6.134. For what integer value of $b$ do the equations $2 x^{2}+(3 b-1) x-3=0$ and $6 x^{2}-(2 b-3) x-1=0$ have a common root?
Solution.

Let $x_{1}$ be the common root. Then

$$
\begin{aligned}
& \left\{\begin{array} { l } 
{ 2 x _ { 1 } ^ { 2 } + ( 3 b - 1 ) x _ { 1 } - 3 = 0 , } \\
{ 6 x _ { 1 } ^ { 2 } - ( 2 b - 3 ) x _ { 1 } - 1 = 0 }
\end{array} \Leftrightarrow \left\{\begin{array}{l}
6 x^{2}+(9 b-3) x-9=0, \\
6 x^{2}-(2 b-3) x_{1}-1=0
\end{array}\right.\right. \\
& \Leftrightarrow(9 b-3) x+(2 b-3) x-9+1=0, \quad x=\frac{8}{11 b-6} .
\end{aligned}
$$

From the first equation we have

$2\left(\frac{8}{11 b-6}\right)^{2}+(3 b-1)\left(\frac{8}{11 b-6}\right)-3=0,99 b^{2}-164 b-68=0$,

$b_{1}=-\frac{34}{99}, b_{2}=2$

$b_{1}=-\frac{34}{99}$ is not an integer value.

Answer: $b=2$.