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Question Number 104062 by bramlex last updated on 19/Jul/20

{ (((x/y) + (y/x) = ((13)/6))),((x+y = 5)) :} find the solution

$$\begin{cases}{\frac{{x}}{{y}}\:+\:\frac{{y}}{{x}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}}\\{{x}+{y}\:=\:\mathrm{5}}\end{cases} \\ $$$${find}\:{the}\:{solution} \\ $$

Commented by Rasheed.Sindhi last updated on 19/Jul/20

An Other Way _↘^↗ ((x^2 +y^2 )/(xy))=((13)/6) 6x^2 −13xy+6y^2 =0 (3x−2y)(2x−3y)=0 ( { ((3x−2y=0)),((x+y=5)) :} )∨ ( { ((2x−3y=0)),((x+y=5)) :}) 3x−2(5−x)=0 ∨ 2x−3(5−x)=0 x=2,y=3 ∨ x=3,y=2

$${An}\:{Other}\:{Way}\:_{\searrow} ^{\nearrow} \:\: \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\mathrm{6}{x}^{\mathrm{2}} −\mathrm{13}{xy}+\mathrm{6}{y}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left(\mathrm{3}{x}−\mathrm{2}{y}\right)\left(\mathrm{2}{x}−\mathrm{3}{y}\right)=\mathrm{0} \\ $$$$\left(\begin{cases}{\mathrm{3}{x}−\mathrm{2}{y}=\mathrm{0}}\\{{x}+{y}=\mathrm{5}}\end{cases}\:\right)\vee\:\left(\begin{cases}{\mathrm{2}{x}−\mathrm{3}{y}=\mathrm{0}}\\{{x}+{y}=\mathrm{5}}\end{cases}\right) \\ $$$$\mathrm{3}{x}−\mathrm{2}\left(\mathrm{5}−{x}\right)=\mathrm{0}\:\vee\:\mathrm{2}{x}−\mathrm{3}\left(\mathrm{5}−{x}\right)=\mathrm{0} \\ $$$${x}=\mathrm{2},{y}=\mathrm{3}\:\:\:\:\:\:\:\:\vee\:\:\:{x}=\mathrm{3},{y}=\mathrm{2} \\ $$

Commented by bemath last updated on 19/Jul/20

cooll

$${cooll} \\ $$

Commented by bramlex last updated on 19/Jul/20

thank you all

$${thank}\:{you}\:{all}\: \\ $$

Commented by Dwaipayan Shikari last updated on 19/Jul/20

Another way (x/y)+1=(5/y)⇒(x/y)=(5/y)−1 and (y/x)=((5/y)−1)^(−1) (x/y)+(y/x)=((13)/6) ((5−y)/y)+(y/(5−y))=((13)/6)⇒((25+2y^2 −10y)/(5y−y^2 ))=((13)/6) ((25+2y^2 −10y)/(10y−2y^2 ))+1=((13)/(12))+1⇒((25)/(10y−2y^2 ))=((25)/(12))⇒2y^2 −10y+12=0→(a) So y=3,2 (from(a)) (x/3)+1=(5/3) or (x/2)+1=(5/2) x=2 or 3 { ((x=2,3)),((y=3,2)) :}

