Question

1/3x <2. Solution is x < 0 or x > 1/6. Normal solving for x does not work. (Openstudy counted my question as answered, when it was not. The answer is not just x > 1/6.)

Answer 1

http://www.wolframalpha.com/input/?i=1/3x%3C2

Answer 2

This is my book's method. I'm pretty sure it is an inefficient way.

Answer 3

I'd guess the book's way is as good as any.
Another way is to move all the terms to one side, so you are comparing to zero.
For example:
\[\frac{1}{3x}-2<0\]
Note that x= 0 is not allowed
put over a common denominator to get
\[ \frac{1-6x}{3x}<0 \]
multiply by 3x
\[ 3x(1-6x)<0 \]
or simply
\[ x(1-6x)<0\]
for this to be true, you need opposite signs: x<0 and (1-6x)>0 or x>0 and (1-6x)<0
the first set x<0 and (1-6x)>0 requires (x<0) and (x< 1/6) so it must be x<0
the second set lead to x>0 and x>1/6, so x>1/6

Answer 4

Here's the books example:
\[\frac{8}{x+5} \le 4\]
note: x can not equal 5 (we will use this fact)
\[ \frac{8}{x+5}-4 \le 0 \]
\[ \frac{8-4(x+5)}{x+5} \le 0 \]
\[ \frac{-4x-12}{x+5} \le 0 \]
multiply through by x+5. factor out -4
\[ -4(x+3)(x+5) \le 0 \]
divide both sides by -4 and reverse the relation
\[ (x+3)(x+5) \ge 0\]
Now both terms must be the same sign. So
(x+3)≥0 and (x+5)≥0 --> x≥-3 and x≥-5, so x≥-3
or
(x+3)<0 and (x+5)<0 ---> x≤ -3 and x ≤ -5, so x< -5 (can't be -5)

x< -5 or x≥-3

Answer 5

Thank you so much.

Answer 6

"multiply through by x+5" Wouldn't we have to multiply the other side by x+5? Since x+5 can be negative, wouldn't we have to switch the inequality sign as well?

Answer 7

Looking at the answer, it doesn't look like we switch the inequality. But I would have to think about it to explain why it works. If I come up with a reason I'll post it here.

Answer 8

OK, here's the right way:
do not multiply through by the denominator.
but the logic is still the same:
\[ \frac{P(x)}{Q(x)} > 0 \]
requires that both P(x) and Q(x) have the same sign, or
P(x)>0 and Q(x)>0 or P(x)<0 and Q(x)<0

Answer 9

Makes sense. Thank you much.