Question

if xy + x + y = 19 find the values of x and y​

Answer 1

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Answer 2

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.