yumi2012 wrote:If 4/x <-1/3, what is the possible range of values for x?

According to the solution page, If x<0, 4/x<-1/3 becomes 12>-x and becomes -12<x

I know that multiplying negative variable cause to flip signs, but in this case, it happened twice in the same equation! (Which equals no sign flips) how does this make sense?

4/x < -1/3

If x < 0,
-4/|x| < -1/3
4/|x| > 1/3
|x| < 12
x > -12

If x > 0,
4/x < -1/3
This can never happen.

Take a simple example,
let x = -4
12 > -(-4)
that is 12 > 4
becomes -12 < -4

Cheers

yumi2012 wrote:If 4/x <-1/3, what is the possible range of values for x?

4/x < -1/3
12/x < -1.

When an inequality involves negative values, many students struggle with putting the <> in the right direction.
Below is a way to avoid the issue.

12/x < -1 implies that x≠0.
Since x is NONZERO, x²>0.
Thus, we can safely multiply each side by x²:
12/x * x² < -1 * x²
12x < -x²
x² + 12x < 0
x(x+12) < 0.

The CRITICAL POINTS are where the lefthand side is EQUAL TO 0: x=0 and x=-12.
To determine where x(x+12) < 0, test one value to the LEFT AND RIGHT OF EACH CRITICAL POINT.

x<-12:
If we plug x=-13 into x(x+12) < 0, we get:
-13(-13+12) < 0
13<0.
Doesn't work.
x < -12 is not a viable range.

-12<x<0:
If we plug x=-1 into x(x+12) < 0, we get:
-1(-1+12) < 0
-11<0.
This works.
-12 < x < 0 is a viable range.

x>0:
If we plug x=1 into x(x+12) < 0, we get:
1(1+12) < 0
13<0.
Doesn't work.
x > 0 is not a viable range.

The only viable range is -12 < x < 0.