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- Thread starter jacks
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- Feb 1, 2012

- 57

I agree with pickslides . . .

[tex]\text{Solve the following system of equations in real numbers:}[/tex]

. . [tex]\begin{array}{ccc}\sqrt{x^2-2x+6}\cdot \log_{3}(6-y) &=&x \\

\sqrt{y^2-2y+6}\cdot \log_{3}(6-z) &=& y \\

\sqrt{z^2-2z+6}\cdot\log_{3}(6-x)&=&z\end{array}[/tex]

Due to the symmetry, I assume that [tex]x = y = z.[/tex]

Then we have: .[tex]\sqrt{x^2-2x+6}\cdot \log_3(6-x) \:=\:x[/tex]

By inspection, we see that: .[tex]x\,=\,3.[/tex]

- Jan 26, 2012

- 890

Hello, jacks!

I agree with pickslides . . .

Due to the symmetry, I assume that [tex]x = y = z.[/tex]

Then we have: .[tex]\sqrt{x^2-2x+6}\cdot \log_3(6-x) \:=\:x[/tex]

By inspection, we see that: .[tex]x\,=\,3.[/tex]

Symmetry only guarantees that any permutation of the values of x, y, z for a solution is also a solution.

Obviously x=y=z=3 is a solution, but symmetry alone does not force us to conclude that it is the only solution.

CB

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