Thread: systems of equations

xy + xz = -1
xy + yz = -9
yz + xz = -4

solve for (x,y,z)

Originally Posted by stasis

xy + xz = -1
xy + yz = -9
yz + xz = -4

solve for (x,y,z)

I'd suggest substituting xy = a, xz = b and yz = c. Solve the three new linear equations simultaneously for a, b and c. NOw use these values to solve for x, y and z.

By the way ..... There are several good reasons NOT delete questions once they've been answered. (I refer to your edit at http://www.mathhelpforum.com/math-he...-question.html)

Originally Posted by stasis

xy + xz = -1 (1)
xy + yz = -9 (2)
yz + xz = -4 (3)

solve for (x,y,z)

Hello Stasis. This problem can easily be solved by doing some basic elimination and substitution.

My approach starts with (1) - (2) to give call this (4)

then (3) + (4) give
substitute into (1) and (3) to get and

so we have:




now (7)*(5) gives (6) gives therefore

putting in (5) gives
and into (7)

so your two solutions are

Bobak