Let us substitute the second linear equation into the first to produce: $x^3 + (kx + d)^3 = 2$; or $x^3 + k^3x^3 + 3dk^2x^2 + 3d^2kx + d^3 = 2$; or $(k^3 + 1)x^3 + (3dk^2)x^2 + (3d^2k)x + (d^3 - 2) = 0$. Since a cubic polynomial is always guaranteed a minimum of one real root, we require $k^3 + 1 = 0 \Rightarrow k = -1$. This now reduces our cubic into the quadratic $3dx^2 - 3d^2x + d^3 - 2 = 0$. The roots will be non-real if and only if the discriminant is negative, or: $9d^4 - 4(3d)(d^3 - 2) < 0$; or $-3d^4 + 24d < 0$; or $-3d(d^3 - 8) < 0$; or $d \in (-\infty, 0) \cup (2, \infty)$. Thus, the system has no real solutions $(x, y)$ for $(k, d) = (-1, t)$, where $t \in (-\infty, 0) \cup (2, \infty)$.