若集合 $\{(x, y) \mid x+y-2=0$ 且 $x-2 y+4=0\}\{(x, y) \mid y=3 x+b\}$, 则 $b=$
首先方程组 $\left\{\begin{array}{l}x+y-2=0, \\ x-2 y+4=0\end{array}\right.$ 的解为 $\left\{\begin{array}{l}x=0, \\ y=2,\end{array}\right.$

代入直线 $y=3 x+b$ 得 $b=2$.