How do you solve an inequality that has a fraction 1/x+3< 0

[Guest]

Hi there,
i was hoping someone would be able to help me with an inequality that i have come across.
1/x+3< 0
Any help would be appreciated as I cannot find out how to solve this anywhere on the internet.

thank you.

[Guest]

Is this the equation?

[tex]\frac1x + 3< 0[/tex]

[Guest]

yes

[Guest]

[tex]\frac{1 + 3x}{x} < 0[/tex]

First we solve [tex]1+3x = 0[/tex]
[tex]3x = -1[/tex]
[tex]x = \frac{-1}{3}[/tex]

The intervals are [tex](-\infty; -\frac{1}{3}), (-\frac{1}{3}, 0)[/tex] and [tex](0, +\infty)[/tex]

Now we choose a random number from the last interval let it be for example 1
the equation
becomes [tex]\frac{1}{1}+3 > 0[/tex]

So numbers in the intervals [tex](0, +\infty)[/tex] and [tex](-\infty; -\frac{1}{3})[/tex] are positive and
solutions are numbers within the interval

[tex](-\frac{1}{3}, 0)[/tex]

[Guest]

Is it clear?

[HallsofIvy]

Another way: The inequality is $\frac{1}{x}+ 3< 0$. Subtracting 3 from both sides, $\frac{1}{x}< -3$.

Now, an important way solving inequalities differs from solving equalities: Multiplying both sides of an inequality by a negative number reverses the direction of the inequality. Multiplying both sides of an inequality by a positive number does not change the direction and, of course, x cannot be 0 here.
So we consider two possibilities. If x> 0, then -3x< 0 so multiplying both sides of $\frac{1}{x}< -3$ by -3x gives $-3> x$ which is the same as $x< -3$. But then x is not positive so there is no solution here. If x< 0 then -3x> 0 so multiplying both sides of $\frac{1}{x}< -3$ bu -3x gives $-3< x$. Since we must have x< 0 for this solution so we must have $-3< x< 0$.

The solution to the inequality is "all x such that -3< x< 0".