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Question Number 178356 by Spillover last updated on 15/Oct/22
Solve simultaneous sinh x+cosh y=5 sinh^2 x+cosh^2 y=13
$$\mathrm{Solve}\:\mathrm{simultaneous} \\ $$$$\mathrm{sinh}\:\mathrm{x}+\mathrm{cosh}\:\mathrm{y}=\mathrm{5} \\ $$$$\mathrm{sinh}\:^{\mathrm{2}} \mathrm{x}+\mathrm{cosh}\:^{\mathrm{2}} \mathrm{y}=\mathrm{13} \\ $$
Answered by Ar Brandon last updated on 16/Oct/22
{ ((sinhx+coshy=5),(...(i))),((sinh^2 x+cosh^2 y=13),(...(ii))) :} (ii) ⇒ 5^2 −2sinhxcoshy=13 ⇒12=2sinhxcoshy ⇒sinhx=(6/(coshy)) ...(iii) (iii) in (i) ⇒ (6/(coshy))+coshy=5 ⇒cosh^2 y−5coshy+6=0 ⇒(coshy−3)(coshy−2)=0 ⇒coshy=3 or coshy=2 (a) coshy=3 ⇒(1/2)(e^y +e^(−y) )=3 ⇒e^(2y) −6e^y +1=0 ⇒e^y =((6±(√(36−4)))/2)=3±2(√2) ⇒y=ln(3±2(√2)) ⇒sinhx=2 ⇒e^(2x) −4e^x −1=0 ⇒e^x =2+(√5)⇒x=ln(2+(√5)) (b) coshy=2 ⇒e^(2y) −4e^y +1=0 ⇒e^y =2±(√3) ⇒y=ln(2±(√3)) ⇒sinhx=3 ⇒e^(2x) −6e^x −1=0 ⇒e^x =3+(√(10)) ⇒x=ln(3+(√(10))) S={(x,y)∣ (x=ln(2+(√5)) ∧ y=ln(3±2(√2))) ∨(x=ln(3+(√(10)))∧y=ln(2±(√3)))}
$$\begin{cases}{\mathrm{sinh}{x}+\mathrm{cosh}{y}=\mathrm{5}}&{…\left({i}\right)}\\{\mathrm{sinh}^{\mathrm{2}} {x}+\mathrm{cosh}^{\mathrm{2}} {y}=\mathrm{13}}&{…\left({ii}\right)}\end{cases} \\ $$$$\left({ii}\right)\:\Rightarrow\:\mathrm{5}^{\mathrm{2}} −\mathrm{2sinh}{x}\mathrm{cosh}{y}=\mathrm{13} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{12}=\mathrm{2sinh}{x}\mathrm{cosh}{y} \\ $$$$\:\:\:\:\:\:\:\:\:\Rightarrow\mathrm{sinh}{x}=\frac{\mathrm{6}}{\mathrm{cosh}{y}}\:\:…\left({iii}\right) \\ $$$$\left({iii}\right)\:\mathrm{in}\:\left({i}\right)\:\Rightarrow \\ $$$$\frac{\mathrm{6}}{\mathrm{cosh}{y}}+\mathrm{cosh}{y}=\mathrm{5}\:\Rightarrow\mathrm{cosh}^{\mathrm{2}} {y}−\mathrm{5cosh}{y}+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{cosh}{y}−\mathrm{3}\right)\left(\mathrm{cosh}{y}−\mathrm{2}\right)=\mathrm{0}\:\Rightarrow\mathrm{cosh}{y}=\mathrm{3}\:\mathrm{or}\:\mathrm{cosh}{y}=\mathrm{2} \\ $$$$\left({a}\right)\:\boldsymbol{\mathrm{cosh}{y}}=\mathrm{3} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{{y}} +{e}^{−{y}} \right)=\mathrm{3}\:\Rightarrow{e}^{\mathrm{2}{y}} −\mathrm{6}{e}^{{y}} +\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{e}^{{y}} =\frac{\mathrm{6}\pm\sqrt{\mathrm{36}−\mathrm{4}}}{\mathrm{2}}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}\:\Rightarrow{y}=\mathrm{ln}\left(\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\mathrm{sinh}{x}=\mathrm{2}\:\Rightarrow{e}^{\mathrm{2}{x}} −\mathrm{4}{e}^{{x}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{e}^{{x}} =\mathrm{2}+\sqrt{\mathrm{5}}\Rightarrow{x}=\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right) \\ $$$$\left({b}\right)\:\boldsymbol{\mathrm{cosh}{y}}=\mathrm{2} \\ $$$$\Rightarrow{e}^{\mathrm{2}{y}} −\mathrm{4}{e}^{{y}} +\mathrm{1}=\mathrm{0}\:\Rightarrow{e}^{{y}} =\mathrm{2}\pm\sqrt{\mathrm{3}}\:\Rightarrow{y}=\mathrm{ln}\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right) \\ $$$$\Rightarrow\mathrm{sinh}{x}=\mathrm{3}\:\Rightarrow{e}^{\mathrm{2}{x}} −\mathrm{6}{e}^{{x}} −\mathrm{1}=\mathrm{0}\:\Rightarrow{e}^{{x}} =\mathrm{3}+\sqrt{\mathrm{10}}\:\Rightarrow{x}=\mathrm{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right) \\ $$$${S}=\left\{\left({x},{y}\right)\mid\:\left({x}=\mathrm{ln}\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\:\wedge\:{y}=\mathrm{ln}\left(\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}\right)\right)\:\vee\left({x}=\mathrm{ln}\left(\mathrm{3}+\sqrt{\mathrm{10}}\right)\wedge{y}=\mathrm{ln}\left(\mathrm{2}\pm\sqrt{\mathrm{3}}\right)\right)\right\} \\ $$
Commented by Spillover last updated on 16/Oct/22
thanks
$${thanks} \\ $$

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