This is a continue from LaTeX Form (http://web2.0calc.com/questions/latex-form)

But the comments have reached over 100 and has alot of lag.

Perhaps you'll learn something from what me and Melody have discussed and (what I) learnt about LaTeX.

(Link's above)

**" This post is to help me with LaTeX so i have set it up with Melody.**

**Anyone can help.**

** **

** **

**But please Do Not Interupt With Unnecessary Things.. "**

Thank you.

MathsGod1 May 9, 2015

#57**+18 **

Longest CODE Yet.

\\1)\;Look\;at\;the\;\textcolor[rgb]{1,0,0}{denominators}\;of\;the\;fractions:\qquad\quad\textcolor[rgb]{0,0,1}{3,\;4\;and\;6}\\\\

2)\;They\;are\;all\;factors\;of\;12\;so\;change\;each\;fraction\;(make\;\textcolor[rgb]{1,0,0}{equivalent}\;fractions)\;so\;the\;\textcolor[rgb]{1,0,0}{denominator\;is\;12}\qquad\quad\frac{5}{6}\;=\;\frac{10}{12}\;,\;\frac{3}{4}\;=\;\frac{9}{12}\;,\;\frac{2}{3}\;=\;\frac{8}{12}\\\\

3)\;Now\;order\;the\;fractions\;by\;comparing\;their\;\textcolor[rgb]{1,0,0}{numerators}.\qquad\quad\frac{8}{12}\;,\;\frac{9}{12}\;,\;\frac{10}{12}\\\\

4)\;Put\;the\;fractions\;back\;into\;their\;\textcolor[rgb]{1,0,0}{original form}.\qquad\quad\textcolor[rgb]{1,0,0}{\frac{2}{3}\;,\;\frac{3}{4}\;,\;\frac{5}{6}}

$$\\1)\;Look\;at\;the\;\textcolor[rgb]{1,0,0}{denominators}\;of\;the\;fractions:\qquad\quad\textcolor[rgb]{0,0,1}{3,\;4\;and\;6}\\\\

2)\;They\;are\;all\;factors\;of\;12\;so\;change\;each\;fraction\;(make\;\textcolor[rgb]{1,0,0}{equivalent}\;fractions)\;so\;the\;\textcolor[rgb]{1,0,0}{denominator\;is\;12}\qquad\quad\frac{5}{6}\;=\;\frac{10}{12}\;,\;\frac{3}{4}\;=\;\frac{9}{12}\;,\;\frac{2}{3}\;=\;\frac{8}{12}\\\\

3)\;Now\;order\;the\;fractions\;by\;comparing\;their\;\textcolor[rgb]{1,0,0}{numerators}.\qquad\quad\frac{8}{12}\;,\;\frac{9}{12}\;,\;\frac{10}{12}\\\\

4)\;Put\;the\;fractions\;back\;into\;their\;\textcolor[rgb]{1,0,0}{original form}.\qquad\quad\textcolor[rgb]{1,0,0}{\frac{2}{3}\;,\;\frac{3}{4}\;,\;\frac{5}{6}}$$

MathsGod1 May 18, 2015

#1**0 **

This is a continue to LaTeX Form (http://web2.0calc.com/questions/latex-form)

But the comments have reached over 100 and has alot of lag.

Perhaps you'll something from what me and Melody have discussed and (I) learnt about LaTeX.

(Link's above)

**" This post is to help me with LaTeX so i have set it up with Melody.**

**Anyone can help.**

** **

** **

**But please Do Not Interupt With Unnecessary Things. "**

Thank you.

TitaniumSoldier1 May 9, 2015

#2**0 **

omg, its the math god again. i need to hide somewhere!(sarcasm)

only german and aussie cowards run and hide.

the troops from the Empire of Royality are far from ordinary, unlike you guys, we can defeat even gods, if we spawn father david and Mother nature

theyre both with us!

TitaniumSoldier1 May 9, 2015

#3**+18 **

I'll start it by continuing off from last time.

$$\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\

=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\

=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\

=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\

=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\

=(9Cr)(z^{((18-3r)/2)}*(-2)^r)

\\The\:constant\:term\:will\:be\:when\\

(18-3r)/2=0\\18-3r=0\\r=6$$

This is the entire coding correct...