$$\mathrm{Another}\:\mathrm{way} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}+\mathrm{1}=\frac{\mathrm{5}}{\mathrm{y}}\Rightarrow\frac{\mathrm{x}}{\mathrm{y}}=\frac{\mathrm{5}}{\mathrm{y}}−\mathrm{1}\:\:\:\mathrm{and}\:\:\:\frac{\mathrm{y}}{\mathrm{x}}=\left(\frac{\mathrm{5}}{\mathrm{y}}−\mathrm{1}\right)^{−\mathrm{1}} \\ $$$$\frac{\mathrm{x}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{x}}=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\frac{\mathrm{5}−\mathrm{y}}{\mathrm{y}}+\frac{\mathrm{y}}{\mathrm{5}−\mathrm{y}}=\frac{\mathrm{13}}{\mathrm{6}}\Rightarrow\frac{\mathrm{25}+\mathrm{2y}^{\mathrm{2}} −\mathrm{10y}}{\mathrm{5y}−\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\frac{\mathrm{25}+\mathrm{2y}^{\mathrm{2}} −\mathrm{10y}}{\mathrm{10y}−\mathrm{2y}^{\mathrm{2}} }+\mathrm{1}=\frac{\mathrm{13}}{\mathrm{12}}+\mathrm{1}\Rightarrow\frac{\mathrm{25}}{\mathrm{10y}−\mathrm{2y}^{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{12}}\Rightarrow\mathrm{2y}^{\mathrm{2}} −\mathrm{10y}+\mathrm{12}=\mathrm{0}\rightarrow\left(\mathrm{a}\right) \\ $$$$\:\:\:\:\:\:\mathrm{So}\:\:\:\:\mathrm{y}=\mathrm{3},\mathrm{2}\:\:\left(\mathrm{from}\left(\mathrm{a}\right)\right) \\ $$$$\frac{\mathrm{x}}{\mathrm{3}}+\mathrm{1}=\frac{\mathrm{5}}{\mathrm{3}}\:\:\:\:\mathrm{or}\:\frac{\mathrm{x}}{\mathrm{2}}+\mathrm{1}=\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\mathrm{x}=\mathrm{2}\:\mathrm{or}\:\mathrm{3} \\ $$$$\begin{cases}{\mathrm{x}=\mathrm{2},\mathrm{3}}\\{\mathrm{y}=\mathrm{3},\mathrm{2}}\end{cases} \\ $$$$ \\ $$$$ \\ $$

Answered by Dwaipayan Shikari last updated on 19/Jul/20

((x^2 +y^2 )/(2xy))=((13)/(12)) ((x^2 +y^2 +2xy)/(x^2 +y^2 −2xy))=((25)/1) (((x+y)/(x−y)))^2 =((25)/1) ((25)/((x−y)^2 ))=((25)/1) x−y=1 or y−x=1 x+y=5 { ((x=3,2 )),((y=2,3)) :}

$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}{xy}}=\frac{\mathrm{13}}{\mathrm{12}} \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}}=\frac{\mathrm{25}}{\mathrm{1}} \\ $$$$\left(\frac{{x}+{y}}{{x}−{y}}\right)^{\mathrm{2}} =\frac{\mathrm{25}}{\mathrm{1}} \\ $$$$\frac{\mathrm{25}}{\left({x}−{y}\right)^{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{1}} \\ $$$${x}−{y}=\mathrm{1}\:\:\:\:{or}\:\:{y}−{x}=\mathrm{1} \\ $$$${x}+{y}=\mathrm{5} \\ $$$$\begin{cases}{{x}=\mathrm{3},\mathrm{2}\:\:\:\:\:\:\:\:}\\{{y}=\mathrm{2},\mathrm{3}}\end{cases} \\ $$

Commented by Rasheed.Sindhi last updated on 19/Jul/20

Everybody is not as intelligent as you are! :) So pl insert some steps between 1st & 2nd lines.

$$\mathcal{E}{verybody}\:{is}\:{not}\:{as}\:{intelligent} \\ $$$$\left.{as}\:{you}\:{are}!\::\right) \\ $$$$\mathcal{S}{o}\:{pl}\:{insert}\:{some}\:{steps}\:{between} \\ $$$$\mathrm{1}{st}\:\&\:\mathrm{2}{nd}\:{lines}. \\ $$

Commented by Dwaipayan Shikari last updated on 19/Jul/20

Sorry sir for your inconvenience

$${Sorry}\:{sir}\:{for}\:{your}\:{inconvenience} \\ $$

Commented by Rasheed.Sindhi last updated on 19/Jul/20

((x^2 +y^2 +2xy)/(x^2 +y^2 −2xy))=((13+12)/(13−12)) This was necessary step to insert.Without this the answer seemed a riddle! Pl don′t mind!

