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Solve: `x^2 +3xy + y^2=81`  and `xy -y^2=0`  x^2+3xy+y^2=81 xy-y^2=0

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alyse1993 | Student, Undergraduate | eNotes Newbie

Posted November 29, 2012 at 4:29 AM via web

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Solve: `x^2 +3xy + y^2=81`  and `xy -y^2=0`

 

x^2+3xy+y^2=81

xy-y^2=0

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durbanville | High School Teacher | (Level 1) Educator Emeritus

Posted November 29, 2012 at 5:30 AM (Answer #1)

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First find a value for x or y in terms of the other :

`xy-y^2=0` becomes

`y(x-y) =0`

`therefore y=0 or (x-y) =0`

`therefore y=x`

Substitute the 2 values for y into the other equation:

when y=0

`x^2 +3x(0) +(0)^2 = 81`

`therefore x^2 = 81`

`therefore x= 9 or -9` when y=0

Alternatively:

when y=x substitute the y values for x:

`x^2 +3x(x) + (x)^2=81`

`therefore x^2 + 3x^2 + x^2 = 81`

`therefore 5x^2=81`

`therefore x^2=81/5`

`therefore x= sqrt(81/5)`

`therefore x= 4.025 or -4.025`

and `therefore y= 4.025 or -4.025`

and x=9 or x=-9 when y = 0

Therefore:

(9;0): (-9;0) or (4.025;4.025): (-4.025;-4.025)

 

 

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