Summer Problem Solving Marathon Solution #41

Adding all three equations, we obtain:

(k+2)x + (k+2)y + (k+2)z = 3k

or:

(k+2)(x+y+z) = 3k

When k = -2, this becomes 0 = -6, which is not true for any values of x, y, and z. So there is no solution to the system of equations when k = -2.

For any other value of k, we have x + y + z = 3k/(k+2). So by setting x = y = z = k/(k+2), we obtain a solution.'

Thus k = -2 is the only value for which there is no solution in x, y, and z to the system of equations.