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?
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$- \times - = +$. – Aryabhata Dec 07 '10 at 20:07
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@Moron:if suppose $(7y-1)(y-7) \ge 0 $ ..does then $y \ge 7 \text { and } y \ge \frac{1}{7}$ would be correct? – Quixotic Dec 07 '10 at 20:13
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@Tretwick Marian: It depends on $y$ ;) – AD - Stop Putin - Dec 07 '10 at 20:21
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I am not getting ... can any body post the rules? Or may be some links might be helpful. – Quixotic Dec 07 '10 at 20:22
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@Tretwick Marian: Do you see how to attack the problem now? – AD - Stop Putin - Dec 07 '10 at 20:26
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2You see, if $a\cdot b\le0$ then either $a\le0$ and $b\ge0$, or $a\ge0$ and $b\le0$. – AD - Stop Putin - Dec 07 '10 at 20:28
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Consider the interval $I=[y_1,y_2]$ between the roots $y_1,y_2$ of $(7y-1)(y-7)$ and $J=\mathbb{R}-I$. The sign of $(7y-1)(y-7)$ in $I$ and $J$ are opposite each other. – Américo Tavares Dec 07 '10 at 23:16
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@Tretwick Marian: The planetmath.org has this entry on Quadratic Inequality http://planetmath.org/encyclopedia/QuadraticInequality.html – Américo Tavares Dec 07 '10 at 23:45
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This does not answer your question “Where exactly am I wrong?,” but you can easily convince yourself that “y≤7 and y≤1/7” is not the correct answer by checking that y=0 satisfies both y≤7 and y≤1/7 but it does not satisfy (7y−1)(y−7)≤0. – Tsuyoshi Ito Dec 08 '10 at 13:18
4 Answers
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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?
Timothy Wagner
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Now I am confused with your nonpositive and nonnegative 's.Can I read them as negative and positive respectively? – Quixotic Dec 07 '10 at 20:11
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@Tretwick: Nonpositive means either negative or zero. Your inequality is not strict. – Timothy Wagner Dec 07 '10 at 20:22
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@Tretwick Marian: Almost, non-negative means it is $0$ or positive, and non-positive means it is $0$ or negative - that is it might be zero which is neither positive or negative. – AD - Stop Putin - Dec 07 '10 at 20:24
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@Timothy Wagner: Never mind, what I meant was that when I posted my clarification of non-negative etc. I did not see your comment... :) – AD - Stop Putin - Dec 07 '10 at 21:44
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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.
Bryan Yocks
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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 $\frac{1}{7}\le y\le 7$.
Isaac
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