TITLE: An elementary doubt on Inequalities
QUESTION [8 upvotes]: Given the following expression:
$$(7y-1)(y-7) \le 0$$
To me this inequality implies $y \le 7$ and $y \le \frac{1}{7}$ but the correct expression (from my module) happens to be $\frac{1}{7} \le y \le 7$ 
Where exactly I am wrong?

REPLY [2 votes]: The Zero-Product Property ($ab=0\implies a=0\text{ or }b=0$) is the mechanics behind going from $(7y-1)(y-7)=0$ to $y=\frac{1}{7}$ or $y=7$.  There is no directly analogous property for inequalities.
The method that I'd suggest for examining inequalities that compare an expression to 0 is the boundary algorithm—find the values that make the expression equal to 0, which are the boundary points of a set of intervals on the number line, then test each interval to see if it satisfies the original equation.
In your specific example, the boundary points are $y=\frac{1}{7}$ and $y=7$, so test some value of $y$ below $y=\frac{1}{7}$ (for example, $y=0$), some value of $y$ between $y=\frac{1}{7}$ and $y=7$ (for example, $y=\frac{1}{2}$), and some value of $y$ above $y=7$ (for example, $y=10$).  The original inequality is false below $y=\frac{1}{7}$ and above $y=7$ and true between $y=\frac{1}{7}$ and $y=7$.  Since the original inequality included = (it was ≤), the solution includes the boundary points that solved the corresponding equation, so the solution is $\frac{1}{7}\le y\le 7$.