Hopefully it shows...My laptop might not be able to take in all this LaTeX.

\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\

=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\

=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\

=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\

=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\

=(9Cr)(z^{((18-3r)/2)}*(-2)^r)

\\The\:constant\:term\:will\:be\:when\\

(18-3r)/2=0\\18-3r=0\\r=6

MathsGod1 May 9, 2015

#5**+18 **

$$\\\left(z-\dfrac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\

=(9Cr)(z)(9-r)\left(\dfrac{z^9}{z^r}\right)\left(\dfrac{(-2)^r}{(\sqrt{z}^r}\right)\\

=(9Cr)\left(\dfrac{z^9*(-2)^r}{z^r*(sqrt{z}^r)}\right)\\

=(9Cr)\left(\dfrac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\

=(9Cr)\left(\dfrac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\

=(9Cr)(z^{((18-3r)/2)}*(-2)^r)

\\The\:constant\:term\:will\:be\:when\\

(18-3r)/2=0\\18-3r=0\\r=6$$

Is there a way to leave larger gaps it seems so cramped.

MathsGod1 May 9, 2015

#6**+13 **

Absolutely :)

// for next line //// for bigger space ////// for bigger again etc.

I like mine well spaced so usually I use //// Occasionally i will use even more. :)

The only thing that I can see wrong with that is one missing backslash in front of a sqrt.

Great job MG. What do you want to do next?

Melody May 9, 2015

#7**+13 **

I didn't even notice thanks i will use \\\\ and i will put in \sqrt

$$\\\\\left(z-\frac{2}{\sqrt{z}}\right)\\\\The\:general\:term\:is\\\\

=(9Cr)(z)(9-r)\left(\frac{z^9}{z^r}\right)\left(\frac{(-2)^r}{(\sqrt{z}^r}\right)\\\\

=(9Cr)\left(\frac{z^9*(-2)^r}{z^r*(\sqrt{z}^r)}\right)\\\\

=(9Cr)\left(\frac{*(-2)^r}{z^(2r/2)*(z)^(r/2)}\right)\\\\

=(9Cr)\left(\frac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\\\

=(9Cr)(z^{((18-3r)/2)}*(-2)^r)

\\\\The\:constant\:term\:will\:be\:when\\\\

(18-3r)/2=0\\18-3r=0\\r=6$$

What to do next.....Hmmmm...

I dont think my math knowledge has exceeded far enough to think of a new topic :)

MathsGod1 May 9, 2015

#9**+13 **

**Hi MathsGod1,**

In response to your private message.

I had to search for the bullet point on the internet. I just googled LaTex, fat dot.

This was useful http://tex.stackexchange.com/questions/47060/placeholder-for-variable-as-in-fx

I also looked through the **forum Latex post** as I know Heureka did a post on dots in LaTex just recently.

You are getting anough knowledge that you can start doing this as well.

This is what I found for you. There are many different version of Latex. Not everything will work here (or anywhere else. Again this is a good reason to know many ways to acheive the same thing.

Using | to give a horizontal line is something I discovered totally by accident. I have never seen it work anywhere other than on this forum but I have found it handy

$$\bullet\: )\qquad $bushes$\\

|\:)\qquad $benches$$$

\bullet\: )\qquad $bushes$\\

|\:)\qquad $benches$

.Melody May 10, 2015

#10**+10 **

**Hi again MG**

One very important command in LaTex that i have not shown you yet is **array**,

here is a little example

$$\begin {array}{rll}\\

4+5+9-7&=&y-19\\

11&=&y-19\\

30&=&y\\

y&=&30\\

\end{array}$$

See how everything is lined up

This is the code. The ampersand **&** is the symbol for the lining up.

**{rll}** means that first part is pushed **right** onto the ampersand.

the = sign (or what ever is next) is pushed **left** on against the &

and the last bit is pushed **left** onto the second &

You can do a great many things with the array function - Heureka makes EXCELLENT use of it !!

