The process of solving the inequality $\frac{8}{19} x\ge -1$ 
Why did he multiply both sides by 19/8 and not 8/19 ? 
Is this a rule when dealing with inequalities that to remove fractions, you have to multiply by the reciprocal ?
 A: Think of it more like a "trick" than a rule.
Everything we do in mathematics should serve a purpose and it just so happens that here, the purpose of what Sal has done was to "get rid" of the $\frac{8}{19}$ that multiplies the x. We say that $\frac{19}{8}$ "cancels" $\frac{8}{19}$.
$$\frac{19}{8} \times \frac{8}{19}x = \frac{19 \times 8}{8 \times 19}x = \frac{19 \times 8}{19 \times 8}x = 1 \times x = x $$
The "price" he pays is that, according to the rules of mathematics, if one alters one side of an equation, one is required to apply the same alteration to both sides, but that's easily fixed with some simplification.
$$\frac{19}{8} \times \frac{8}{19}x \ge \frac{19}{8} \times (-1)$$
$$\implies x \ge -\frac{19}{8}$$
A: If he had multiplied both sides by $8/19$, he would have obtained
\begin{align*}
&\left( \frac{8}{19} \right) \cdot \left( \frac{8}{19} x \right)  \ge \left( \frac{8}{19} \right) \cdot (-1) \\
\iff & \frac{64}{361} x \ge - \frac{8}{19},
\end{align*}
which is true but not helpful.  It doesn't bring us any closer to figuring out what $x$ is.
A: We want to solve for $x$.So:
$$\frac{8}{19}x \geq -1 \Rightarrow 19 \cdot \frac{8}{19}x \geq (-1) \cdot 19 \Rightarrow 8x \geq -19 \Rightarrow \frac{1}{8} \cdot 8x \geq \frac{1}{8} \cdot (-19) \Rightarrow x \geq -\frac{19}{8}$$
A: Just a little clarification: if the expression had been 2*x <= -1 you'd divide both sides by 2 and get x <= -1/2. Same thing here: divide by 8/19 to get the answer. Dividing by a fraction is the same as multiplying by the reciprocal of the fraction, so multiplying by 19/8 is the same as dividing by 8/19.
A: We need to solve for x. So, to get rid of the coefficient of x in LHS we need to multiply it by 19/8 - such a thing won't come out by multiplying with 8/19.
A: No, it's not a rule, but in general you would want to remove the fraction as coefficient of $x$ to be isolated.
In this case:
$\dfrac{8}{19}x\ge-1$
$\dfrac{8}{19}\cdot\dfrac{19}{8}x\ge-\dfrac{1}{1}\cdot\dfrac{19}{8}$
On the left hand side the fraction cancels and we are left with $x$.
$x\ge-2\dfrac{3}{8}$
