Question-43822

Question Number 43822 by peter frank last updated on 15/Sep/18

Answered by Joel578 last updated on 17/Sep/18

(b)

4 \cosh x + 3 \cosh y = 10 \dots (i)

4 \sinh x + 3 \sinh y = 7 \dots (i i)

{16 \cosh}^{2} x + 24 \cosh x \cosh y + {9 \cosh}^{2} y = 100 \dots (i i i)

{16 \sinh}^{2} x + 24 \sinh x \sinh y + {9 \sinh}^{2} y = 49 \dots (i v)

(i i i) - (i v)

16 (\cosh^{2} x - \sinh^{2} x) + 24 (\cosh x \cosh y - \sinh x \sinh y) + 9 (\cosh^{2} y - \sinh^{2} y) = 51

16 + 24 \cosh (x - y) + 9 = 51

\cosh (x - y) = \frac{26}{24} = \frac{13}{12}

\frac{e^{x - y} + e^{y - x}}{2} = \frac{13}{12}

\frac{e^{x}}{e^{y}} + \frac{e^{y}}{e^{x}} = \frac{13}{6}

(i) + (i i)

4 (\cosh x + \sinh x) + 3 (\cosh y + \sinh y) = 17

4 e^{x} + 3 e^{y} = 17

Let e^{x} = a, e^{y} = b

\frac{a}{b} + \frac{b}{a} = \frac{13}{6} \to \frac{a^{2} + b^{2}}{a b} = \frac{13}{6} \to a^{2} + b^{2} = \frac{13}{6} a b

4 a + 3 b = 17 \to a = \frac{17 - 3 b}{4}

{(\frac{17 - 3 b}{4})}^{2} + b^{2} = \frac{13}{6} (\frac{17 - 3 b}{4}) b

\frac{289 - 102 b + 25 b^{2}}{2} = \frac{221 b - 39 b^{2}}{3}

3 (289 - 102 b + 25 b^{2}) = 2 (221 b - 39 b^{2})

153 b^{2} - 748 b + 867 = 0

9 b^{2} - 44 b + 51 = 0

b = 3 \lor b = \frac{17}{9}

b = 3 \to a = 2

\Rightarrow x = \ln 2, y = \ln 3

b = \frac{17}{9} \to a = \frac{17}{6}

\Rightarrow x = \ln (\frac{17}{6}), y = \ln (\frac{17}{9})

Commented by peter frank last updated on 16/Sep/18

f i n d x a n d y