\begin {array}{rll} EDITED

4+5+9-7&=&y-19\\

11&=&y-19\\

30&=&y\\

y&=&30\\

\end{array}

Melody May 10, 2015

#11**+10 **

"The thing i messaged you doesn't work..."

What doesnt work MG ?

what you have done looks really good to me :)

Soon you will be teaching me - I am being serious, if you keep learning this quickly I will be asking you to research and then show me how to do things in LaTex

I really need helpers on this forum. NinjaDevo is the only non-moderator who has really helped.

I am very grateful to him for everything that he did.

I am hoping that one day you will be able and willing to help here too :)

Melody May 10, 2015

#12**+13 **

Suree I'd love too help here!

The question involves 16 benches and 9 bushes

benches (4x4) so a huge square

And imagine tinier squares inside made up of 4 benches and in the centre of those benches there will be a huge bush and if i do it for all of the 4 benches there'd be 9 bushes.

I used Imgur to show you

I found a soloution without Latex Lol.

MathsGod1 May 10, 2015

#13**+5 **

I think that you have posted this twice MathsGod1.

You know I get annoyed when people do that.

Melody May 10, 2015

#15**+8 **

You have posted this here

http://web2.0calc.com/questions/tricky-maths-question

I suppose this version on this thread is not really finished, you haven't actually stated the question here.

Melody May 10, 2015

#16**+13 **

I actually wanted to post the question on the forum but I didn't know how to display the question so I asked you for help .

MathsGod1 May 10, 2015

#18**+13 **

Yes you can do an enormous amount of things with the array function.

It may be time for you to really practice the LaTex that you have already acquired for a while before you try too hard to add to it. You have learned an enormous amount in a very short space of time.

Have a look through the LaTex thread or just look at mine and Heureka's answers if you want to add more to your knowledge base.

Also there are the automated functions that admin added in the first place - like colours, italics, bold and others. Have a look at what is easily available to you in the LaTex ribbon.

Melody May 10, 2015

#20**+10 **

There is a ribbon at the top of the open LaTex box.

There is some functions that have already been loaded in there that you can use :)

Melody May 10, 2015

#22**+10 **

Hi Melody,

In LaTeX when you use $ for spaces in my coding it doesn't work..For the first line it does but for the rest it doesn't leave spaces between my words so i had to stick with \; , however I'd prefer to just use the dollar sign as it's easier...

E.g. recent code.

\\\\$Let's hope Steve has high stamina\\\\

\\So 40m every two minutes (2\times30=60 minutes) so\;40m\times30= a\;hour = 1200\;meters\\

\\$1200 meters\times\;24\;for\;a\;day\;=\;28000\;meters$\\

\\$28000\times\;365\;for\;a\;year\;=\;10512000\;meters$\\

\\10512000m\;is\;a\;year,\;a\;century\;is\;100\;years\;\\

\\10512000m\times\;100=1051200000m\\

\\\\Convert\;1051200000m\;into\;centermeters\;=\;105120000000 cm\\\\

Your\;answer\;is\;105120000000 cm

MathsGod1 May 12, 2015

#23**+10 **

Hi MG,

I never use a $ for spaces.

But LaTex does a lot of strange things and it behaves slightly differently in different forums too. This is because there are many different versions of it around. It is best to know a lot of methods for doing the same thing because if one method doesn't work another probably will.

I saw that Latex of yours on the forum - it looked great.

I think it is time for you to start really practicing the array functions.

That can be extended to do a lot of things but I think that you need some practice with it first.

How good are you at solving equations?

Can you solve this one and present it in latex using the array function?

(5x+6)/12=-2

Melody May 13, 2015

#24**+13 **

The equations where you have to make it as small as possible.

I recently learnt that and I'm not doing the best.

Well, honestly i think it's the teacher...I have two, one that (like you) has a lot of passion in me being a good a mathematician

The other **teacher's** **teaching technique** in my opinion, is bad... He's foreign, and he just gives the equations and shows the step, but he doesn't explain well.

Talking of maths teachers, my math teacher told me I'm getting invited to this University for an award on June 3rd for winning this maths competition i was a few out of 40,000+ !

