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Solve the given inequality by interpreting it as a statement about distances on the real line.

$$\lvert x+1\rvert > \lvert x-3\rvert$$

I am confused on what this question is asking. Can anyone give me a quick explanation and answer?

Please show numerically and graphically.

Thanks

Thomas
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madman75
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  • Please show some work of what you know about this problem. – NasuSama May 12 '13 at 18:50
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    Draw a number line. On it, put the numbers $-1$ and $3$. Then $|x+1|=|x-(-1)|$ is the distance from $x$ to $-1$. Similarly, $|x-3|$ is the distance from $x$ to $3$. So our inequality says that $x$ is farther from $-1$ than it is from $3$, or equivalently that $x$ is closer to $3$. Now the picture should tell you who these $x$ are. – André Nicolas May 12 '13 at 18:55

3 Answers3

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If $x$ and $y$ are real numbers, $|x-y|$ is the distance between the points $x$ and $y$ on the real line. Thus, $|x-3|$ is the distance between $x$ and $3$. The other absolute value expression in your problem is a little trickier, because it’s written as the absolute value of a sum rather than a difference, but you can rewrite it: $|x+1|=|x-(-1)|$ is the distance between $x$ and $-1$. To solve the inequality $|x+1|>|x-3|$, therefore, you’re looking for the real numbers $x$ that are farther from $-1$ than they are from $3$. It’s probably easier to think about if you turn that around: you’re looking for the real numbers that are closer to $3$ than they are to $-1$.

      --------------|------|-----|-------------|--------|----|----------------  
                    a     -1     b             c        3    d

Take a look at the numbers $a,b,c$, and $d$ in the sketch above; which are closer to $3$ than they are to $-1$? Can you find a way to tell whether a real number $x$ will be closer to $3$ than to $-1$ just by comparing it with a single real numbers?

Brian M. Scott
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They want you to find all values of $x$ for which $|x+1| > |x - 3|$.

The method used to solve the problem should involve interpreting $|x+1|$ as the distance from $x$ to $-1$ on the real number line, and interpreting $|x - 3|$ as the distance from $x$ to $3$ on the real number line.

Zarrax
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Ultimately, we want to solve the inequality. However, dealing with multiple absolute value signs is a headache to do algebraically, so the question statement is asking you to consider a geometric approach.

Remember that $|x|$ means the distance of $x$ from zero, and that $|x-A|$ can be interpreted as meaning the distance of $x$ away from $A$.

So, for example, $|x+1|>B$ means that $x$ is a distance of more than $B$ away from $-1$. In other words, to get from $-1$ to $x$, you have to travel at least $B$ units (maybe forward or maybe backward).

Can you use this to help settle your question?

Eric Stucky
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