over 12} \\ & = 3.6837 \text{ or } -2.8503 \\ & \approx 3.68 \text{ or } -2.85 \end{align}

(d)

\begin{align} {7 \over x^2 - x - 6} - {x \over x + 2} & = 0 \\ {7 \over (x + 2)(x - 3)} & = {x \over x + 2} \\ {7 \over (x + 2)(x - 3)} & = {x(x - 3) \over (x + 2)(x - 3)} \\ \\ 7 & = x (x - 3) \\ 7 & = x^2 - 3x \\ 0 & = x^2 - 3x - 7 \\ \\ x & = {-b \pm \sqrt{b^2 - 4ac} \over 2a} \\ & = {-(-3) \pm \sqrt{(-3)^2 - 4(1)(-7)} \over 2(1)} \\ & = {3 \pm \sqrt{37} \over 2} \\ & = 4.5413 \text{ or } -1.5413 \\ & \approx 4.54 \text{ or } -1.54 \end{align}

Question 7

Question 8

Question 9