I haven't learnt a lot but i'll try.

MathsGod1 May 13, 2015

#25**+13 **

$$\begin array{rrl}

(5x+6)\div12&=&-2\\

(5x)\div12&=&4\\

5x&=&48

\end{array}$$

\begin array{rrl}

(5x+6)\div12&=&-2\\

(5x)\div12&=&4\\

5x&=&48

\end{array}

*Puff*

MathsGod1 May 13, 2015

#26**+13 **

Try putting the first 'array' in parentheses. :/

also you forget the +6 on the second line. So your answer is not going to be correct. :(

Melody May 13, 2015

#27

#28**+13 **

Ok ill try when I get home.

But, when I took out the six its because I had done +6 I think. But as it's in brackets you can't do that you have to also take out the x.

Sorry, like I said I was never taught this well, I have never understood it properly .

MathsGod1 May 14, 2015

#29**+13 **

Ok ill try when I get home.

But, when I took out the six its because I had done +6 I think. But as it's in brackets you can't do that you have to also take out the x.

Sorry, like I said I was never taught this well, I have never understood it properly .

MathsGod1 May 14, 2015

#30**+13 **

(5x+6)/12=-2

$$\frac{5x+6}{12}=-2$$

When you are solving equations you need to constantly ask yourself - "What do I want to get rid of"

[hint: you can sort of thing of PEDMAS and do it BACKWARDS]

The aim here is to eventually get the x all by itself.

But first you will need to get rid of the 12.

You can do this by multiplying BOTH sides by 12.

It is important to keep the equation balanced. What you do to one side you need to do to the other side as well.

so

$$\begin{array}{rll}\\

\frac{5x+6}{12}&=&-2\\\\

5x+6&=&-24\\\\

\end{array}$$

now you need to get rid of the tacked on 6

The opposite of + is minus so minus it from both sides

$$\begin{array}{rll}\\

\frac{5x+6}{12}&=&-2\\

5x+6&=&-24\\

5x+6-6&=&-24-6\\

5x&=&-30\\

5*x&=&-30\\

\end{array}$$

Now you need to get rid of the 5 and the opposite of * is divide (which is the same as a fraction)

$$\begin{array}{rll}\\

\frac{5x+6}{12}&=&-2\\

5x+6&=&-24\\

5x+6-6&=&-24-6\\

5x&=&-30\\

5*x&=&-30\\

\frac{5*x}{5}&=&\frac{-30}{5}\\

x&=&-6\\

\end{array}$$

And that is how it is done.

Always make sure that your equations balance that is really important!

Here is another for you to do

**3x-7=2*-4**

Hint: Simplify the right hand side first.

Melody May 14, 2015

#31**+18 **

Thanks for your great explanation!

$$\begin {array}{rrl}

3x-7&=&2\times-4\

3x-3&=&2\times\

3x&=&-6\

-6/div3&=&x

x&=&-2

\end{array}$$

\begin {array}{rrl}3x-7&=&2\times-4\3x-3&=&2\times\3x&=&-6\-6/div3&=&xx&=&-2\end{array}

.....

MathsGod1 May 14, 2015

#32**+13 **

Hi Mg,

for starters you have the slash in front of 'div' the wrong way around.

fix that and I will take another look :)

I think you also need \\ after {rll}

Plus you have (rrl } instead of {rll}

Ok this is how I wanted it to look

**3x-7=2*-4**

**$$\begin{array}{rll}\\ 3x-7&=&2*-4\\ 3x-7&=&-8\\ 3x-7+7&=&-8+7\\ 3x&=&-1\\ \frac{3x}{3}&=&\frac{-1}{3}\\ x&=&\frac{-1}{3}\qquad or \\ x&=&-\frac{1}{3}\\ \end{array}$$**

**this is my coding**

\begin{array}{rll}\\

3x-7&=&2*-4\\

3x-7&=&-8\\

3x-7+7&=&-8+7\\

3x&=&-1\\

\frac{3x}{3}&=&\frac{-1}{3}\\

x&=&\frac{-1}{3}\qquad or \\

x&=&-\frac{1}{3}\\

\end{array}

**Here is another one to try**

(y/3)+5=-9

hint: Subtract 5 from both sides, THEN multiply both sides by 3

If you want to concentrate on the maths instead of the coding you can use the maths formula display and just do it one line at the time, that will be easier

