Homework

[drainpipe32]

Try again guys. The answer in the book says

1
__ - x(x+3)
x

and yeah, I have nfi how to use my Cas as I only just got it a couple of weeks ago.

Didn't you use one in Year 10?

Gazza added 15 Minutes and 41 Seconds later...

I obv stuffed up my Cas work, it says

9(x^2-9)(3x+6)
---------------
x(2x)(x-3)

 

[Insomniac]

The book can eff off.

I did it right _|_

Insomniac added 5 Minutes and 7 Seconds later...

Actually, I'm gonna try to simplify this:

[ 27 (x^3 + 2x^2 - 9x - 18) ] / [ x (x^2 - x - 6) ]

Now, lets work from there.

That's the same thing as this.

 

[King Cricket]

I don't know what method you guys are using to solve this problem, but my method tells me that the answer is

27(x + 3)
----------
x

I guess it's an extremely stupid method, but here you go:

So, the problem is this:

x^2-9........3x+6.......x
-------..X.---------../----
x+2............x-3..........9

Well, the problem can be written as

(x+3)(x-3)(3)(x+2)(9)
------------------
(x+2)(x-3)(x)

That means we can cancel x+2 and x-3, as they are present in both the numerator and the denominator.

So, we are left with

(3)(9)(x+3)
-----------
x

ie, 27(x+3)
--------
x

Silly response, I know. But meh....

 

[Insomniac]

In summary, your book is stupid.

We all have different answers (which essentially equal the same thing).

 

[drainpipe32]

Ok. I've figured it out.

The answer is window

 

[Gilly Fan]

Thanks but no body actually told me how to do it. :P I'll give a simpler one.

3x^2..............y
-------....x..------
4y...............6x

and also, when you add algebraic fractions, how do you do it when there's 3? 2 is okay, but 3 i find hard.

 

[drainpipe32]

3x^2y
-------
24xy

=
x^3y^2
---------
8

I'l get that question soon.

 

[mattfb]

These last 2 pages make my head hurt.

 

[Gilly Fan]

These last 2 pages make my head hurt.

Enjoy year 10 while it lasts Matthew. Year 11 is a ████.

 

[Neo95]

Try do that with this.

x^2-9........3x+6.......x
-------..X.---------../----
x+2............x-3..........9

OMG what sort of problems are those I am now in 10th grade and I did those sort of problems in 8th grade.Those are easy as piss

 

[MuchMore]

Thanks but no body actually told me how to do it. :P I'll give a simpler one.

3x^2..............y
-------....x..------
4y...............6x

and also, when you add algebraic fractions, how do you do it when there's 3? 2 is okay, but 3 i find hard.

Things to note:

1) Believe the question is easy. It helps.

2) Don't take fractions separately. Just combine all numerators together and all denominators too.

In this case,

3.x^2.y
-------
4y.6x

3) Reduce the term you get i.e. Simply strike off the common expressions in numerator and denominator.
In this case, strike of
i)y in numerator for y in denominator
ii)3 from numerator and make the 6 in denominator as 2 (as 3x2 = 6 )
iii)reduce the power of x in numerator (which is equivalent to striking one of the two 'x' as x^2=x.x) for the only 'x' in denominator

4) Finally, after striking off the common expressions, what you get is:
x
--
4.2
which is equal to
x
--
8
or x/8

PS: I have tried to explain in the simplest form possible. Please ignore the excessive detail in case it sounds obvious to you. Otherwise, learn it 'coz its actually easy

 

[Funknath]

Thanks but no body actually told me how to do it. :P I'll give a simpler one.

3x^2..............y
-------....x..------
4y...............6x

and also, when you add algebraic fractions, how do you do it when there's 3? 2 is okay, but 3 i find hard.

3x^2y
------
24xy

= 6x ( strike of x from numerator and denomator and do same with y)
---
24

= x
--(6x4=24 ..hope you got what I mean )
8

 

[drainpipe32]

OMG what sort of problems are those I am now in 10th grade and I did those sort of problems in 8th grade.Those are easy as piss

Care to solve them then?

And for some people, piss isn't very easy :laugh

 

[Neo95]

Care to solve them then?

And for some people, piss isn't very easy :laugh

LOL I didn't bother to solve them because 2 guys had already solved them.Didn't see any point in solving what's already solved

To your second statement -Feel sympathy for such guys

 

[King Cricket]

Stuck with two algebra problems. Found them while randomly flicking through the pages of my Maths book. Anyone knows the solution?

1. If a^2-4a-1= 0, then find the value of a^2 + 1/a^2?
2. What's the value of x^3+y^3+xy when x+y= 1/3?