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Solve-for-real-numbers-x-y-5-x-y-5-5-y-x-y-5-5-y-5-x-

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Question Number 161744 by HongKing last updated on 21/Dec/21
Solve for real numbers: (x/y) + (5/x) + ((y - 5)/5) = ((y + x)/(y + 5)) + ((5 + y)/(5 + x))
$$\mathrm{Solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}\:+\:\frac{\mathrm{5}}{\mathrm{x}}\:+\:\frac{\mathrm{y}\:-\:\mathrm{5}}{\mathrm{5}}\:=\:\frac{\mathrm{y}\:+\:\mathrm{x}}{\mathrm{y}\:+\:\mathrm{5}}\:+\:\frac{\mathrm{5}\:+\:\mathrm{y}}{\mathrm{5}\:+\:\mathrm{x}} \\ $$
Commented by Rasheed.Sindhi last updated on 22/Dec/21
Not unique solution.Because only one equation is given in two variables.
$${Not}\:{unique}\:{solution}.{Because} \\ $$$${only}\:\boldsymbol{{one}}\:{equation}\:{is}\:{given}\:{in}\:\boldsymbol{{two}} \\ $$$${variables}. \\ $$
Answered by Rasheed.Sindhi last updated on 22/Dec/21
y=5: (x/5)+(5/x)=((5+x)/(10))+((10)/(5+x)) ((x^2 +5^2 )/(5x))=(((5+x)^2 +10^2 )/(10(5+x))) ((x^2 +25)/x)=((25+10x+x^2 +100)/(2(5+x))) (x^2 +25)(2x+10)=x^3 +10x^2 +125x 2x^3 +10x^2 +50x+250−x^3 −10x^2 −125x x^3 −75x+250=0 x=5,−10 (5,5),(−10,5) are two of many solutions.
$$\mathrm{y}=\mathrm{5}: \\ $$$$\frac{\mathrm{x}}{\mathrm{5}}+\frac{\mathrm{5}}{\mathrm{x}}=\frac{\mathrm{5}+\mathrm{x}}{\mathrm{10}}+\frac{\mathrm{10}}{\mathrm{5}+\mathrm{x}} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{5}^{\mathrm{2}} }{\mathrm{5x}}=\frac{\left(\mathrm{5}+\mathrm{x}\right)^{\mathrm{2}} +\mathrm{10}^{\mathrm{2}} }{\mathrm{10}\left(\mathrm{5}+\mathrm{x}\right)} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{25}}{\mathrm{x}}=\frac{\mathrm{25}+\mathrm{10x}+\mathrm{x}^{\mathrm{2}} +\mathrm{100}}{\mathrm{2}\left(\mathrm{5}+\mathrm{x}\right)} \\ $$$$\left(\mathrm{x}^{\mathrm{2}} +\mathrm{25}\right)\left(\mathrm{2x}+\mathrm{10}\right)=\mathrm{x}^{\mathrm{3}} +\mathrm{10x}^{\mathrm{2}} +\mathrm{125x} \\ $$$$\mathrm{2x}^{\mathrm{3}} +\mathrm{10x}^{\mathrm{2}} +\mathrm{50x}+\mathrm{250}−\mathrm{x}^{\mathrm{3}} −\mathrm{10x}^{\mathrm{2}} −\mathrm{125x} \\ $$$$\mathrm{x}^{\mathrm{3}} −\mathrm{75x}+\mathrm{250}=\mathrm{0} \\ $$$$\mathrm{x}=\mathrm{5},−\mathrm{10} \\ $$$$\left(\mathrm{5},\mathrm{5}\right),\left(−\mathrm{10},\mathrm{5}\right)\:{are}\:{two}\:{of}\:{many} \\ $$$${solutions}. \\ $$
Commented by HongKing last updated on 25/Dec/21
thank you so much my dear Sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$

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