Melody

Welcome to our web2.0calc forum Mischay:)

Ask your questions and we will help you as much as we can :))

Good work Rubenhh   

And welcome to the forum :))

      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 :)

$${\frac{\left({\frac{{\mathtt{3}}}{{\mathtt{4}}}}\right)}{\left({\frac{{\mathtt{8}}}{{\mathtt{9}}}}\right)}} = {\frac{{\mathtt{27}}}{{\mathtt{32}}}} = {\mathtt{0.843\: \!75}}$$

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 :)

No, their combined ages is 4 times Gemma age - 27

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 :)

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

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.

$$\\(x-1)(x-6)/(x-6)(x-2)\\\\
=\frac{(x-1)(x-6)}{(x-6)}\times(x-2)\\\\
=\frac{(x-1)(x-6)(x-2)}{(x-6)}\\\\
whereas\\\\
((x-1)(x-6))\;/\;((x-6)(x-2))\\\\
=\frac{(x-1)(x-6)}{(x-6)(x-2)}\\\\$$

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