$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}}=\frac{\mathrm{13}+\mathrm{12}}{\mathrm{13}−\mathrm{12}} \\ $$$$\mathcal{T}{his}\:{was}\:{necessary}\:{step}\:{to} \\ $$$${insert}.\mathcal{W}{ithout}\:{this}\:{the}\:{answer} \\ $$$${seemed}\:{a}\:{riddle}! \\ $$$${Pl}\:{don}'{t}\:{mind}! \\ $$

Commented by Rasheed.Sindhi last updated on 19/Jul/20

Nice Solution!

$$\mathcal{N}{ice}\:\mathcal{S}{olution}! \\ $$

Commented by 1549442205PVT last updated on 19/Jul/20

applying the property: (a/m)=(b/n)=((a+b)/(m+n))=((a−b)/(m−n)) ((x^2 +y^2 )/(13))=((2xy)/(12))=((x^2 +y^2 +2xy)/(25))=((x^2 +y^2 −2xy)/1) ⇒((x^2 +2xy+y^2 )/(x^2 −2xy+y^2 ))=((25)/1)

$$\mathrm{applying}\:\mathrm{the}\:\mathrm{property}:\:\frac{\mathrm{a}}{\mathrm{m}}=\frac{\mathrm{b}}{\mathrm{n}}=\frac{\mathrm{a}+\mathrm{b}}{\mathrm{m}+\mathrm{n}}=\frac{\mathrm{a}−\mathrm{b}}{\mathrm{m}−\mathrm{n}} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} }{\mathrm{13}}=\frac{\mathrm{2xy}}{\mathrm{12}}=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{2xy}}{\mathrm{25}}=\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} −\mathrm{2xy}}{\mathrm{1}} \\ $$$$\Rightarrow\frac{\mathrm{x}^{\mathrm{2}} +\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }{\mathrm{x}^{\mathrm{2}} −\mathrm{2xy}+\mathrm{y}^{\mathrm{2}} }=\frac{\mathrm{25}}{\mathrm{1}} \\ $$

Answered by bobhans last updated on 19/Jul/20

set x+y = u , xy = v ((x^2 +y^2 )/(xy)) = ((13)/6) ⇒(((x+y)^2 −2xy)/(xy)) = ((13)/6) 25−2v = ((13v)/6) ⇒150 = 25v { ((v = 6→y = (6/x))),((u=5→ x+(6/x) = 5 ; x^2 −5x+6=0)) :} { ((x= 3→y= 2)),((x = 2→y=3)) :} solution set {(3,2);(2,3)}

$${set}\:{x}+{y}\:=\:{u}\:,\:{xy}\:=\:{v}\: \\ $$$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{{xy}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}\:\Rightarrow\frac{\left({x}+{y}\right)^{\mathrm{2}} −\mathrm{2}{xy}}{{xy}}\:=\:\frac{\mathrm{13}}{\mathrm{6}} \\ $$$$\mathrm{25}−\mathrm{2}{v}\:=\:\frac{\mathrm{13}{v}}{\mathrm{6}}\:\Rightarrow\mathrm{150}\:=\:\mathrm{25}{v}\: \\ $$$$\begin{cases}{{v}\:=\:\mathrm{6}\rightarrow{y}\:=\:\frac{\mathrm{6}}{{x}}}\\{{u}=\mathrm{5}\rightarrow\:{x}+\frac{\mathrm{6}}{{x}}\:=\:\mathrm{5}\:;\:{x}^{\mathrm{2}} −\mathrm{5}{x}+\mathrm{6}=\mathrm{0}}\end{cases} \\ $$$$\begin{cases}{{x}=\:\mathrm{3}\rightarrow{y}=\:\mathrm{2}}\\{{x}\:=\:\mathrm{2}\rightarrow{y}=\mathrm{3}}\end{cases}\:{solution}\:{set}\:\left\{\left(\mathrm{3},\mathrm{2}\right);\left(\mathrm{2},\mathrm{3}\right)\right\} \\ $$

