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30
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Solve

xy + x + y = 23

yz + y + z = 31

zx + z + x = 47

in real numbers.

 
 Jan 9, 2022
 #1
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There's bound to be something easier, but this works.

From the first equation, y = (23 - x)/(1 + x),

and from the third, z = (47 - x)/(1 + x).

Substitute into the middle eqation and tidy up.

That gets you the quadratic x^2 + 2x -35 = 0, from which x = 5 or -7.

Taking x = 5 and back substituting yields y = 3 and z = 7.

Either of the other two pairings from the original set of equations could be used instead, and a second set of solutions can be obtained from x = -7,  (-7, -5, -9).

 
 Jan 9, 2022

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