# Inequalities, when does the sign change here?

I have encountered a problem with inequalities. I have been looking at examples provided by two websites which 'solve' inequalities, however when I try using my own method, the extremely simple 'addition, subtraction, multiply, divide', I end up with the wrong inequality sign.

What I have done is as follows:

$$\frac{x-3}{x+3} < 4$$ $$x-3 < 4(x+3)$$ $$x-3 < 4x+12$$ $$x < 4x+15$$ $$-3x < 15$$ $$3x > -15$$ $$x > -5$$

I have guessed that that problem is somewhere in the first 2 lines, embarrassingly, however I am unsure what I am doing wrong; the 'calculators' are telling me the answer is $x < -5$ or $x > -3$.

What am I doing wrong with my solution, to get the $x > -5$ the wrong way around? What is this other path I can take to find the $x > -3$?

It'd be preferred if I did not have to start using the 'table' to test all solutions, but if that is the only way, it's fine.

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You need to deal with two cases: $x+3>0$ and $x+3<0$. Remember that when you multiply both sides of an inequality by a negative number the inequality sign must be "flipped." – vociferous_rutabaga Jul 21 '14 at 22:53

Your problem happens when you multiply both sides by $(x+3)$, which can be a negative number, in which case the inequalitites would be reversed.

You need to deal with two cases: x+3>0 and x+3<0.

You use $x+3>0$ (or $x>-3$) to get $x>-5$, combining both give $x>-3$, which is part of the right answer.

Now use $x+3<0$ (or $x<-3$), change the direction of inequality while multiplying on both sides, and get the other direction, $x<-5$. Combining $x<-3$ and $x<-5$ gives just $x<-5$.

Best of luck.

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I considered that, but since x is unknown, how should I know whether to reverse the sign or not? – Destrotant Jul 21 '14 at 22:53
You mean $x^2 - 9 < 4(x+3)^2$? – Mathmo123 Jul 21 '14 at 22:55
To anonymous voter: Why the negative vote? :O – Juanito Jul 21 '14 at 22:55
You can take 2 cases, once $(x+3)$ being positive, once negative. – Juanito Jul 21 '14 at 22:57
@Destrotant to expound on Juanito's comment a bit, it's not about "whether to reverse the sign or not". It's "reverse the sign AND not reverse the sign". $x+3$ can be both negative and positive, depending on x. So you solve two equations - one where you treat $x+3$ as if it were negative and one where you treat $x+3$ as if it were positive. If the equation means $x+3$ can't be one of those, you'll encounter an issue later one. It's a bit like absolute value. $|x| = 3$ means $x=3$ or $x=-3$. You have to take both into account. – Duncan Jul 21 '14 at 23:09

Your error is at the second line. If $(x+3)$ is negative, then you would need to switch the inequality there. So you have to consider two cases: first when $(x+3)$ is positive (i.e. $x \geq -3$) and when $(x+3)$ is negative (i.e. $x \leq -3$). The rest of your calculations were good

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When you multiply both sides by $(x+3)$ if $x < -3$, this value will be negative, and you will reverse the inequality.

The correct method is to multiply by $(x+3)^2$ which is always positive, and then solve the resulting $$(x-3)(x+3) <4(x+3)^2$$as a quadratic inequality.

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When you want to solve such inequalities, the standard way is to check how they compare with zero. In example, you would have $$\frac{x-3}{x+3}-4<0$$ and thus $$\frac{-3x-15}{x+3}<0$$ You will have to use that table as you said, but it is the safest way to avoid mistakes.

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