When $-4(y+2)<36$ becomes $y+2>-9$, why does $<$ become $>$? I found this question in a mock exam. My answer was mostly correct, but the inequality sign was flipped in the answer. Why?

Solve for $y$ in $$\frac{-4(y+2)}{3}<12$$

Part of the solution is dividing the $-4$ on both sides i.e $-4(y+2)<36$ becomes $y+2>-9$. Why does the $<$ sign flip to a $>$ sign, especially if the RHS of the equation is becoming negative anyway?
 A: When you divide by a negative number, the inequality reverses. The term $-4$ is not "moved over" to the right side of the inequality. You divide both sides of the inequality by $-4$.
For example, take
$$
-4x < 0.
$$
What values are allowed for $x$? Try inserting $x=1$ and $x=-1$ and seeing what happens.
First try $x=-1$: $$-4\times-1=4>0,$$ so the inequality is not true. Now try $x=1$:
$$-4\times 1=-4<0,$$ so the inequality is true. Thus, $-4x<0$ is the same thing as $x>0$. We conclude that if you divide both sides by $-4$ you do not get $x<0$ but $x>0$.
A: Consider the strictly decreasing function $f(x)=-\dfrac1ax,$ where $a$ is a positive number.
Notice that when inputs are fed to this function in ascending order, it returns outputs that are in descending order.
So, applying this function to both sides of an inequality will flip the inequality sign.
Well, dividing an inequality by the negative number $-a$ is essentially applying the function $f$ to the inequality.
Hence, dividing an inequality by a negative number flips the inequality sign.
