Inequalities with the Unknown Denominator. (1 Viewer)

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I am having trouble solving these type of problems for some reason. I cant get to the answer. If anyone could care to explain to me it would be greatly appreciated.

Example 1: 1/x < 1/4

Example 2: 1/x-3 > 2
 

matt grime

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The only thing you need to remember is that when multiplying an inequality through you keep the same > or < if it's a postivie number and reverse it for negatives.

If Y is some unknown and you're multiplying through then you need to do it case by case for when Y is positive and when Y is negative. Remeber that at the end you must put together all of the information, that is if you assume Y is postive do some manipulation and determine that this implies Y is greater than -1 then you can't have a positive number between -1 and 0 so in fact all positive Y is the answer, ie you must combine the fact that Y>0 and Y>-1 to get Y>0, similarly if you assume Y is positve do the manpulation and get that Y<4 then it must be that 0<Y<4.
 
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Hybr!d said:
I am having trouble solving these type of problems for some reason. I cant get to the answer. If anyone could care to explain to me it would be greatly appreciated.
Example 1: 1/x < 1/4
Example 2: 1/x-3 > 2
Try cross-multiplying.

1/x < 1/4
x*4 < 1*1
4x < 1
etc...

1/x-3 > 2
1/x-3 > 2/1
2*(x-3) > 1*1
etc...

I never did cross-multiplying in inequalities, but it should work out.
 

NateTG

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nazgjunk said:
Try cross-multiplying.
1/x < 1/4
x*4 < 1*1
4x < 1
etc...
1/x-3 > 2
1/x-3 > 2/1
2*(x-3) > 1*1
etc...
I never did cross-multiplying in inequalities, but it should work out.
This does not work

For example:
[tex]\frac{1}{x} < \frac{1}{4}[/itex]
has the solution:
[tex]x>4[/tex]
or
[tex]x<0[/tex]

Instead do something like this:
[tex]\frac{1}{x} < \frac{1}{4}[/tex]
Muliply both sides by [itex]4[/itex]
[tex]\frac{4}{x}<1[/tex]
Split into three cases:
[itex]x>0[/itex], [itex]x<0[/itex] and [itex]x=0[/itex].
Clearly the [itex]x=0[/itex] case is no solition since the left side is undefined.
Now, assuming that [itex]x>0[/itex] we can multiply both sides by x to get:
[tex]4<x \rm{and} x>0 \Leftrightarrow x>4[/tex]
Assuming that [itex]x<0[/itex] we get
[tex]4>x \rm{and} x<0 \Leftrightarrow x<0[/tex]
 
Oh. In that case, sorry. As i said, i never combined cross-multiplying and inequalities, but since it works in equalities, i reckoned it would. I am working it out on paper now, however.
 
OH GOD, ****.

My maths teacher will have to spank me for this. I said cross-multiplying, which means in
[tex]\frac{1}{x} < \frac{1}{4}[/tex]
that x*1>1*4. What i did was multiplying both denominators etc. That was indeed one hell of a stupid mistake. It also turns out that when cross-multiplying in a case like this, i need to swap the < to a >. Geez. OK, my teacher never told me *that*, but now i do know.

And of course it is true x can also be smaller than 0.

I guess i should just stop thinking about things my teacher never told me.

Yuk.

/me slaps himself around a bit with a large trout
/me slaps himself around a bit with his maths textbook

I apologise,

>Nazgjunk
 
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Thanx for the information guys. Now I clearly get it.
 

uart

Science Advisor
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In addition to the "splitting into cases" method that Matt mentions there is another method that is often usful for this type of problem. This method involves always multiplying by a square so that you know that it's non-negative.

As an example take the inequality : x/(3-x) > 2

Mult both sides by (x-3)^2 : 3x - x^2 > 18 - 12x + 2x^2

Rearrange all terms to one side of the inequation: 3x^2 -15x + 18 < 0

Simplify : x^2 - 5x + 6 < 0

Now factorize (or any other method) to find the zeros and note that since this is an "upward" parabola then the region between to two zeros will be where the curve is less than zero.

(x - 2) (x - 3) < 0, therefore 2 < x < 3 is the solution.

This is a good method to use if have covered (or are currently studying) parabolas and quadratic equations. Just be a little careful about the multiply by zero case, that is x=3 in the above example. Be aware that x=3 wouldn't be in the solution even if the original inequation had a "greater than or equal" instead of the straight "greater than" as given.
 
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