An elementary doubt on Inequalities 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?
 A: Hint: If the product of two numbers is nonpostive, one of them must be nonpositive and one of them must be nonnegative.
Edit: $(7y-1)(y-7)\leq 0$. So either, $7y-1 \leq 0$ and $y-7 \geq 0$ or $7y-1 \geq 0$ and $y-7 \leq 0$. Now, can you eliminate one of these two cases?
A: Another way is to analyze as a quadratic function $f(y)=7y^2-50y+7$.
Note that you want solve : $f(y)=7y^2-50y+7 \leq 0$, whose graph is: 
The intersections with the x-axis are : $x=\frac{1}{7}$ and $x=7$, this graph we say that $f(y) \leq 0$ in $[\frac{1}{7},7]$, this  wanted to see.
A: 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 $\frac{1}{7}\le y\le 7$.
A: HINT $\ $ If $\rm\ a < b\ $ then $\rm\  (x-a)\ (x-b) < 0\ \iff\ a < x < b $
