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

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

Commented by infinityaction last updated on 11/Jun/22

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

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

$$\mathcal{GOOD}!$$

Commented by udaythool last updated on 11/Jun/22

$$super$$

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

$$\overline{)\begin{array}{c}\frac{x}{y}+\frac{y}{x}=\frac{26}{5};\frac{x+y}{x-y}=?\end{array}}$$$$a+\frac{1}{a}=\frac{26}{5}$$$$5a^2-26a+5=0$$$$(a-5)(5a-1)=0$$$$a=5\mid a=\frac{1}{5}$$$$\frac{x}{y}=5\mid \frac{x}{y}=\frac{1}{5}$$$$\bullet \frac{x+y}{x-y}=\frac{\cancel{y}(\frac{x}{y}+1)}{\cancel{y}(\frac{x}{y}-1)}=\frac{\frac{x}{y}+1}{\frac{x}{y}-1}$$$$=\frac{5+1}{5-1}\mid =\frac{1/5+1}{1/5-1}=\frac{1+5}{1-5}=\frac{6}{-4}$$$$=\frac{3}{2}\mid =-\frac{3}{2}$$

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

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

Commented by mathlove last updated on 11/Jun/22

$$thanks$$

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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