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Question Number 171244 by mathlove last updated on 11/Jun/22

(x/y)+(y/x)=((26)/5) ((x+y)/(x−y))=? (/)

$$\frac{{x}}{{y}}+\frac{{y}}{{x}}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\frac{{x}+{y}}{{x}−{y}}=? \\ $$$$\frac{}{} \\ $$

Commented by infinityaction last updated on 11/Jun/22

((x^2 +y^2 )/(2xy)) = ((26)/(10)) ((x^2 + y^2 +2xy)/(x^2 +y^2 −2xy)) = ((36)/(16)) (((x+y)/(x−y)))^2 = ((36)/(16)) ((x+y)/(x−y)) = ±(3/2)

$$\frac{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} }{\mathrm{2}{xy}}\:=\:\frac{\mathrm{26}}{\mathrm{10}} \\ $$$$\frac{{x}^{\mathrm{2}} \:+\:{y}^{\mathrm{2}} \:+\mathrm{2}{xy}}{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}}\:\:=\:\frac{\mathrm{36}}{\mathrm{16}} \\ $$$$\left(\frac{{x}+{y}}{{x}−{y}}\right)^{\mathrm{2}} \:=\:\frac{\mathrm{36}}{\mathrm{16}} \\ $$$$\frac{{x}+{y}}{{x}−{y}}\:=\:\pm\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Commented by Rasheed.Sindhi last updated on 11/Jun/22

GOOD!

$$\mathcal{GOOD}! \\ $$

Commented by udaythool last updated on 11/Jun/22

super

$${super} \\ $$

Answered by Rasheed.Sindhi last updated on 11/Jun/22

determinant ((((x/y)+(y/x)=((26)/5) ; ((x+y)/(x−y))=?))) a+(1/a)=((26)/5) 5a^2 −26a+5=0 (a−5)(5a−1)=0 a=5 ∣ a=(1/5) (x/y)=5 ∣ (x/y)=(1/5) • ((x+y)/(x−y))=((y((x/y)+1))/(y((x/y)−1)))=(((x/y)+1)/((x/y)−1)) =((5+1)/(5−1)) ∣ =((1/5+1)/(1/5−1))=((1+5)/(1−5))=(6/(−4)) =(3/2) ∣ =−(3/2)

$$\begin{array}{|c|}{\frac{{x}}{{y}}+\frac{{y}}{{x}}=\frac{\mathrm{26}}{\mathrm{5}}\:\:;\:\:\frac{{x}+{y}}{{x}−{y}}=?}\\\hline\end{array} \\ $$$${a}+\frac{\mathrm{1}}{{a}}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\mathrm{5}{a}^{\mathrm{2}} −\mathrm{26}{a}+\mathrm{5}=\mathrm{0} \\ $$$$\left({a}−\mathrm{5}\right)\left(\mathrm{5}{a}−\mathrm{1}\right)=\mathrm{0} \\ $$$${a}=\mathrm{5}\:\mid\:\:{a}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\frac{{x}}{{y}}=\mathrm{5}\:\:\mid\:\frac{{x}}{{y}}=\frac{\mathrm{1}}{\mathrm{5}} \\ $$$$\bullet\:\frac{{x}+{y}}{{x}−{y}}=\frac{\cancel{{y}}\left(\frac{{x}}{{y}}+\mathrm{1}\right)}{\cancel{{y}}\left(\frac{{x}}{{y}}−\mathrm{1}\right)}=\frac{\frac{{x}}{{y}}+\mathrm{1}}{\frac{{x}}{{y}}−\mathrm{1}} \\ $$$$=\frac{\mathrm{5}+\mathrm{1}}{\mathrm{5}−\mathrm{1}}\:\:\mid\:=\frac{\mathrm{1}/\mathrm{5}+\mathrm{1}}{\mathrm{1}/\mathrm{5}−\mathrm{1}}=\frac{\mathrm{1}+\mathrm{5}}{\mathrm{1}−\mathrm{5}}=\frac{\mathrm{6}}{−\mathrm{4}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{2}}\:\mid\:=−\frac{\mathrm{3}}{\mathrm{2}} \\ $$

Answered by Rasheed.Sindhi last updated on 11/Jun/22

(x/y)+(y/x)=((26)/5)_((i)) ; ((x+y)/(x−y))=? ((x+y)/(x−y))=(p/q) (say) (x/y)=((p+q)/(p−q)) (i) : ((p+q)/(p−q))+((p−q)/(p+q))=((26)/5) (((p+q)^2 +(p−q)^2 )/((p−q)(p+q)))=((26)/5) ((2(p^2 +q^2 ))/(p^2 −q^2 ))=((26)/5) ((p^2 +q^2 )/(p^2 −q^2 ))=((26)/(10)) (p^2 /q^2 )=((36)/(16))=(9/4) (p/q)=±(3/2) ((x+y)/(x−y))=±(3/2) Formula Used: determinant ((( determinant ((((a/b)=(c/d)⇒((a+b)/(a−b))=((c+d)/(c−d)))))_ ^ _() ^(•) )))

$$\underset{\left({i}\right)} {\underbrace{\frac{{x}}{{y}}+\frac{{y}}{{x}}=\frac{\mathrm{26}}{\mathrm{5}}}}\:;\:\frac{{x}+{y}}{{x}−{y}}=?\: \\ $$$$\frac{{x}+{y}}{{x}−{y}}=\frac{{p}}{{q}}\:\left({say}\right) \\ $$$$\frac{{x}}{{y}}=\frac{{p}+{q}}{{p}−{q}} \\ $$$$\left({i}\right)\::\:\:\:\frac{{p}+{q}}{{p}−{q}}+\frac{{p}−{q}}{{p}+{q}}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\frac{\left({p}+{q}\right)^{\mathrm{2}} +\left({p}−{q}\right)^{\mathrm{2}} }{\left({p}−{q}\right)\left({p}+{q}\right)}=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\frac{\mathrm{2}\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)}{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\frac{\mathrm{26}}{\mathrm{5}} \\ $$$$\frac{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} }{{p}^{\mathrm{2}} −{q}^{\mathrm{2}} }=\frac{\mathrm{26}}{\mathrm{10}} \\ $$$$\frac{{p}^{\mathrm{2}} }{{q}^{\mathrm{2}} }=\frac{\mathrm{36}}{\mathrm{16}}=\frac{\mathrm{9}}{\mathrm{4}} \\ $$$$\frac{{p}}{{q}}=\pm\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\frac{{x}+{y}}{{x}−{y}}=\pm\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\mathcal{F}{ormula}\:\mathcal{U}{sed}: \\ $$$$\begin{array}{|c|}{\underset{} {\overset{\bullet} {\:\begin{array}{|c|}{\frac{{a}}{{b}}=\frac{{c}}{{d}}\Rightarrow\frac{{a}+{b}}{{a}−{b}}=\frac{{c}+{d}}{{c}−{d}}}\\\hline\end{array}_{} ^{} }}}\\\hline\end{array} \\ $$

Commented by mathlove last updated on 11/Jun/22

thanks

$${thanks} \\ $$

Commented by haladu last updated on 11/Jun/22

awesome solution

Commented by Rasheed.Sindhi last updated on 11/Jun/22

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