# System

• Apr 1st 2009, 04:18 PM
linda2005
System
Hello, you can help me please ?

Prove that $(x,y) = (2,1)$ is the only solution of the system : $\quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 - y^3) = 7 \\
y(x + y)^2 = 9 \end{array} \right.$
• Apr 2nd 2009, 07:40 AM
NonCommAlg
Quote:

Originally Posted by linda2005

Hello, you can help me please ?

Prove that $(x,y) = (2,1)$ is the only solution of the system : $\quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 - y^3) = 7 \\
y(x + y)^2 = 9 \end{array} \right.$

clearly from the equations we have $x > y> 0.$ let $y=t^2, \ t > 0.$ so from the second equation we get: $x=\frac{3}{t} - t^2$ which with the first equation gives us: $2t^9 - 9t^6 + 27t^3 + 7t - 27=0. \ \ \ (*)$

obviously $t=1$ satisfies $(*).$ we'll show that that's the only real solution of $(*).$ to see this, let $f(z)=2z^9 - 9z^6 + 27z^3 + 7z - 27, \ \ z \in \mathbb{R}.$ then:

$f'(z)=18z^8 - 54z^5 + 81z^2 + 7=9z^2((z^3 - 3)^2 + z^6) + 7 \geq 7.$ thus $f'(z) \neq 0, \ \forall z \in \mathbb{R},$ and thereofore it is not possible for $f$ to have more than one real zero by Rolle's theorem. Q.E.D.
• Apr 2nd 2009, 09:21 AM
DeMath
Quote:

Originally Posted by linda2005
Hello, you can help me please ?

Prove that $(x,y) = (2,1)$ is the only solution of the system : $\quad \large (x,y) \in {\mathbb R}^2 ,\quad \left\{\begin{array}{l} y(x^3 - y^3) = 7 \\
y(x + y)^2 = 9 \end{array} \right.$

View the solution of your problems, for example, here (Problem 12, pages 29-30). You need a special program to view this file.

http://www.lix.polytechnique.fr/Labo...ardi/M00-06.ps

These two pages of evidence from the file.

• Apr 2nd 2009, 05:46 PM
NonCommAlg
Quote:

Originally Posted by DeMath
View the solution of your problems, for example, here (Problem 12, pages 29-30). You need a special program to view this file.

http://www.lix.polytechnique.fr/Labo...ardi/M00-06.ps

These two pages of evidence from the file.