Solve System of Equations X+YZ=6 Y+XZ=6 Z+XY=6

Research LInks

Find the solution for the system of equations:

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)

(X+Y)=6

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+3)(X-2)=0 giving the solutions

X = 2,-3

Categories: Math