We have the following:
[tex]\begin{gathered} -138\ge-6\cdot(6b-7) \\ \end{gathered}[/tex]
solving for b:
[tex]\begin{gathered} -6\cdot(6b-7)\le-138 \\ -\frac{6}{6}\cdot(6b-7)\le-\frac{138}{6} \\ -(6b-7)\le-23 \\ (6b-7)\ge23 \\ 6b-7+7\ge23+7 \\ 6b\ge30 \\ b\ge\frac{30}{6} \\ b\ge5 \end{gathered}[/tex]
Therefore, the interval is:
[tex]\lbrack5,\infty)[/tex]
