I am trying to solve the following inequality
$$\frac{\lvert x+2\rvert}{\lvert x-1\rvert} \gt 1$$
I include a picture of two of my wrong solutions….i don’t know what i did wrong and more importantly WHY i am not allowed to do it…..
Some of this inequalities i do get right, some i get absolutely wrong, and i don’t get where i make the mistake what am i doing wrong.
With the second try i know that i should not do it but instead Substrat 1 first, but i don’t get why, why is it wrong if i multiply…. if i change the greater or smaller sign
I hope you can give me some tips, íve spent the last two days trying to solve inequalities as this one, and its hit or miss…
Many thanks
Best Answer
The first solution is correct: the inequality is solved if and only if $x\in (-1/2,1)\cup (1,+\infty)$. In the second solution when $x<1$ you should not change the sign!
A faster way. The inequality is equivalent to $$|x+2|>|x-1|$$ with $x\not=1$. Now we consider three cases:
i) if $x<-2$ then we solve $-x-2>-x+1$ getting no solutions;
ii) if $-2\leq x<1$ then we solve $x+2>-x+1$ getting $x\in (-1/2,1)$;
iii) if $x>1$ then we solve $x+2>x-1$ getting $x\in (1,+\infty)$.
P.S. The point $x\in\mathbb{R}\setminus \{1\}$ is a solution iff its distance from $-2$, i.e. $|x+2|$, is greater than its distance from $1$, i.e. $|x-1|$. Note that the midpoint of the segment $[-2,1]$ is just $(-2+1)/2=-1/2$.