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

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

Commented byRasheed.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

Commented bybemath last updated on 19/Jul/20

cooll

Commented bybramlex last updated on 19/Jul/20

thank you all

Commented byDwaipayan 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)) :}

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)) :}

Commented byRasheed.Sindhi last updated on 19/Jul/20

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

Commented byDwaipayan Shikari last updated on 19/Jul/20

Sorry sir for your inconvenience

Commented byRasheed.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!

Commented byRasheed.Sindhi last updated on 19/Jul/20

Nice Solution!

Commented by1549442205PVT 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)

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)}

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