MHB Solve System of 2 Variables: $x^5+y^5=33,\,x+y=3$

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The system of equations $x^5+y^5=33$ and $x+y=3$ leads to two possible values for the product $xy$: either 2 or 7. The first case, where $xy=2$, yields real solutions $(x=1,y=2)$ and $(x=2,y=1)$. The second case, with $xy=7$, results in no real solutions. Additionally, complex solutions are identified as $x=(3+i\sqrt{19})/2$ and $y=(3-i\sqrt{19})/2$, along with their reverse. The discussion emphasizes both real and complex solutions to the original system.
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Solve the system $x^5+y^5=33,\,x+y=3$.
 
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[sp]$(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4=33+5xy(x^3+y^3)+10x^2y^2(x+y)=

=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[/sp]
 
Last edited:
Thanks for participating, solakis! Ah, the question is meant to ask for complex solutions too! (Nod)
 
solakis said:
[sp]$(x+y)^5=x^5+y^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4=33+5xy(x^3+y^3)+10x^2y^2(x+y)=

=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[/sp]
[sp]The complex solutions are:
[x=(3+i$\sqrt 19$)/2, y=(3-i$\sqrt 19$)/2]...[x=(3-i$\sqrt 19$)/2 , y=( 3+i$\sqrt 19$)/2][/sp]
 

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