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6-1-c-x-1-1-c-x-y-Express-11-1-c-x-1-1-c-x-in-terms-of-y-

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Question Number 176785 by Ar Brandon last updated on 26/Sep/22
(6/(1+c^x ))+(1/(1+c^(−x) ))=y Express ((11)/(1+c^x ))+(1/(1+c^(−x) )) in terms of y
$$\frac{\mathrm{6}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }={y} \\ $$$$\mathrm{Express}\:\:\frac{\mathrm{11}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:{y} \\ $$
Commented by Ar Brandon last updated on 26/Sep/22
Ans. 2y−1
$$\mathrm{Ans}.\:\mathrm{2}{y}−\mathrm{1} \\ $$
Answered by a.lgnaoui last updated on 26/Sep/22
y=6(1/(1+c^x ))+(1/(1+(1/c^x )))=6(1/(1+c^x ))+(c^x /(1+c^x )) y=6(1/(1+c^x ))+((1+c^x −1)/(1+c^x ))=5(1/(1+c^x ))+1 ((11)/(1+c^x ))+(1/(1+c^(−x) ))= ((11)/(1+c^x ))+1−(1/(1+c^x ))=10(1/(1+c^x ))+1 ((11)/(1+c^x ))+(1/(1+c^(−x) ))=2y−1
$${y}=\mathrm{6}\frac{\mathrm{1}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{{c}^{{x}} }}=\mathrm{6}\frac{\mathrm{1}}{\mathrm{1}+{c}^{{x}} }+\frac{{c}^{{x}} }{\mathrm{1}+{c}^{{x}} } \\ $$$${y}=\mathrm{6}\frac{\mathrm{1}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}+{c}^{{x}} −\mathrm{1}}{\mathrm{1}+{c}^{{x}} }=\mathrm{5}\frac{\mathrm{1}}{\mathrm{1}+{c}^{{x}} }+\mathrm{1} \\ $$$$\frac{\mathrm{11}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }=\:\frac{\mathrm{11}}{\mathrm{1}+{c}^{{x}} }+\mathrm{1}−\frac{\mathrm{1}}{\mathrm{1}+{c}^{{x}} }=\mathrm{10}\frac{\mathrm{1}}{\mathrm{1}+{c}^{{x}} }+\mathrm{1} \\ $$$$\frac{\mathrm{11}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }=\mathrm{2}{y}−\mathrm{1} \\ $$
Commented by Ar Brandon last updated on 26/Sep/22
Thank you, Sir!
Answered by mr W last updated on 26/Sep/22
((6+c^x )/(1+c^x ))=y (5/(1+c^x ))=y−1 ((11)/(1+c^x ))+(1/(1+c^(−x) )) =(6/(1+c^x ))+(1/(1+c^(−x) ))+(5/(1+c^x )) =y+y−1 =2y−1
$$\frac{\mathrm{6}+{c}^{{x}} }{\mathrm{1}+{c}^{{x}} }={y} \\ $$$$\frac{\mathrm{5}}{\mathrm{1}+{c}^{{x}} }={y}−\mathrm{1} \\ $$$$ \\ $$$$\frac{\mathrm{11}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} } \\ $$$$=\frac{\mathrm{6}}{\mathrm{1}+{c}^{{x}} }+\frac{\mathrm{1}}{\mathrm{1}+{c}^{−{x}} }+\frac{\mathrm{5}}{\mathrm{1}+{c}^{{x}} } \\ $$$$={y}+{y}−\mathrm{1} \\ $$$$=\mathrm{2}{y}−\mathrm{1} \\ $$
Commented by Tawa11 last updated on 26/Sep/22
Great sirs
$$\mathrm{Great}\:\mathrm{sirs} \\ $$
Commented by Ar Brandon last updated on 26/Sep/22
Thank you, Sir!

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