juriemagic

ElectricPavlov,

thank you very much!..much appreciated!

Melody,

yes, I understand..brrr...okay, it's a lot right now, I wil have a look at all the replies a bit later on...my brain is smoking!....but..I will check it and I hereby also say a BIIIG thank you!..

Hi Omi67,

I see your answers are different to ElectricPavlov's?...Thanx for your time..

It's like a flower blooming...it really is not that difficult after all. In a certain sense I feel I have wasted your time on this...I should have been able to see this..anyways, as always, you are appreciated!!..Thank you..

Hi melody,

oops, yes it is mentioned here that the range of f is $y>=-2$

sorry..

Melody,

mmm, yes, that makes all the sense in the world!...thank you..

Hey hey!!!..what do you know!!!!...whoop whoop!..I am really excited!!..Thank you sir!

Melody!!!...,

I do understand everything you had shared with me....really I do..thank you also for helping..again....

I do however need to ask something....with quadratic sequences we have a 1st and 2nd difference, okay, I know the second difference is ALWAYS Arithmetic by nature...The 1st one however, is that aslo always Aritmetic or can that be of a geometric nature as well?

ElectricPavlov..

gosh it really makes soo much sense now!!..I do however have a question but will post that with my response to Melody..please feel free to also comment...thanx a million!!!

Alan, Thanx for walking me through this...honestly, you will never realise how much this means!!..okay, x:y = 3:2

that gives $2x=3y$

which can go to $2x-3y=0$

Nothing can be factored..soooo, mabe then I should look at simultaneous equations???, meaning...

$2x-3y=0.....(1)$

$x+y=2$

$x=2-y.....(2)$

(2) in (1)

$2(2-y)-3y=0$

$4-2y-3y=0$

$-5y=-4$

$y= {4 \over 5}...(3)$

(3) in (2)

$x=2- {4 \over5}$

$x={6 \over5}$

So x then is 6\5 and y is 4\5

Is this correct maybe??..