Answered by Rasheed.Sindhi last updated on 19/Jul/20

Using formula: (x+y)^2 −(x−y)^2 =4xy _(−) { (((x/y) + (y/x) = ((13)/6).....(i))),((x+y = 5..............(ii))) :} (i)⇒(x/y) +1+ (y/x)+1 = ((13)/6)+2 ((x+y)/y) + ((x+y)/x) =((25)/6) (5/y) + (5/x) =((25)/6) (1/y) + (1/x) =(5/6)................(iii) (ii)⇒(x/(xy))+(y/(xy))=(5/(xy)) (1/y) + (1/x)=5((1/x))((1/y)).......(iv) (iii) & (iv): ((1/x))((1/y))=(1/6)⇒xy=6 x+y=5 ∧ xy=6 (x+y)^2 −(x−y)^2 =4xy (5)^2 −(x−y)^2 =4(6) x−y=±1 { ((x+y=5)),((x−y=1 )) :} ∨ { ((x+y=5)),((x−y=−1)) :} x=3,y=2 ∨ x=2,y=3

$$\mathrm{Using}\:\mathrm{formula}: \\ $$$$\underset{−} {\:\:\:\:\:\:\:\left({x}+{y}\right)^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{4}{xy}\:\:\:\:\:\:\:\:\:} \\ $$$$\begin{cases}{\frac{{x}}{{y}}\:+\:\frac{{y}}{{x}}\:=\:\frac{\mathrm{13}}{\mathrm{6}}.....\left({i}\right)}\\{{x}+{y}\:=\:\mathrm{5}..............\left({ii}\right)}\end{cases}\:\:\:\:\:\:\:\: \\ $$$$\left({i}\right)\Rightarrow\frac{{x}}{{y}}\:+\mathrm{1}+\:\frac{{y}}{{x}}+\mathrm{1}\:=\:\frac{\mathrm{13}}{\mathrm{6}}+\mathrm{2} \\ $$$$\:\:\:\:\:\frac{{x}+{y}}{{y}}\:+\:\frac{{x}+{y}}{{x}}\:=\frac{\mathrm{25}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\frac{\mathrm{5}}{{y}}\:+\:\frac{\mathrm{5}}{{x}}\:=\frac{\mathrm{25}}{\mathrm{6}} \\ $$$$\:\:\:\:\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{x}}\:=\frac{\mathrm{5}}{\mathrm{6}}................\left({iii}\right) \\ $$$$\left({ii}\right)\Rightarrow\frac{{x}}{{xy}}+\frac{{y}}{{xy}}=\frac{\mathrm{5}}{{xy}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:\frac{\mathrm{1}}{{y}}\:+\:\frac{\mathrm{1}}{{x}}=\mathrm{5}\left(\frac{\mathrm{1}}{{x}}\right)\left(\frac{\mathrm{1}}{{y}}\right).......\left({iv}\right) \\ $$$$\left({iii}\right)\:\&\:\left({iv}\right): \\ $$$$\:\:\:\:\:\:\:\:\:\left(\frac{\mathrm{1}}{{x}}\right)\left(\frac{\mathrm{1}}{{y}}\right)=\frac{\mathrm{1}}{\mathrm{6}}\Rightarrow{xy}=\mathrm{6} \\ $$$${x}+{y}=\mathrm{5}\:\wedge\:{xy}=\mathrm{6} \\ $$$$\:\:\:\:\left({x}+{y}\right)^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{4}{xy} \\ $$$$\:\:\:\:\:\left(\mathrm{5}\right)^{\mathrm{2}} −\left({x}−{y}\right)^{\mathrm{2}} =\mathrm{4}\left(\mathrm{6}\right) \\ $$$$\:\:\:\:\:\:\:\:{x}−{y}=\pm\mathrm{1} \\ $$$$\begin{cases}{{x}+{y}=\mathrm{5}}\\{{x}−{y}=\mathrm{1}\:\:}\end{cases}\:\:\:\:\vee\:\:\:\begin{cases}{{x}+{y}=\mathrm{5}}\\{{x}−{y}=−\mathrm{1}}\end{cases} \\ $$$${x}=\mathrm{3},{y}=\mathrm{2}\:\:\:\:\:\:\vee\:\:\:{x}=\mathrm{2},{y}=\mathrm{3} \\ $$

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