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# Anime Soc Chat XIV: where bad memes come to die watch

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1. (Original post by Cooro)
I know that. Just, skippy said she didn't like windmills and clogs, and that's kind of what they're famous for....
I think liquid was making fun of me too
2. (Original post by skipp)
I think liquid was making fun of me too
Fair enough.

I am going to go watch a suitably scary Doctor Who episode before bed ('Blink') so goodnight and happy halloween guys
3. As with fairies and elves, I have my own ideas of what these things should look like.

Werewolf:

Vampire:

Zombie:

(Okay, so she's actually a golem. It's not my fault all the real zombies in the series are bishonen).
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4. (Original post by skipp)
Alright...want to outline what it is you're tackling?
Linear equations. I'm not sure how far to simplify.

Here's one I've been confused by.

6(x-y) + 3(y-2x)

I'm not sure exactly how that can be simplified using distributive law.

At first I was thinking 6x-6y + 3y-5x. That doesn't seem to work though.
5. (Original post by Cooro)
Fair enough.

I am going to go watch a suitably scary Doctor Who episode before bed ('Blink') so goodnight and happy halloween guys
You too coo...enjoy doctor who
6. Night Rolo, don't let the angels bite
7. (Original post by Liquidus Zeromus)
Linear equations. I'm not sure how far to simplify.

Here's one I've been confused by.

6(x-y) + 3(y-2x)

I'm not sure exactly how that can be simplified using distributive law.

At first I was thinking 6x-6y + 3y-5x. That doesn't seem to work though.
With simlification you've got to group like terms...by value and polynomial order so in basic terms group the x's the y's and if there's any squared's or cubed's sub-group those too

If that makes any sense

If that looks confusing I can give an example by the way I've read it through and I have made it look a little confusing

You're nearly there but you've got a mistake in your expansion and you've got one more step to group...if you look you'v got +'s and -'s only there hint hint
8. (Original post by skipp)
With simlification you've got to group like terms...by value and polynomial order so in basic terms group the x's the y's and if there's any squared's or cubed's sub-group those too

If that makes any sense

If that looks confusing I can give an example by the way I've read it through and I have made it look a little confusing

You're nearly there but you've got a mistake in your expansion and you've got one more step to group...if you look you'v got +'s and -'s only there hint hint
That would be helpful I can generally get the hang of it just by looking at an example. There are some in this book, but they don't seem to point to solving this.

It's not uni work so it's fine to just simplify this one.

Oh, and am I on the right track if I simplify (5-2z)z by going 5z - 2z^2 ?

9. Noe
10. (Original post by Liquidus Zeromus)
That would be helpful I can generally get the hang of it just by looking at an example. There are some in this book, but they don't seem to point to solving this.

It's not uni work so it's fine to just simplify this one.

Oh, and am I on the right track if I simplify (5-2z)z by going 5z - 2z^2 ?
The one with the z's is perfect

My explanation looked a mess; so I'm going to cheat and give you an already laid out internet example:

http://www.math.com/school/subject2/.../S2U2L5GL.html

Remember expand then group like terms

my example looks like this:

(x+y) + (2 x^2 + y) (y+4)

(x+y) + (2y(x^2) + 8 (x^2) +(y^2) + 4y)

....

I'll continue if you like but I reckon the internet's clearer

Oneday I'll learn LaTex
11. (Original post by skipp)
The one with the z's is perfect

My explanation looked a mess; so I'm going to cheat and give you an already laid out internet example:

http://www.math.com/school/subject2/.../S2U2L5GL.html

Remember expand then group like terms

my example looks like this:

(x+y) + (2 x^2 + y) (y+4)

(x+y) + (2y(x^2) + 8 (x^2) +(y^2) + 4y)

....

I'll continue if you like but I reckon the internet's clearer

Oneday I'll learn LaTex
Ok, so...

For the original example,

6(x-y)+3(y-2x)

Could that be simplified into 9y-11x?

