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

Answered by Joel578 last updated on 17/Sep/18

$$\left(b\right)$$$$4\cosh x+3\cosh y=10...\left(i\right)$$$$4\sinh x+3\sinh y=7...\left(ii\right)$$$$$$$${16\cosh }^{2}x+24\cosh x\cosh y+{9\cosh }^{2}y=100...\left(iii\right)$$$${16\sinh }^{2}x+24\sinh x\sinh y+{9\sinh }^{2}y=49...\left(iv\right)$$$$$$$$\left(iii\right)-\left(iv\right)$$$$16({\cosh }^{2}x-{\sinh }^{2}x)+24(\cosh x\cosh y-\sinh x\sinh y)+9({\cosh }^{2}y-{\sinh }^{2}y)=51$$$$16+24\cosh (x-y)+9=51$$$$\cosh (x-y)=\frac{26}{24}=\frac{13}{12}$$$$\frac{e^{x-y}+e^{y-x}}{2}=\frac{13}{12}$$$$\frac{e^x}{e^y}+\frac{e^y}{e^x}=\frac{13}{6}$$$$$$$$\left(i\right)+\left(ii\right)$$$$4(\cosh x+\sinh x)+3(\cosh y+\sinh y)=17$$$$4e^x+3e^y=17$$$$$$$$\mathrm{Let}{e}^x=a,e^y=b$$$$\frac{a}{b}+\frac{b}{a}=\frac{13}{6}\to \frac{a^2+b^2}{ab}=\frac{13}{6}\to a^2+b^2=\frac{13}{6}ab$$$$4a+3b=17\to a=\frac{17-3b}{4}$$$$$$$${\left(\frac{17-3b}{4}\right)}^2+b^2=\frac{13}{6}\left(\frac{17-3b}{4}\right)b$$$$\frac{289-102b+25b^2}{2}=\frac{221b-39b^2}{3}$$$$3(289-102b+25b^2)=2(221b-39b^2)$$$$153b^2-748b+867=0$$$$9b^2-44b+51=0$$$$b=3\vee b=\frac{17}{9}$$$$$$$$b=3\to a=2$$$$\Rightarrow x=\ln 2,y=\ln 3$$$$$$$$b=\frac{17}{9}\to a=\frac{17}{6}$$$$\Rightarrow x=\ln \left(\frac{17}{6}\right),y=\ln \left(\frac{17}{9}\right)$$

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

$$findxandy$$

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