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

In summary, the system of equations $x^5+y^5=33$ and $x+y=3$ has two real solutions: $(x=1,y=2)$ and $(x=2,y=1)$, and two complex solutions: $[x=(3+i\sqrt{19})/2, y=(3-i\sqrt{19})/2]$ and $[x=(3-i\sqrt{19})/2, y=(3+i\sqrt{19})/2]$.
  • #1
anemone
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Solve the system $x^5+y^5=33,\,x+y=3$.
 
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  • #2
[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:
  • #3
Thanks for participating, solakis! Ah, the question is meant to ask for complex solutions too! (Nod)
 
  • #4
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]
 

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

What is a system of 2 variables?

A system of 2 variables is a set of two equations with two unknown variables. The goal is to find the values of the variables that satisfy both equations simultaneously.

How do you solve a system of 2 variables?

To solve a system of 2 variables, you can use substitution, elimination, or graphing methods. In substitution, you solve one equation for one variable and substitute it into the other equation. In elimination, you manipulate the equations to eliminate one of the variables. In graphing, you plot both equations on a graph and find the intersection point.

What is the solution to the system of equations $x^5+y^5=33,\,x+y=3$?

The solution to this system of equations is (1,2). When you substitute x=1 and y=2 into both equations, they both equal 3.

Can a system of 2 variables have more than one solution?

Yes, a system of 2 variables can have infinitely many solutions. This occurs when the two equations represent the same line, meaning they have an infinite number of intersection points.

What if there is no solution to a system of 2 variables?

If the two equations are parallel and do not intersect, there is no solution to the system of 2 variables. This means that the equations are not consistent and there is no possible value for the variables that satisfy both equations simultaneously.

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