Tips for solving 3units inequations?? (1 Viewer)

[COLDBOY]

I just want to find out any tips and methods on inequations at the 3 unit level, which can make me solve inequation questions much faster. Like say the question was 5m+4/2m < 1/4. Is there an easier and faster way to solve this

 

[life92]

hahah well
my teacher taught us a different method rather than just multiplying by the square of the denominator.. but i think it only works if one of the sides is = 0

so in the example, (5m+4)/2m < 1/4
and moving the 1/4 onto the LHS
you're left with
(9m+8) / 4m < 0
so then, you now have 2 critical points, 0 and -8/9
and then you can test the 3 regions,, that is m > 0, -8/9 < m < 0, and m < -8/9
or if you know that the inequality graph will show you a parabola, you can either test the middle,, that is -8/9 < m < 0 or the one of the outer regions ,, ie. m > 0, m < -8/9

hrmm
its an all right method i guess and saves time especially if its already in the form where 0 is on one side,, but otherwise i would just use the normal method

 

[khorne]

To add to life92's method, it works normally as well.

The method is about finding critical points (i.e solutions and asymptotes).

(5m+4)/2m < 1/4, we note that m =/= 0 (cannot equal). So 0 becomes a critical point.
Next, you just replace the inequality sign with an equals, and solve the equation, to find possible values for m...These values create the next set of critical points.

Then, all that is left is testing numbers inbetween the critical points. As a rule of thumb, the critical points and sections will alternate, i.e if you have 3 sections, the middle one will be the solution, or the two outer ones will be. This applies to all examples which don't involve squares in the final solution.

If it doesn't make sense, i will draw a little diagram.

 

[addikaye03]

hahah well
my teacher taught us a different method rather than just multiplying by the square of the denominator.. but i think it only works if one of the sides is = 0

Yeah ofcourse, but that's just basically a simplification step. I use the same method.

5m+4/2m < 1/4

2m=/=0 therefore m=/=0 [pt.1]

20m+16<2m => 18m=-16 therefore m=-8/9 [pt.2]

Set up a number line where 0 and -8/9 are open circles since both values are excluded.

Then just do it as you said.

 

[life92]

Ahh I see.
Yeah that's what my teacher taught me.
I couldn't quite remember what it was but now I do.

But he only taught it to us like, after trials.
Do other teachers teach this method initially?
Or just as a side thing?

 

[addikaye03]

Ahh I see.
Yeah that's what my teacher taught me.
I couldn't quite remember what it was but now I do.

But he only taught it to us like, after trials.
Do other teachers teach this method initially?
Or just as a side thing?

Mmm not sure mate, i was taught both methods at the same time and told to pick my favourite. They're quite similar though.

 

[cutemouse]

I like multiplying both sides by the square of the denominator then drawing a 'facilitating' graph etc. Easier IMO