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- Feb 14, 2012

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Solve the system $x^5+y^5=33,\,x+y=3$.

- Thread starter anemone
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- Thread starter
- Admin
- #1

- Feb 14, 2012

- 3,812

Solve the system $x^5+y^5=33,\,x+y=3$.

=33+5xy((x+y)^3-3xy(x+y))+30x^2y^2= 33+5xy(27-9xy)+30x^2y^2=243$

or

$15x^2y^2-135xy+210=0$

or

$x^2y^2-9xy+14=0$

And xy=7 or xy=2 impling the following 2 systems of equations :

x+y=3. (A)

xy=2

x+y=3 (B)

xy=7

And (A) gives (x=1,y=2),(x=2,y=1) (B) has no real solutions

Last edited:

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=33+5xy((x+y)^3-3xy(x+y))+30x^2y^2= 33+5xy(27-9xy)+30x^2y^2=243$

or

$15x^2y^2-135xy+210=0$

or

$x^2y^2-9xy+14=0$

And xy=7 or xy=2 impling the following 2 systems of equations :

x+y=3. (A)

xy=2

x+y=3 (B)

xy=7

And (A) gives (x=1,y=2),(x=2,y=1) (B) has no real solutions

[x=(3+i$\sqrt 19$)/2, y=(3-i$\sqrt 19$)/2]............[x=(3-i$\sqrt 19$)/2 , y=( 3+i$\sqrt 19$)/2]