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I apologize if this question in general, but I've been having trouble finding solutions as Google discards absolute value signs and inequality symbols.

I am looking for a way to eliminate absolute value functions in $|a| < |b|$.

I can solve $|a| < b$ and $|a| > b$, but I am unsure what method / combination of methods to use to eliminate absolute value signs from both sides.

Thank you!

An example problem:

$|x + 2| < |x - 4|$

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    Since you mentioned trying Google, maybe you should try [Wolfram Al$p$ha](http://www.wolframalpha.com/input/?i=|x%2B2|<|x−4|) instead2012-08-11

5 Answers 5

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We have $|a| \lt |b|\,$ if any of these is true:

(i) $\,a$ and $b$ are $\gt 0$ and $a \lt b$

(ii) $a\lt 0$ and $b\ge 0$ and $-a \lt b$

(iii) $b \lt 0$ and $a \gt 0$ and $a \lt -b\,$

(iv) $\,a\lt 0$ and $b\lt 0$ and $-a\lt -b$. We can rewrite this as $b \lt a$.

Four cases! Not surprising, since eliminating a single absolute value sign often involves breaking up the problem into $2$ cases.

Sometimes, one can exploit the simpler $|a| \lt| b|\,$ iff $\,a^2\lt b^2$. But squaring expressions generally makes them substantially messier.

Added: With your new sample problem, squaring happens to work nicely. We have $|x+2| \lt |x-4|$ iff $(x+2)^2 \lt (x-4)^2$. Expand. We are looking at the inequality

$x^2+4x+4 \lt x^2-8x+16.$

The $x^2$ cancel, and after minor algebra we get the equivalent inequality $12x \lt 12$, or equivalently $x\lt 1$. The squaring strategy works well for any inequality of the form $|ax+b| \lt |cx+d|$.

But the best approach for this particular problem is geometric. Draw a number line, with $-2$ and $4$ on it. Our inequality says that we are closer to $-2$ than we are to $4$. The number $1$ is halfway between $-2$ and $4$, so we must be to the left of $1$.

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    Ha! I never would of thought of this. This is brilliant! Thank you so much for your help!2012-08-11
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There are different approaches; one is to look at the zeroes of the expressions inside the absolute values, and split up $\mathbb R$ into intervals accordingly. In your example $|x + 2| < |x - 4|$, the points of interest are at $x=-2$ and $x=4$. You can therefore consider three cases:

1. If $x \in (-\infty,-2)$, then $x+2 < 0$ and $x-4 < 0$, so the absolute values will reverse the signs of both. This gives: $\begin{align} -(x+2) &< -(x-4) \\ x+2 &> x-4 \\ 2 &> -4 \end{align}$ This is true for all $x$ in the interval.

2. If $x \in [-2,4)$, then $x+2 \geq 0$ so its sign is unaffected by the absolute value, but $x-4 <0$ so its sign will be reversed: $\begin{align} x+2 &< -(x-4) \\ 2x+2 &< 4 \\ x &< 1 \end{align}$ Combining this last inequality with the assumption that $x \in [-2,4)$, we see that any $x$ in $[-2,1)$ is valid.

3. Finally, if $x \in [4,\infty)$, neither expression's sign is reversed: $\begin{align} x+2 &< x-4 \\ 2 &< -4 \end{align}$ This is false for all $x$ in the interval.

Putting all the information together from the above three cases, we have $x \in (-\infty, 1)$.


Note: as some other contributors have mentioned, there are simpler ways to deal with your problem, such as viewing it geometrically. The method that I have shown above is more useful when the expressions are more complicated or when you have several absolute values; for example, an inequality like $3 |x^2-1|+|x-2|+|x^2-3x| > 5$.

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    @BenMillwood: Fair enough! I've changed i$t$.2012-08-11
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We can see $|x-a|$ as a distance point $x$ from point $a$.

Now, the question with the above "definition" would be like:

For which $x$ values distance point $x$ from $-2$ be less than distance point $x$ from $4$?

Clearly, by drawing it maybe, you can observe that the answer is for all $x<1$.

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We can divide by $|b|$ to get $|a/b|<1$. Let $x=a/b$ then $|x|<1$ so $-1.

Now multiply through by $b$.

If $b>0$ then $-b.

If $b<0$ then $-b>a>b$

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$|a| < |b| \iff -|b| < a < |b| \iff \begin{cases} -b < a < b & \text{If $b > 0$} \\ \text{No solution.} &\text{If $b=0$} \\ b < a < -b & \text{If $b < 0$} \end{cases}$

E. G. $|x + 2| < |x - 4|$

\begin{align} x > 4 &\implies-x+4 < x+2 < x-4 \\ &\implies -2x+4 < 2 < -4 \\ &\implies \text{No solution.} \end{align}

\begin{align} x > 4 &\implies -x+4 < x+2 < x-4 \\ &\implies 4 < 2 < 2x-4 \\ &\implies x < 1 \end{align}

\begin{align} x < 4 &\implies x-4 < x+2 < -x+4 \\ &\implies -4 < 2 < -2x+4 \\ &\implies x < 1 \end{align}

Hence $x \in (-\infty, 1)$