Well as you say it's obvious that x must be so that will be . But i don't understand why it should be also . For every the fraction is positive...
Hi all, I've got a inequality question. Should be basic but it's stuffing me up for some reason. I'll explain why in a moment. I'll go through a sample question, to explain how I usually solve it. Then solve for each case... Then plot these on a number line and determine the range etc. What stuffs me up with the original question is the > 0. I've got nothing on the other side, to bring over and essentially get the x's up the top. The correct answer is -2/3 < x < 1/3. I can get the first half, however I'm not sure how to get the 1/3 answer. P.s. Sorry but the formatting seems to be stuffing up for some reason. I had this nicely spaced out but posting seemed to destroy it.
LaTeX doesn't recognise space.
Notice that since you're saying the left hand side is greater than 2, you're saying the left hand side is positive.
Since the numerator is positive, so must be the denominator.
So multiplying both sides by the denominator WILL NOT change the sign of the inequality.
.
Thus .
Ok - Thanks for the feedback guys. I'll try and restate it all now I understand it was latex causing the problems with the spacing. Sorry about that.
1st - The problem I was trying to solve is the following.
I was unsure of how to do it. As a result, I tried to show some working as to how I usually do the problem, with a similar number.
The similar (but not required problem) was the following.
This is how I go to solve this problem.
1.
2.
3.
4. Solving top
5. Solving bottom
Therefore critical values occur at these two locations, so we sub in 0, and find that x satisfies this point. Therefore,
Prove it - Small mistake with your solution, 8 / 6 = 4/3 not 2/3, which is one bound of the solution. However x cannot be simply less than 4/3 , because for instance -1, which is less than 4/3 does not satisfy the equation. Doing it via your method loses a solution.
Saying that, it seems that perhaps the answers in the book were wrong, because it doesn't seem as though there is an upper bound. The answer is given as -2/3 < x < 1/3. I.e. I can sub in 1 and still satisfy the equation, so the answer given must be wrong.
Thanks for your help, sorry for the unclear OP.
Sorry Stroodle - Have to disagree (although I understand I was coming on here asking for help how to do these).
If Solve It's method was correct, then every value of x less than should satisfy the equation...
Clearly, because -1 does not satisfy this, his solution must be incorrect.
I'm not sure what you mean in regards to the solving of my values, they are simply to find the critical points at which a change occurs on the number line, not trying to definitively solve those particular equations.