Question

how to solve the inequality in terms of intervals and illustrate the solution set on the number line.
-6<1/x<=1

Answer 1

Find the reciprocal of each, and flip each of the inequality signs. You therefore have: \[-\dfrac{1}{6}>x\ge 1\] The reason you flip the inequality signs is the same reason that \[3<5 \text{ but } \dfrac{1}{3}>\dfrac{1}{5}\]For more solutions to textbook problems, be sure to check out http://www.slader.com/s/eWFrZXlnbGVl . I submit a lot of content there as well.

Answer 2

Shoot wow...that's TOTALLY off...one sec.

Answer 3

you have to flip the inequality so you get two intervals. there is no such interval
\[-\frac{1}{6}>x>1\] the two intervals are
\[x<-\frac{1}{6}\] or
\[x>1\]

Answer 4

Sorry, that rule isn't always true for negatives. For the part dealing with 1 it was correct. Take the left side:\[-6<\dfrac{1}{x}\]x can be positive or negative. When it is positive, solving for x gives the following (remembering to flip the sign when you divide a negative):\[x>-\dfrac{1}{6}\]When it is negative, solving for x gives the following (remembering to flip the sign when you divide a negative):\[x<-\dfrac{1}{6}\]Take note that restricting to positive values and negative values really implies the following about the first equation in general:\[\left(x > -\dfrac{1}{6} \text{ and } x > 0\right)\text{ or }\left(x < -\dfrac{1}{6} \text{ and } x < 0\right)\]Take note that the left hand "and" part can't have any value. Therefore, we just have \[x > -\dfrac{1}{6} \text{ and } x > 0\]Sketching the intersection of this and the interval from before \[x\ge 1\]gives the following. This is a very difficult problem. Don't be discouraged by it. It fooled me certainly. For more solutions to textbook problems, be sure to check out http://www.slader.com/s/eWFrZXlnbGVl . I submit a lot of content there as well.