#4**+15 **

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]

CPhill
Aug 1, 2015

#1**+15 **

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
Aug 1, 2015

#4**+15 **

Best Answer

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]

CPhill
Aug 1, 2015

#5**+5 **

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
Aug 2, 2015

#6**+5 **

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?

sabi92
Aug 2, 2015

#7**+10 **

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

Melody
Aug 3, 2015