NvrEvrGvUp wrote:
Gotta bump this back up because I'm stuck on what should be a very simple inequalities problem.
Given: \(\frac{4}{x}\) < -\(\frac{1}{3}\), I simplified the equation to: 12 < -x by cross-multiplying
Now, we have 2 scenarios:
1. x > 0 : no sign changes in 12 < -x, so x < -12. However, since we know x > 0, this scenario is impossible.
2. x < 0 : 12 < -x should become 12 < -(-x) so wouldn't this just be 12 < x? However, given x < 0, this scenario doesn't appear possible either.
Can someone please point out the obvious? It's driving me crazy...
First of all, the actual question is \(\frac{4}{x}\) < \(\frac{1}{3}\) (there is no negative with 1/3)
Also, you know what you have to do but you probably do not understand why you have to do it. That is why you are facing problem in this question.
Given: \(\frac{4}{x}\) < -\(\frac{1}{3}\), I simplified the equation to: 12 < -x by cross-multiplyingThere is a problem here. You don't cross multiply and then take cases. You take cases and then cross multiply. Why? Because you CANNOT cross multiply till you know (or assume) the sign of x. The result of the cross multiplication depends on whether x is positive or negative. So you need to take cases and then cross multiply.
Case 1: x > 0
12 < x
Case 2: x < 0
12 > x (note that the sign has flipped here because you are multiplying by a negative number)
x should be less than 12 AND less than 0 so the range in x < 0.
Hence, two cases: x > 12 or x < 0.
Also because x is negative, you cannot just multiply it by -1 to make -x = x. That is certainly not correct. -x is positive and x is negative. They are not equal if x is not 0.
I still dont understand when x < -12, how can we say x < 0. Since if x = -7, -6 it is not valid for x < -12. ( I understand the other range x > 12)