Fractional inequalites can be turned into quadratic or linear ones.
It requires care to do so though.
You have to "keep an eye out" for the possibilty of
multiplying or dividing by a negative number (the variable x can be + or -).
This is irrelevent for "equalities" since if x = 5 then -x = -5
but if x > 5 then -x < -5 ( 6>5 but -6<-5 as it's a lower temperature).
In the quadratic case you see where the graph (or chart) is above and below the x axis.
You see that nicely illustrated by Soroban.
In the rational case, you must carefully rearrange it if you have to.
How would we go from -x>7 to x<-7 ?
If we change the sign, we are in fact multiplying or dividing by -1.
If we do that without reversing the inequality, then we have gone wrong.
Of course, you can avoid that also by simply adding and subtracting terms from
both sides of the inequality in non-rational cases.
When you have x under the line in an inequality fraction, be careful about multiplying by x
due to what is mentioned.
Here is an example 2/x + x/3 > -4
2/x is asymptotic and x/3 is a line.
Their values are negative when x<0 and positive when x>0,
2/x has no value that we can define when x is zero,
because it starts "shooting for the stars" as it gets near zero, it evades us.
The sum of the fractions is negative only when x is negative,
positive only when x is positive. So, the sum is >-4 for x>0 and some negative x possibly.
Even though 2/x is undefined at x=0, you could say it's infinite and has to be >0.
If you choose to multiply both sides by x, you have 2 cases....
If x is negative, we reverse the inequality,
if not we don't so you could proceed to examine both cases.
It turns that if we imagine x is positive, our solutions for x are both negative
for the resulting quadratic x[sup:2oy0auit]2[/sup:2oy0auit] +12x +6 = 0.
If you then say the answers correspond to the range of x that make this >0
you'd be wrong!! If you sketch the curve, you'd notice the graph >0 for x< about -11 and x > about -1/2.
But we know that values of x much more negative than this will cause the initial sum
of two fractions to be much less than -4.
What you'd in fact have to do is begin with 2/x + x/3 >-4 so 2/x + x/3 + 4 >0
Now you should write....
if x is negative, then 2 + (x[sup:2oy0auit]2[/sup:2oy0auit])/3 + 4x <0
if x is positive, then 2 + (x[sup:2oy0auit]2[/sup:2oy0auit])/3 + 4x >0
So if x is negative, you can multiply both sides by x "while reversing the inequality".
You just need to "account for" x being + or -
and check the solutions you get, as they must be + if x should be +
and - if x should be -
therefore you are looking for the solutions of x[sup:2oy0auit]2[/sup:2oy0auit] + 12x + 6 < 0 for negative x
which is the "cup" part of the graph under the x axis
and x[sup:2oy0auit]2[/sup:2oy0auit] + 12x + 6 > 0 for positive x which is the "branch" extending from the y axis crossing point at y = 6
As that curve is U-shaped, it is >0 when x is positive for all x>=0.
It's also + for x < about -11 or so but x cannot be - for this part of the solution.
It takes a bit of getting used to but it's worth the effort.
Sorry, I have no graphs, I need to figure out how to include them here (.pdf files won't upload).
