Let α, β ∈ ℝ be such that the system of linear equations x + 2y + z...

The given system of linear equations is:
$x + 2y + z = 5$
$2x + y + \alpha z = 5$
$8x + 4y + \beta z = 18$

For the system to have no solution, the determinant of the coefficient matrix $D$ must be zero.

$D = \begin{vmatrix} 1 & 2 & 1 \\ 2 & 1 & \alpha \\ 8 & 4 & \beta \end{vmatrix} = 0$

Expanding along the first row:

$1(\beta - 4\alpha) - 2(2\beta - 8\alpha) + 1(8 - 8) = 0$

$\beta - 4\alpha - 4\beta + 16\alpha = 0$

$12\alpha - 3\beta = 0$

$\beta = 4\alpha$

To verify the no solution condition, substitute $\beta = 4\alpha$ into the third equation:

$8x + 4y + 4\alpha z = 18$

$4(2x + y + \alpha z) = 18$

$2x + y + \alpha z = \dfrac{9}{2}$

However, the second equation is $2x + y + \alpha z = 5$. This results in a contradiction since $\dfrac{9}{2} \neq 5$. Thus, the system has no solution when $\beta = 4\alpha$.

Therefore, $\dfrac{\beta}{\alpha} = 4$.

Answer: $4$