Thus we set $k_{\alpha, \beta}(e) = b$, $k_{\alpha, \beta}(t) = y$ and derive $(\alpha, \beta)$:
$(\alpha e + \beta = b, \alpha t + \beta = y)$, this means $(4\alpha + \beta = 1, 19\alpha + \beta = 24)$, and hence
$15\alpha = 23$, $\alpha = 7 \times 23 = 161 = 5$, $\beta = 7$; thus the key is very likely to be $k = (5, 7)$
