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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

enter image description here

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  • $\begingroup$ Start by rewriting the first equation as $|x+1| > |x-2|$ instead. This would be true for all numbers if we took the absolute values off but eventually some negative $x$ will falsify the inequality and any number smaller than that will as well. Can you find such an $x$? $\endgroup$ – CyclotomicField Feb 23 at 16:07
  • $\begingroup$ @CyclotomicField that would be the value in the middle of -2 and 1 so -1/2, but why x+1 and x-2? $\endgroup$ – Ang Feb 23 at 17:45
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    $\begingroup$ because I made a mistake. It should be $|x+2| > |x-1|$ $\endgroup$ – CyclotomicField Feb 23 at 19:29
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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 $(-2+1)/2=-1/2$.

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Because $|x-1|$ is always positive, you can multiply it to the top without having to flip the inequality, so:

$$|x+2|>|x-1|$$ The most important points are when one of the $|*|$ expressions equals $0$. So here, $|x+2|=0$ when $x=-2$

On account of this, are you happy that for $x<2; |x+2|=-x-2$, and $|x-1|=1-x$?

We then solve this inequality:

$$-x-2>1-x \implies 0>3$$ which is obviously false, so there are no solutions for $x<2$.

Apply the same process for the other side, remembering to find solutions for $-2<x<1$ and $x>1$ separately.

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For $$x>1$$ we get $$x+2>x-1$$ and we obtain $$3>0$$ which is true. For $$-2\le x<1$$ we get $$x+2>-x+1$$ so.. And for $$x<-2$$ we get $$-x-2>-x+1$$ Can you proceed?

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The common thread in both your solutions is that you split the absolute value into two cases, which is a sensible way to treat most absolute value equations/inequalities. The problem is, you look at $|x|$ in two cases: $x > 0$ and $x < 0$. What about when $x = 0$? You neglect this case, rightfully so in the $|x - 1|$ case, as it cannot be $0$ in order for the fraction to make sense, but you should also consider what happens when $x = -2$. It's clearly not a solution to the inequality, but you should address this.

Other than that, your first attempt is sound up to the conclusion. You split up into four cases based on the two absolute values, and derived that two of the cases were impossible. The other two cases produced solution sets $(-1/2, 1)$ and $(1, \infty)$. To get the complete solution, you need to union these solution sets: $$(-1/2, 1) \cup (1, \infty) = (-1/2, \infty) \setminus \{1\}.$$ This now covers all solutions from all four cases.

In your second attempt, when you consider $x - 1 < 0$, you multiply both sides by $|x - 1| = -(x - 1)$. The problem arises from you flipping the sign on the inequality. Remember, $-(x - 1)$ is a positive number, and thus flipping the sign is wrong. If you change the signs, you'll find it will agree with the first solution.

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  • $\begingroup$ Sorry I did not mention that in my post, that is the solution Wolfram alpha gave me, and I tried to understand..... $\endgroup$ – Ang Feb 23 at 17:57
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The first solution is correct, except that you need to put together the solution sets for the disjoint cases.

In the second solution you incorrectly invert the inequality sign: if $x<1$, then the inequality becomes $$ \frac{|x+2|}{1-x}>1 $$ which is equivalent to $$ |x+2|>1-x $$ You may better realize this if you do the multiplication before separating cases: $$ \frac{|x+2|}{|x-1|}>1 \iff |x+2|>|x-1| \quad(x\ne1) $$ By the way, there's a slicker solution: square both sides, which is allowed because both are nonnegative numbers: $$ x^2+4x+4>x^2-2x+1 $$ that becomes $6x>-3$, so $x>-1/2$ (but $x\ne1$).

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