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I am asked to find the solution set fot $|x-7|<-4$.

I arrived at $(-\infty, 3)\cup(11, \infty)$

For $x - 7 > 0$:

$x-7<-4$

=> $x<3$

For $x-7 < 0$:

$-(x-7)<-4$

=> $-x+7<-4$

=> $-x<-11$

=> $x>11$

So, I arrive at a solution set of:

$(-\infty, 3)\cup(11, \infty)$

However, my textbook says "no solution". Why is there no solution?

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    $\begingroup$ The magnitude of $x-7$ is always $\ge0$ $\endgroup$ – Ak19 Jun 12 at 14:02
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    $\begingroup$ You wrote: "for $x-7>0:\ x<3$" and "for $x-7<0:\ x>11$", but then you forgot about the conditions $x-7>0$ and $x-7<0$. Can it be that $x-7>0$ AND $x<3$ at the same time? Can it be that $x-7<0$ AND $x>11$ at the same time? $\endgroup$ – Giuseppe Negro Jun 12 at 14:03
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    $\begingroup$ Bacuase absolute value is a number $\ge 0$. $\endgroup$ – Mauro ALLEGRANZA Jun 12 at 14:04
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    $\begingroup$ Take any of your alleged solutions, e.g. $x=12$, and substitute it into the original inequality. Does it satisfy $|x-7|<-4$? Then repeat this check for each step in your argument until you find the error. $\endgroup$ – Martin R Jun 12 at 14:05
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    $\begingroup$ There is a solution set which is the empty set. $\endgroup$ – drhab Jun 12 at 14:33
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When the absolute value is on the "less than" side, the conjunction is "and", not "or." You've discovered that $x<3$ AND $x>11$, which is impossible.

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If you graph the solution on a x-y plane you get...this plane As you can see if you graph all of these according to y, $|x-7|$ can never reach -4, hence, no solution. :))

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The absolute value, or the magnitude, of a number or an expression will always be non-negative ( $ \geq 0 $). That is why $ |x-7| < -4 $ has no solution.

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  • $\begingroup$ This sounds very intuitive now that you put it like that, thanks for the tip $\endgroup$ – Doug Fir Jun 14 at 16:48

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