Research LInks

Find the solution for the system of equations:

        X+YZ=6            Y+XZ=6             Z+XY=6  

Multiply each equation by the term missing to make each second term a triple

 X^2+XYZ=6X             Y^2+XYZ=6Y             Z^2+XYZ=6Z  

So now subtract the second equation from the first: 

 X^2 -Y^2=6(X-Y)    

(X+Y)(X -Y)=6(X-Y)    


Since the equations are symmetric we should have

(X+Y)=6        (X+Z)=6           (Y+Z)=6 

Subtracting the 3rd from the 1st:

X - Z =0   or 

 X = Z     and again since the equations are symmetric  X = Y = Z  


The very first equation at the top  X+YZ=6   becomes

 X^2+X - 6=0  

 X^2+X - 6=0  

 (X+3)(X-2)=0   giving the solutions

 X = 2,-3 

Categories: Math

Fudgy McFarlen

Woo hoo.

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