+364 ok..
i don't know if this is possible..
x can equal 1 to infinity and it works
+2668 Note that \(1 \over {1 \over3}\) is 3. When \(x\) gets bigger, the value of the expression gets smaller.
Likewise, \(1 \over -{1 \over 4}\) is -4. As \(x\) gets smaller than \( -{1 \over 4}\), the value of the equation gets bigger, but never is greater than 0.
So, the 2 ranges that work are \(\color{brown}\boxed{{x<-{1\over4} \space \text {and} \space{x>{1\over 3}}}}\)
+23252 Rewrite -4 < 1/x < 3 into its two parts: -4 < 1/x and 1/x < 3
Divide the problem into two parts:
Part 1: x < 0:
(Remember that when you multiply both sides of an inequality by a negative number,
you must change the sense of the inequality.
Also remember that, for this part, x is negative.)
-4 < 1/x ---> -4x > 1 ---> x < -1/4
1/x < 3 ---> 1 > 3x ---> 3x < 1 ---> x < 1/3
Since these two inequalities are combined by the word and, the solution to this
part is the more restrictive of the two answers: x < -1/4
Part 2: x > 0
-4 < 1/x ---> -4x < 1 ---> x > -1/4
1/x < 3 ---> 1 < 3x ---> 3x > 1 ---> x > 1/3
The more restrictive of these two is: x > 1/3
Since the answer consists of two parts, either Part 1 or Part 2, the final answer is the
region: x < -1/4 or x > 1/3
which is: (-inf, -1/4) union (1/3, inf)
