thats whta ive done" note:x cannot be 6,1, and 2

(x-1)(x-6)/(x-6)(x-2)>0

(x-1)/(x-2)>0 /*(x-2)^2

(x-1)(x-2)>0 the cofficent of is positive

the root at x=1 x=2

answer:x<1 or x>2

but its a wrong anwer:(

sabi92
Aug 6, 2015

#7**+10 **

Oh, i see what you are saying - yes you are right. :)

I didn't bother about this because I knew the final answer would take care of it.

You just needed to work our qwhere the 'holes' might be by looking at the denominator at the very beginning.

Oh, and the top of the fraction is called the numerator :)

Melody
Aug 6, 2015

#1**+5 **

Hi Sabi,

Firstly x cannot be 2 or 6 because you cannot divide by 0 BUT it can be 1.

Your working and your answer is correct except you forgot that $$x\ne6$$

So the answer should be

$$\\x<1,\quad2 6\qquad x\in R\\\\

True for\;\;\;(-\infty,1),\;\;(2,6)\;\;(6,\infty)$$

Does this answer agree with yours now?

I will also take you to task for your setting out. You need more brackets!

I have added squar brackets where more brackets were needed :)

[(x-1)(x-6)]/[(x-6)(x-2)]>0

Melody
Aug 6, 2015

#3**+5 **

and i dont understand why i need more brackets

so i need to do (x-1)(x-6)(x-6)(x-2)>0 ?

sabi92
Aug 6, 2015

#4**+5 **

Yes, you are right, x cannot be 1 in the final answer becaus it will not make the inequality true but you didn't know that at the very beginning.

At the beginning you knew that x could not be 6 or 2 because it would make the denominator 0 and you cannot divide by zero.

$$\\(x-1)(x-6)/(x-6)(x-2)\\\\

=\frac{(x-1)(x-6)}{(x-6)}\times(x-2)\\\\

=\frac{(x-1)(x-6)(x-2)}{(x-6)}\\\\

whereas\\\\

((x-1)(x-6))\;/\;((x-6)(x-2))\\\\

=\frac{(x-1)(x-6)}{(x-6)(x-2)}\\\\$$

Melody
Aug 6, 2015

#5**+5 **

but when i did (x-1)(x-6)/(x-6)(x-2)>0 we can see immediately that x cant be 1 and 6 cause the nemirator cant be 0

sabi92
Aug 6, 2015

#6**+5 **

if

$$\\f(x)=\frac{(x-1)(x-6)}{(x-6)(x-2)}\\\\

f(1)=\frac{(0)(-5)}{(-5)(-1)}=0$$

There is no 0 in the denominator so x can be 1

Melody
Aug 6, 2015

#7**+10 **

Best Answer

Oh, i see what you are saying - yes you are right. :)

I didn't bother about this because I knew the final answer would take care of it.

You just needed to work our qwhere the 'holes' might be by looking at the denominator at the very beginning.

Oh, and the top of the fraction is called the numerator :)

Melody
Aug 6, 2015

#8**+5 **

and the next step after (x-1)(x-6)/(x-6)(x-2)>0 is (x-1)/(x-2)>0 ?

so if i get it right even though x cant be some number i need to put this number on the graph?

sabi92
Aug 6, 2015

#9**+5 **

Mine is finished. Did you need to graph it? Yes there will be a hole in the graph.

And you are still leaving out those brackets :)

Melody
Aug 6, 2015

#10**+5 **

i think i am ok with the graph but i am a bit confused with the brackets/

after this [(x-1)(x-6)]/[(x-6)(x-2)]>0 can i simlify it so i can get (x-1)(x-2)>0

or i need to do [(x-1)(x-6)]*[(x-6)(x-2)]>0?

sabi92
Aug 6, 2015

#11**+5 **

It is not multiply. It is divide.

after this [(x-1)(x-6)]/[(x-6)(x-2)]>0 can i simlify it so i can get (x-1)(x-2)>0

this statement is fine :)

Melody
Aug 6, 2015

#12

#13**+5 **

You are welcome Sabi.

You answer was almost perfect.

You just have to be careful about those pesky holes :)

Melody
Aug 6, 2015

#14**+5 **

I want to play with this one, too

[x^2 -7x + 6] / [x^2 - 8x + 12] > 0 factor top and bottom

[(x - 6) (x - 1)] / [ (x - 6) ( x - 2) ] > 0 Note that this simplifies to :

(x - 1) / ( (x - 2) > 0 however, we have to be aware that our grpah will have a "hole" at x = 6

The intervals of interest are (-∞, 1) (1, 2) (2, ∞ )

If you check, the two "outside" intervals make the equation the original eqaution true....again, we have to be aware that the last interval will have the "hole" where x = 6......here's the graph.....

https://www.desmos.com/calculator/yyh7iwefu0

Even though, the "hole" doesn't show up on the graph, you can hold down the left button on your mouse and drag along the curve....at x =6, you will see that Desmos displays "undefined" at that point...!!!

So ......the "true" answer to the problem is (-∞, 1) U (2, 6) U (6, ∞ )

CPhill
Aug 6, 2015