Solve the following equations: $xy + x + y = 23$, $xz + x + z = 41$, $yz + y + z = 27$.
- $x=4, -2; y=2; 6; z=6, -5$
- $x=2, -4; y=2,4 ;z=2, -6$
- $x = 5, -7; y = 3, -5; z = 6, -8$
- $x=3, 4; y=2, -5; z=2,-7$
Given xy+x+y=23, xz+x+z=41, yz+y+z=27. Adding 1 to each side: (x+1)(y+1)=24, (x+1)(z+1)=42, (y+1)(z+1)=28. Let X=x+1, Y=y+1, Z=z+1. Then XY=24, XZ=42, YZ=28. Multiplying gives (XYZ)^2 = 24*42*28 = 28224, so XYZ = 168. Then Z = 168/24 = 7, Y = 168/42 = 4, X = 168/28 = 6. Thus x=5, y=3, z=6. The negative roots follow from the quadratic nature.
Rewrite the equations by adding 1 to both sides of each to use the factorization identity xy + x + y + 1 = (x + 1)(y + 1). This gives (x + 1)(y + 1) = 24, (x + 1)(z + 1) = 42, and (y + 1)(z + 1) = 28. Multiplying all three equations gives ((x + 1)(y + 1)(z + 1))^2 = 24 * 42 * 28 = 28224, so (x + 1)(y + 1)(z + 1) = 168. Dividing this product by each pairwise equation yields z + 1 = 7, y + 1 = 4, and x + 1 = 6, meaning x = 5, y = 3, and z = 6. The corresponding symmetric negative roots from the factor pairs are x = -7, y = -5, and z = -8, giving the result x = 5, -7; y = 3, -5; z = 6, -8.