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)