# Why is there no solution set for $|x-7|<-4$?

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?

• The magnitude of $x-7$ is always $\ge0$ – Ak19 Jun 12 at 14:02
• 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? – Giuseppe Negro Jun 12 at 14:03
• Bacuase absolute value is a number $\ge 0$. – Mauro ALLEGRANZA Jun 12 at 14:04
• 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. – Martin R Jun 12 at 14:05
• There is a solution set which is the empty set. – drhab Jun 12 at 14:33

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