inequalities with x in the denominator (1 Viewer)

[x jiim]

For stuff like |x+1| + |2x-1| < 5 I'd rather rearrange and graph than use case. Case is too fiddly for my liking, and I always screw up halfway through ><

 

[x jiim]

Re: solving inequalities with x in the denominator

you just have to multiply both sides by the denominator squared, can''t remember why squared

Because as long as everything's real, the square will be positive so you don't have to reverse the sign as you would when multiplying by a negative.

 

[cutemouse]

|5x-2|/x > 0, automatically x > 0.

??

And for |x+1| + |2x-1| < 5, I just do ++, +-, -+, -- and then substitute back into the equation to see if it works.

That is the case method.

By definition, |x+1| = x+1 if x+1>= 0 Therefore x>-1
and |x+1|=-(x+1) if (x+1)<0, therefore x>-1

etc

 

[jet]

??



That is the case method.

By definition, |x+1| = x+1 if x+1>= 0 Therefore x>-1
and |x+1|=-(x+1) if (x+1)<0, therefore x>-1

etc

For the first bit it's quite obvious. Since |5x - 2| > 0 for all x in R, The only condition required for |5x - 2|/x > 0 is x > 0

 

[gurmies]

Firstly, this is NOT a 2 unit topic.



Secondly, for those who don't like dealing with complicated algebra or graphs, there is a much easier method:



1. Make the inequality an equation, and solve.

2. Find the values of x that make the denominator(s) zero.

3. Plot the answers from (1) and (2) on a number line, call them the boundary points.

4. Test a point in every region between consecutive boundary points, and also test the boundar points themselves.



Then just write down the solution.



This always works.

The method I have used since the very beginning - this can't be beaten.

 

[shady145]

Could also use the case method... although it appears that alot of people don't get it.

I remember in Year 11 the teacher of another class taught this method to his students. Only one of them (out of 30) got it in the test, whereas in our class everyone got the correct answer using the above method.

:O ive never been taught the other method, only the case method XD
and its simple, don't know why "Only one of them (out of 30) got it in the test,"

 

[The Nomad]

Re: solving inequalities with x in the denominator

For that method you have to either use a set of rules or multiply by the square of the denominator (realizing that this can in fact create more answers), and then substitute the answers back in or simply evaluate the answers in your head.

Can you please show me an example where it creates more answers? Assuming that the denominator is linear, ofcourse...

 

[jet]

Re: solving inequalities with x in the denominator

Can you please show me an example where it creates more answers? Assuming that the denominator is linear, ofcourse...

I'm too lazy to get one now, but sometimes it will mislead you into thinking that one of the boundaries is actually part of the domain. That's what I've seen before (though this was two years ago).

 

[The Nomad]

Re: solving inequalities with x in the denominator

I'm too lazy to get one now, but sometimes it will mislead you into thinking that one of the boundaries is actually part of the domain. That's what I've seen before (though this was two years ago).

Are you sure you're not just talking about the case where one of the boundaries gives the original denominator a value of 0?

1/(x - 4) >= 0.
x - 4 >= 0.
x >= 4.

But x <> 4 since x - 4 <> 0?

Or there's something else?

 

[jet]

Re: solving inequalities with x in the denominator

Are you sure you're not just talking about the case where one of the boundaries gives the original denominator a value of 0?

1/(x - 4) >= 0.
x - 4 >= 0.
x >= 4.

But x <> 4 since x - 4 <> 0?

Or there's something else?

I honestly don't remember exactly what my teacher said to me. You might be right, I'm not too sure.

 

[The Nomad]

Re: solving inequalities with x in the denominator

I honestly don't remember exactly what my teacher said to me. You might be right, I'm not too sure.

Ahh it's ok then. I just don't remember seeing something like this.

 

[cutemouse]

For the first bit it's quite obvious. Since |5x - 2| > 0 for all x in R, The only condition required for |5x - 2|/x > 0 is x > 0

Well, you're forgetting that x =/= 2/5 since it's just a ">" (no equality). You answer however would be correct if it was "greater than or equal to".

 

[cutemouse]

:O ive never been taught the other method, only the case method XD
and its simple, don't know why "Only one of them (out of 30) got it in the test,"

A bad teacher may've been a contributing factor. I would know... this same teacher taught me for Extension 2... hahaha