$\frac{5}{3x + 2} = 2x - 1$
$⇔ (2x - 1)(3x + 2) = 5$
$⇔ 6x^{2} + x - 2 - 5 = 0$
$⇔ 6x^{2} + x - 7 = 0$
$⇔6x^{2} - 6x + 7x - 7 = 0$
$⇔ 6x(x - 1) + 7(x - 1) = 0$
$⇔ \left[ \begin{array}{l}x - 1 = 0\\6x + 7 = 0\end{array} \right.$
$⇔ \left[ \begin{array}{l}x = 1\\x = -\frac{7}{6}\end{array} \right.$