help please

Here's another way to approach this just using some logic.....[although I have no quibble with Melody's method]

Note that , x cannot be 1 because the fraction would = 0 .  Also, if x were less than 1, the numerator would be positive and denominator negative. Tuis, the fraction would be negative.....no good, either.

And x cannot equal 2, because then, the fraction is undefined.

Likewise, if x were greater than 2, the numerator would be negative and the denominator positive, resulting in the same situation as before - a negative fraction.

Thus, x must lie between 1 and 2, because this makes numerator and denominator negative [i.e., the fraction positive and > 0]

Hi Sabi,  

There are a couple of ways to do this but I will show you the way that I think is usually easiest.

You want to get rid of the fraction.  If this was an equal sign then you would note that x cannot be 2 and then you would multiply both sides by x-2.

It is a little different with an inequality because you would have to have 2 different scenarios, one for when x-2 is positive and one for when x-2 is negative.   SO

it is much better to multiply both sides by something that you KNOW is positive.

Multiply both sides by (x-2)^2   that HAS to be positive

$$\\\frac{1-x}{x-2}>0\\\\
\frac{1-x}{x-2}\times \frac{(x-2)^2}{1}>0\times \frac{(x-2)^2}{1}\\\\
\frac{1-x}{1}\times \frac{(x-2)}{1}>0\\\\
(1-x)(x-2)>0\\\\\\
--------------\\
$The only way 2 expressions multiply to be positive is if they are both positive or they are both negative.$\\\\
1-x>0\quad and \quad x-2>0 \qquad OR \qquad 1-x<0 \quad and \quad x-2<0\\\\
$ I can continue this if you want but there is a better way.$\\\\
------------------\\\\$$

$$\\(1-x)(x-2)>0\\
$If you let $ y=(1-x)(x-2)\\
$then the statement will be true when $y>0\\
y=(1-x)(x-2)\\
y=-1x^2+3x-2\\
$this is a parabola- the coefficient of $ x^2$ is -1$\\
$So it is concave down.$\\
$The middle bit will be above the x axis where y is positive.$\\
$The roots are x=1 and x=2$\\
SO\\
(1-x)(x-2)>0\quad when\;\;1

This is not as hard as it may at first look.

If you need me to explain anything please ask :)

melody thanks i think i've got it:)

That is good :))

Thanks Chris,  you are right, this question can be tackled very easily your way :)

This is a fairly simple example and it can be worked out just as CPhill says and it is easier that way too. :)

Melody i have a question about the first step you've done.

i know that in problems like this we cant "cancel" the denominator so how did you get (1-x)(x-2)

the only reason i can think about is that we can get rid of the denominator by reducing x-2 and one of the brackets?is that what you've done?

Hi Sabi,

Yes i guess I missed a couple of steps, :)

$$\frac{(1-x)}{(x-2)}>0$$

FIRST:   I must state that x-2 cannot be zero so  x cannot equal 2

Now I am allowed to multiply both sides by whatever I want and in so doing I will cancel the denpominator out.

If this was an equation (not an inequality) then I would multiply both sides by (x-2)  BUT that causes some problems here because x-2 could be negative or positive.

It would make life easier if we multiply by something that HAS to be positive.

So I multiplied by   $$(x-2)^2$$

$$\\\frac{(1-x)}{(x-2)}\times \frac{(x-2)^2}{1}>0\times \frac{(x-2)^2}{1}\\\\
\frac{(1-x)}{1}\times \frac{(x-2)}{1}>0\\\\
\frac{(1-x)(x-2)}{1}>0\\\\
(1-x)(x-2)>0$$

Now  the rest is the same as before :)

If you need more explanation - just ask :)

$$\\(1-x)(x-2)>0\\
$If you let $ y=(1-x)(x-2)\\
$then the statement will be true when $y>0\\
y=(1-x)(x-2)\\
y=-1x^2+3x-2\\
$this is a parabola- the coefficient of $ x^2$ is -1$\\
$So it is concave down.$\\
$The middle bit will be above the x axis where y is positive.$\\
$The roots are x=1 and x=2$\\
SO\\
(1-x)(x-2)>0\quad when\;\;1