It's the second half which presents a problem, I think. I don't know how to simplify 3(2x).

Latex
12. (Original post by Liquidus Zeromus)
Ok, so...

For the original example,

6(x-y)+3(y-2x)

Could that be simplified into 9y-11x?

It's the second half which presents a problem, I think. I don't know how to simplify 3(2x).

Latex
Not quite...you're having a bit of trouble with adding and subtracting methinks;

B rackets
I ndices (to the power of..)
D ivide
M ultiply
A dd
S ubtract

expand the first bracket;
6 multiplied by x plus 6 multiplied by - y
put the answer in a bracket to make it clearer
call it expanded bracket one

ignore the +...that comes later

expand the second set of brackets
3 multiplied by y plus 3 multiplied by - 2x
put this answer in a new bracket
call this expanded bracket two

putting the two answers back either side of the plus you should now have:

(expanded bracket 1) + (expanded bracket 2)

This is the first step...see if you can work from there
13. (Original post by skipp)
Not quite...you're having a bit of trouble with adding and subtracting methinks;

B rackets
I ndices (to the power of..)
D ivide
M ultiply
A dd
S ubtract

expand the first bracket;
6 multiplied by x plus 6 multiplied by - y
put the answer in a bracket to make it clearer
call it expanded bracket one

ignore the +...that comes later

expand the second set of brackets
3 multiplied by y plus 3 multiplied by - 2x
put this answer in a new bracket
call this expanded bracket two

putting the two answers back either side of the plus you should now have:

(expanded bracket 1) + (expanded bracket 2)

This is the first step...see if you can work from there
Oh, BIDMAS. I know of that rule, but in some it didn't really seem to apply, I thought it meant working out the figures inside the bracket, with multiplication between the bracket and the number next to it after.

(6x + 6(-y)) + (3y+3(-2x)

=

6x - 2x + 3y - 6y?
14. (Original post by Liquidus Zeromus)
Oh, BIDMAS. I know of that rule, but in some it didn't really seem to apply, I thought it meant working out the figures inside the bracket, with multiplication between the bracket and the number next to it after.

(6x + 6(-y)) + (3y+3(-2x)

=

6x - 2x + 3y - 6y?
So close....

6x + 6(-y) + 3y + 3(-2x)

is correct but what is 3 times -2?

you'll kick yourself when you see the answer...
15. (Original post by skipp)
So close....

6x + 6(-y) + 3y + 3(-2x)

is correct but what is 3 times -2?

you'll kick yourself when you see the answer...
-6

Does that mean it can be simplified further than that? What was I doing wrong? Hmm, it's probably bleedingly obvious <_<
16. Hahaha, I've got it I think.

Let's say x=2

2x=4

4 x 3 = 12

6x2= 12

SO, 6x is the same as 3(2x)...

Brilliant!
17. (Original post by Liquidus Zeromus)
-6

Does that mean it can be simplified further than that? What was I doing wrong?
well now you have:

6x + 6(-y) + 3y -6x

= 6x - 6y +3y -6x

grouping like terms;

6x-6x+3y-6y

I'll use brakets to make it clear;

(6x - 6x) + (3y - 6y)

=0x - 3y

i.e. -3y

and bidmas always applies...it's the order you do things in

And yes...3(2x) is 6x from your example below
18. Wow...........maths
19. (Original post by caraniel)
Wow...........maths
It is 1am...and Kanamemo froze
20. (Original post by skipp)
well now you have:

6x + 6(-y) + 3y -6x

= 6x - 6y +3y -6x

grouping like terms;

6x-6x+3y-6y

I'll use brakets to make it clear;

(6x - 6x) + (3y - 6y)

=0x - 3y

i.e. -3y

and bidmas always applies...it's the order you do things in
Thank you alot, you've really cleared that up for me Cookie?

That removes an obstacle in my way.

I think we've scared everyone off again

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Updated: December 8, 2009
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