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


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

${\begin{cases} \frac{x}{y} + \frac{y}{x} = \frac{13}{6} \\ x + y = 5 \end{cases}$ $f i n d t h e s o l u t i o n$
Commented by Rasheed.Sindhi last updated on 19/Jul/20
$An Other Way _↘^↗ ((x^2 +y^2 )/(xy))=((13)/6) 6x^2 −13xy+6y^2 =0 (3x−2y)(2x−3y)=0 ( { ((3x−2y=0)),((x+y=5)) :} )∨ ( { ((2x−3y=0)),((x+y=5)) :}) 3x−2(5−x)=0 ∨ 2x−3(5−x)=0 x=2,y=3 ∨ x=3,y=2$
$A n O t h e r W a y_{↘}^{↗}$ $\frac{x^{2} + y^{2}}{x y} = \frac{13}{6}$ $6 x^{2} - 13 x y + 6 y^{2} = 0$ $(3 x - 2 y) (2 x - 3 y) = 0$ $({\begin{cases} 3 x - 2 y = 0 \\ x + y = 5 \end{cases}) \lor ({\begin{cases} 2 x - 3 y = 0 \\ x + y = 5 \end{cases})$ $3 x - 2 (5 - x) = 0 \lor 2 x - 3 (5 - x) = 0$ $x = 2, y = 3 \lor x = 3, y = 2$
Commented by bemath last updated on 19/Jul/20

$c o o l l$
Commented by bramlex last updated on 19/Jul/20

$t h a n k y o u a l l$
Commented by Dwaipayan Shikari last updated on 19/Jul/20
$Another way (x/y)+1=(5/y)⇒(x/y)=(5/y)−1 and (y/x)=((5/y)−1)^(−1) (x/y)+(y/x)=((13)/6) ((5−y)/y)+(y/(5−y))=((13)/6)⇒((25+2y^2 −10y)/(5y−y^2 ))=((13)/6) ((25+2y^2 −10y)/(10y−2y^2 ))+1=((13)/(12))+1⇒((25)/(10y−2y^2 ))=((25)/(12))⇒2y^2 −10y+12=0→(a) So y=3,2 (from(a)) (x/3)+1=(5/3) or (x/2)+1=(5/2) x=2 or 3 { ((x=2,3)),((y=3,2)) :}$
$Another way$ $\frac{x}{y} + 1 = \frac{5}{y} \Rightarrow \frac{x}{y} = \frac{5}{y} - 1 and \frac{y}{x} = {(\frac{5}{y} - 1)}^{- 1}$ $\frac{x}{y} + \frac{y}{x} = \frac{13}{6}$ $\frac{5 - y}{y} + \frac{y}{5 - y} = \frac{13}{6} \Rightarrow \frac{25 + {2 y}^{2} - 10 y}{5 y - y^{2}} = \frac{13}{6}$ $\frac{25 + {2 y}^{2} - 10 y}{10 y - {2 y}^{2}} + 1 = \frac{13}{12} + 1 \Rightarrow \frac{25}{10 y - {2 y}^{2}} = \frac{25}{12} \Rightarrow {2 y}^{2} - 10 y + 12 = 0 \to (a)$ $So y = 3, 2 (from (a))$ $\frac{x}{3} + 1 = \frac{5}{3} or \frac{x}{2} + 1 = \frac{5}{2}$ $x = 2 or 3$ ${\begin{cases} x = 2, 3 \\ y = 3, 2 \end{cases}$
	Answered by Dwaipayan Shikari last updated on 19/Jul/20
	$((x^2 +y^2 )/(2xy))=((13)/(12)) ((x^2 +y^2 +2xy)/(x^2 +y^2 −2xy))=((25)/1) (((x+y)/(x−y)))^2 =((25)/1) ((25)/((x−y)^2 ))=((25)/1) x−y=1 or y−x=1 x+y=5 { ((x=3,2 )),((y=2,3)) :}$
	$\frac{x^{2} + y^{2}}{2 x y} = \frac{13}{12}$ $\frac{x^{2} + y^{2} + 2 x y}{x^{2} + y^{2} - 2 x y} = \frac{25}{1}$ ${(\frac{x + y}{x - y})}^{2} = \frac{25}{1}$ $\frac{25}{{(x - y)}^{2}} = \frac{25}{1}$ $x - y = 1 o r y - x = 1$ $x + y = 5$ ${\begin{cases} x = 3, 2 \\ y = 2, 3 \end{cases}$
	Commented by Rasheed.Sindhi last updated on 19/Jul/20

	$E v e r y b o d y i s n o t a s i n t e l l i g e n t$ $a s y o u a r e! :)$ $S o p l i n s e r t s o m e s t e p s b e t w e e n$ $1 s t & 2 n d l i n e s .$
	Commented by Dwaipayan Shikari last updated on 19/Jul/20

	$S o r r y s i r f o r y o u r i n c o n v e n i e n c e$
	Commented by Rasheed.Sindhi last updated on 19/Jul/20

