Prove if $x > 100$ then $\frac{100}{3-2x} > -1$ If $x &gt; 100$ then $\frac{100}{3-2x} &gt; -1$

Pf: 


Assume $x&gt;100$


$-2x &gt; -20$
$-2x + 3 &gt; -200$
$\frac{1}{3-2x} &gt; \frac{1}{-200}$



This is what I have done so far, however I'm not sure how to proceed from here on out. If anyone has any tips, or possibly and example to further push me in the right direction I'd appreciate it greatly!
Be careful! From $2&gt;1$ you cannot deduce that $-200&gt;-100$.

From $x&gt;100$ you can deduce that $2x&gt;200$ and so that
$$
2x-3&gt;200-3=197
$$
It follows that
$$
\frac{1}{2x-3}&lt;\frac{1}{197}
$$
and so
$$
\frac{100}{2x-3}&lt;\frac{100}{197}&lt;1
$$
Changing sign reverses inequalities, so
$$
-\frac{100}{2x-3}&gt;-1
$$
By plugging in $x = 100$ in the denominator, we have

$$ \frac{100}{3-2x} &gt; -\frac{100}{197} &gt; -1$$