Question Number 100954 by bobhans last updated on 29/Jun/20

 { (((1/(2x−y)) + (√y) = 1)),((((√y)/(2x−y)) = −6)) :}

Commented byDwaipayan Shikari last updated on 29/Jun/20

         ((−6)/(√y))=(1/(2x−y))      →(1)            ((−6)/(√y))+(√y)=1    ⇒t^2 −t−6=0⇒t=  3  or  −2     {suppose (√y)=t}  ((−6)/3)=(1/(2x−9 ))⇒−4x+18=1   ⇒x=((17)/4)     { ((x=((17)/4))),((y=9)) :}

Answered by ajfour last updated on 29/Jun/20

s=(1/(2x−y))  ,  t=(√y)  s+t=1  ,   st=−6    ⇒  t>0 ,  s<0  s, t = (1/2)±(√((1/4)+6))   t=(1/2)+(5/2) = 3   ,  s= −2  ⇒  y= t^2  = 9 ,   s=(1/(2x−y))= (1/(2x−9)) = −2  ⇒  2x=9−(1/2)  ⇒   x=((17)/4) .

Answered by john santu last updated on 29/Jun/20

⇒(1/(2x−y )) = −(6/(√y)) (1)  ⇒−(6/((√y) )) +(√y) = 1 ; ((√y))^2 −(√y) −6 =0  ((√y) −3)((√y) +2 ) = 0 ; (√y) = 3  y= 9 ⇒(1/(2x−9)) = −2 ; 2x−9 = −(1/2)  2x = ((17)/2) , x = ((17)/4)