halp!

We break the inequality down into two inequalities: \(\frac{1}{3}t-5 . We solve for  \(\frac{1}{3}t-5  first.

$\frac{1}{3}t-5 -\frac{9}{2}$

Then we solve for $t-2\le-3t+7$.

$t-2\le-3t+7\\ = 4t\le9\\ = t\le\frac{9}{4}$

Then t must be \(-\frac{9}{2} . In interval notation this is  $\boxed{(-\frac{9}{2}, \frac{9}{4}]}$.