Find the value of x
√x-y=3√x+y = √13
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y+(x)^1/2 =13 …………….(1)
or (x)^1/2 =(13-y)
or x=(13-y)^2 , on putting x=(13-y)^2 in eq.(2)
(13-y)^2 +(y)^1/2 =19
169 +y^2 -26 y + y^1/2=19
y^2–26y+y^1/2 +150 =0
Let y=9 , R =81–234+3+150 =234–234=0
(y-9) is a factor.
y^2–26y+y^1/2+150 = 0
or y(y-9)-17y+y^1/2+150=0
or y(y-9)-17(y-9)-3+y^1/2 =0
or y(y^1/2–3)(y^1/2+3)-17(y^1/2–3)(y^1/2+3)+(y^1/2–3)=0
or (y^1/2 -3)[y(y^1/2+3)-17(y^1/2+3) +1] =0
Either (y^1/2 -3 )=0 => y^1/2= 3
or y=9 , but x= (13-y)^2
x=(13–9)^2 =(4)^2 =16.
x =16 , y = 9 , Answer.
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