Solving an Inequality: -9 < 1/x

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To solve the inequality -9 < 1/x, it is essential to consider the sign of x, as multiplying both sides by a negative value reverses the inequality. The correct approach involves splitting the problem into two cases: x > 0 and x < 0. For x < 0, the inequality becomes -9x < 1, leading to x > -1/9. This confirms that the solution is x < -1/9 when considering the conditions for x. Understanding these rules is crucial for correctly solving inequalities.
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Homework Statement



Solve the inequality -9 < 1/x

A simple inequality, I can see the solution is just x < -1/9 but I can't prove it at all.

The Attempt at a Solution



-9 < 1/x

-9x < 1

x > -1/9

Any helpful rules I am forgetting about inequalities? This was a problem in a review from high school set provided by my instructor for my introductory math class. Just curious about a solution, it's a calculus course that doesn't really test on this sort of thing but I figure I should actually try and figure out these basic things.
 
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brownman said:

Homework Statement



Solve the inequality -9 < 1/x

A simple inequality, I can see the solution is just x < -1/9 but I can't prove it at all.

The Attempt at a Solution



-9 < 1/x

-9x < 1

x > -1/9

Any helpful rules I am forgetting about inequalities? This was a problem in a review from high school set provided by my instructor for my introductory math class. Just curious about a solution, I have no

You are forgetting if you multiply both sides by x and x is negative you have to reverse the inequality. Split into two cases x>0 and x<0.
 


Oh okay, that makes sense, thanks for the help :).
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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