If $-8<-\frac{3}{5}m+1<-\frac{16}{5}$, what is one possible value of $m$ ?

First, work on the left side of the equation: $-8<-\frac{3}{5}m+1$.
$5(-8)<5\left(-\frac{3}{5}m+1\right)$
$-40<-3m+5$
$-40-5<-3m+5-5$
$-45<-3m$
$\frac{-45}{-3}>\frac{-3m}{-3}$
$15>m$
Then, work on the right side of the equation: $-\frac{3}{5}m+1\le -\frac{16}{5}$.
$-\frac{3}{5}m+1-1\le -\frac{16}{5}-1$
$-\frac{3}{5}m\le -\frac{21}{5}$
$5\left(-\frac{3}{5}m\right)\le 5\left(-\frac{21}{5}\right)$
$-3m\le -21$
$\frac{-3m}{-3}\ge \frac{-21}{-3}$
$m\ge 7$
Once you have both pieces of the inequality simplified, you just need to put them back together.
If $15>m$ and $m\ge 7$, then $15>m\ge 7$, but this inequality doesn't make logical sense. Generally, inequalities are written with the smaller number on the left and the larger number on the right, so when you solve an inequality like this, you may need to rearrange the equation. This isn't difficult; just make sure the "arrows" are still pointing at the same numbers when you change the order.

So, a correct answer to this question would be any number between 7 and 15 , which includes 7 , but does NOT include 15.