# Inequality - x on denominator

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• Jun 21st 2009, 12:41 AM
Peleus
Inequality - x on denominator
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. $\displaystyle \frac{12}{3x + 2} > 0$ I'll go through a sample question, to explain how I usually solve it. $\displaystyle \frac{12}{3x + 2} > 2 \frac{12}{3x + 2} - 2 > 0 \frac{8 - 6x}{3x + 2} > 0$ Then solve for each case... $\displaystyle 8 - 6x = 0 x = \frac{4}{3} 3x + 2 = 0 x = \frac{-2}{3}$ 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.
• Jun 21st 2009, 01:19 AM
SENTINEL4
Well as you say it's obvious that x must be $\displaystyle x>\frac{-2}{3}$ so that will be $\displaystyle \frac{12}{3x+2}>0$. But i don't understand why it should be also $\displaystyle x<\frac{1}{3}$. For every $\displaystyle x>\frac{-2}{3}$ the fraction is positive...
• Jun 21st 2009, 01:19 AM
Stroodle
I don't quite understand your working, but the way I do it is:

As you can't divide a number by $\displaystyle 0$ either $\displaystyle 3x+2>0$ or $\displaystyle 3x+2<0$ so in the first case $\displaystyle x>-\frac{2}{3}$ and in the second case $\displaystyle x<-\frac{2}{3}$ so $\displaystyle x$ is any real number excluding $\displaystyle -\frac{2}{3}$

Does that help? I may have misunderstood your question...
• Jun 21st 2009, 01:25 AM
TheAbstractionist
You want to solve the inequality $\displaystyle \frac{12}{3x+2}>0.$

Since the numberator 12 is positive, the denominator must be positive as well. Hence

$\displaystyle \frac{12}{3x+2}\ >\ 0.$

$\displaystyle \implies\ 3x+2\ >\ 0$

$\displaystyle \implies\ 3x\ >\ -2$

$\displaystyle \implies\ x\ >\ \frac{-2}3$
• Jun 21st 2009, 01:26 AM
SENTINEL4
Quote:

Originally Posted by Stroodle
... so $\displaystyle x$ is any real number excluding $\displaystyle -\frac{2}{3}$
...

But if we take x=-1 then it would be $\displaystyle \frac{12}{3(-1)+2} = \frac{12}{-3+2} = \frac{12}{-1} = -12 < 0$ but we want the fraction to be >0.
• Jun 21st 2009, 01:38 AM
Prove It
Quote:

Originally Posted by Peleus
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. $\displaystyle \frac{12}{3x + 2} > 0$ I'll go through a sample question, to explain how I usually solve it. $\displaystyle \frac{12}{3x + 2} > 2 \frac{12}{3x + 2} - 2 > 0 \frac{8 - 6x}{3x + 2} > 0$ Then solve for each case... $\displaystyle 8 - 6x = 0 x = \frac{4}{3} 3x + 2 = 0 x = \frac{-2}{3}$ 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.

$\displaystyle \frac{12}{3x + 2} > 2$

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.

$\displaystyle 12 > 2(3x + 2)$

$\displaystyle 12 > 6x + 4$

$\displaystyle 8 > 6x$

$\displaystyle \frac{2}{3} > x$.

Thus $\displaystyle x < \frac{2}{3}$.
• Jun 21st 2009, 01:38 AM
Stroodle
Makes sense.
I completely misunderstood the question.
• Jun 21st 2009, 01:43 AM
SENTINEL4
I still didn't understand what you want to prove...
You want $\displaystyle \frac{12}{3x+2}>0$ or $\displaystyle \frac{12}{3x+2}>2$ ???
• Jun 21st 2009, 02:52 AM
Peleus
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.

$\displaystyle \frac{12}{3x + 2} > 0$

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.
$\displaystyle \frac{12}{3x + 2} > 2$

This is how I go to solve this problem.

1. $\displaystyle \frac{12}{3x + 2} - 2 > 0$

2. $\displaystyle \frac{12}{3x+2} - 2(3x + 2) > 0$

3. $\displaystyle \frac{8 - 6x}{3x + 2}$

4. Solving top $\displaystyle x = \frac{4}{3}$

5. Solving bottom $\displaystyle x = \frac{-2}{3}$

Therefore critical values occur at these two locations, so we sub in 0, and find that x satisfies this point. Therefore, $\displaystyle \frac{-2}{3} < x < \frac{4}{3}$

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.
• Jun 21st 2009, 04:01 AM
Stroodle
The thing with solving the top and the bottom is that if $\displaystyle 8-6x<0$ and $\displaystyle 3x+2<0$ then the fraction will still be $\displaystyle >0$ as a negative divided by a negative gives a positive.

Prove It's method is correct. (except it's $\displaystyle \frac{4}{3}$ not $\displaystyle \frac{2}{3}$ )
• Jun 21st 2009, 04:13 AM
Peleus
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 $\displaystyle \frac{2}{3}$ should satisfy the equation...

$\displaystyle \frac{12}{3x + 2} > 2$

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.
• Jun 21st 2009, 04:34 AM
SENTINEL4
Quote:

Originally Posted by TheAbstractionist
You want to solve the inequality $\displaystyle \frac{12}{3x+2}>0.$

Since the numberator 12 is positive, the denominator must be positive as well. Hence
$\displaystyle \frac{12}{3x+2}\ >\ 0.$
$\displaystyle \implies\ 3x+2\ >\ 0$

$\displaystyle \implies\ 3x\ >\ -2$

$\displaystyle \implies\ x\ >\ \frac{-2}3$

TheAbstractionist didi it very well... this is the solution, it makes sense
• Jun 21st 2009, 04:44 AM
Peleus
Agreed, a question with the "Because 12 is positive, x must be positive as well", does this apply in all cases when it's > or < 0?

For instance, saying that x must be positive when we have > 2 for instance is incorrect, and x can be negative.
• Jun 21st 2009, 04:48 AM
SENTINEL4
In your fraction over here, because the numerator is always positive, if the denominator is positive too then the whole fraction will be positive. So it must be 3x+2>0 not just x>0.
• Jun 21st 2009, 04:52 AM
Stroodle
I don't get this at all. I'll ask my maths teacher tomorrow if there's a good method to use and post it here, if no-one else does by then.

Using the Abstractionists method wouldn't it be $\displaystyle 3x+2<6$ and $\displaystyle 3x+2>0$ ??
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