	$\frac{x^{2} + y^{2} + 2 x y}{x^{2} + y^{2} - 2 x y} = \frac{13 + 12}{13 - 12}$ $T h i s w a s n e c e s s a r y s t e p t o$ $i n s e r t . W i t h o u t t h i s t h e a n s w e r$ $s e e m e d a r i d d l e!$ $P l {d o n}^{'} t m i n d!$
	Commented by Rasheed.Sindhi last updated on 19/Jul/20

	$N i c e S o l u t i o n!$
	Commented by 1549442205PVT last updated on 19/Jul/20

	$applying the property : \frac{a}{m} = \frac{b}{n} = \frac{a + b}{m + n} = \frac{a - b}{m - n}$ $\frac{x^{2} + y^{2}}{13} = \frac{2 xy}{12} = \frac{x^{2} + y^{2} + 2 xy}{25} = \frac{x^{2} + y^{2} - 2 xy}{1}$ $\Rightarrow \frac{x^{2} + 2 xy + y^{2}}{x^{2} - 2 xy + y^{2}} = \frac{25}{1}$
	Answered by bobhans last updated on 19/Jul/20
	$set x+y = u , xy = v ((x^2 +y^2 )/(xy)) = ((13)/6) ⇒(((x+y)^2 −2xy)/(xy)) = ((13)/6) 25−2v = ((13v)/6) ⇒150 = 25v { ((v = 6→y = (6/x))),((u=5→ x+(6/x) = 5 ; x^2 −5x+6=0)) :} { ((x= 3→y= 2)),((x = 2→y=3)) :} solution set {(3,2);(2,3)}$
	$s e t x + y = u, x y = v$ $\frac{x^{2} + y^{2}}{x y} = \frac{13}{6} \Rightarrow \frac{{(x + y)}^{2} - 2 x y}{x y} = \frac{13}{6}$ $25 - 2 v = \frac{13 v}{6} \Rightarrow 150 = 25 v$ ${\begin{cases} v = 6 \to y = \frac{6}{x} \\ u = 5 \to x + \frac{6}{x} = 5; x^{2} - 5 x + 6 = 0 \end{cases}$ ${\begin{cases} x = 3 \to y = 2 \\ x = 2 \to y = 3 \end{cases} s o l u t i o n s e t {(3, 2); (2, 3)}$
	Answered by Rasheed.Sindhi last updated on 19/Jul/20
	$Using formula: (x+y)^2 −(x−y)^2 =4xy _(−) { (((x/y) + (y/x) = ((13)/6).....(i))),((x+y = 5..............(ii))) :} (i)⇒(x/y) +1+ (y/x)+1 = ((13)/6)+2 ((x+y)/y) + ((x+y)/x) =((25)/6) (5/y) + (5/x) =((25)/6) (1/y) + (1/x) =(5/6)................(iii) (ii)⇒(x/(xy))+(y/(xy))=(5/(xy)) (1/y) + (1/x)=5((1/x))((1/y)).......(iv) (iii) & (iv): ((1/x))((1/y))=(1/6)⇒xy=6 x+y=5 ∧ xy=6 (x+y)^2 −(x−y)^2 =4xy (5)^2 −(x−y)^2 =4(6) x−y=±1 { ((x+y=5)),((x−y=1 )) :} ∨ { ((x+y=5)),((x−y=−1)) :} x=3,y=2 ∨ x=2,y=3$
	$Using formula :$ $\underline{{(x + y)}^{2} - {(x - y)}^{2} = 4 x y}$ ${\begin{cases} \frac{x}{y} + \frac{y}{x} = \frac{13}{6} . . . . . (i) \\ x + y = 5 . . . . . . . . . . . . . . (i i) \end{cases}$ $(i) \Rightarrow \frac{x}{y} + 1 + \frac{y}{x} + 1 = \frac{13}{6} + 2$ $\frac{x + y}{y} + \frac{x + y}{x} = \frac{25}{6}$ $\frac{5}{y} + \frac{5}{x} = \frac{25}{6}$ $\frac{1}{y} + \frac{1}{x} = \frac{5}{6} . . . . . . . . . . . . . . . . (i i i)$ $(i i) \Rightarrow \frac{x}{x y} + \frac{y}{x y} = \frac{5}{x y}$ $\frac{1}{y} + \frac{1}{x} = 5 (\frac{1}{x}) (\frac{1}{y}) . . . . . . . (i v)$ $(i i i) & (i v) :$ $(\frac{1}{x}) (\frac{1}{y}) = \frac{1}{6} \Rightarrow x y = 6$ $x + y = 5 \land x y = 6$ ${(x + y)}^{2} - {(x - y)}^{2} = 4 x y$ ${(5)}^{2} - {(x - y)}^{2} = 4 (6)$ $x - y = \pm 1$ ${\begin{cases} x + y = 5 \\ x - y = 1 \end{cases} \lor {\begin{cases} x + y = 5 \\ x - y = - 1 \end{cases}$ $x = 3, y = 2 \lor x = 2, y = 3$
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