Melody May 14, 2015

#33**+13 **

$$\begin{array}{rrl}\\

(y/3)+5&=&-9\qquad(take\;away\;5)\\

(y/3)&=&-4\qquad(times\;3)\\

y&=&-12\\

\end{array}$$

MathsGod1 May 14, 2015

#34**+8 **

Your logic and your LaTex is great MG, unfortunately -9-5 does not equal -4

Could you correct it please.

I am not sure if it is a careless mistake of if you have problems with directed numbers.

If you have problems tell me and I'll teach you

This post might help you :)

http://web2.0calc.com/questions/it-just-doesn-t-make-sense-to-me#r2

Melody May 15, 2015

#35**+18 **

$$\begin{array}{rrl}\\

(y/3)+5&=&-9\qquad(take\;away\;5)\\

(y/3)&=&-14\qquad(times\;3)\\

y&=&-42\\

\end{array}$$

MathsGod1 May 15, 2015

#36**+13 **

http://web2.0calc.com/questions/33-2x-5-9x

First use of array.

My question is, when i give the extra info by the side in bracket can you also allign the brackets?

MathsGod1 May 16, 2015

#37**+13 **

YES YOU CAN MG

I have changed you second list of arguments to {rlll}

This means

push the first things to the right this is right align BEFORE the first ampersand (&)

then allign the next left that is the equal sign this is left align AFTER the first ampersand

then align the next left that is the answer this is left align AFTER the second ampersand

then align the next left that is the comment this is left align AFTER the third ampersand

See how i have added a third ampersand.

Heureka makes fantastic use of aligning in many of his posts!

$$\begin{array}{rlll}\\

(y/3)+5&=&-9\qquad&(take\;away\;5)\\

(y/3)&=&-14\qquad&(times\;3)\\

y&=&-42\\

\end{array}$$

\begin{array}{rlll}\\

(y/3)+5&=&-9\qquad&(take\;away\;5)\\

(y/3)&=&-14\qquad&(times\;3)\\

y&=&-42\\

\end{array}

Melody May 16, 2015

#38**+13 **

So when you do that. I've been staring at the code for a long time now but I GET IT :D.

You add rrlll instead of rrl and after qquad (before the brackets) you add an ampersand.

I've done it on the same post (first comment).

\begin{array}{rrlll}

33-2x&=&5-9x\qquad&(- 5)\\

28-2x&=&-9x\qquad&(+9x)\\

28+7x&=&0\qquad&(which\;means)\\

28&=&7x\qquad&(28\div7=4)\\

x&=&4\\

\end{array}

MathsGod1 May 16, 2015

#39**+13 **

http://web2.0calc.com/questions/33-2x-5-9x

I asked a question for Alan on the post but he hasn't repsonded I'm wondering in the equation why does the -7 all of a sudden turn negative?

thanks

MathsGod1 May 16, 2015

#40**+13 **

I can't comment unless I see it MG.

Please put the output and the code in the same post - Like I did in my last post :)

I've done it on the same post (first comment). What same post are you talking about MG ?

\begin{array}{rrlll}

33-2x&=&5-9x\qquad&(- 5)\\

28-2x&=&-9x\qquad&(+9x)\\

28+7x&=&0\qquad&(which\;means)\\

28&=&7x\qquad&(28\div7=4)\\

x&=&4\\

\end{array}

$$\begin{array}{rrlll}

33-2x&=&5-9x\qquad&(- 5)\\

28-2x&=&-9x\qquad&(+9x)\\

28+7x&=&0\qquad&(which\;means)\\

28&=&7x\qquad&(28\div7=4)\\

x&=&4\\

\end{array}$$

**NO MG It is still not right :/**

**I shall do it for you**

$$\begin{array}{rlll}

33-2x&=&5-9x\qquad&\\

33-5-2x&=&5-5-9x\qquad&$(- 5 on both sides)$\\

28-2x&=&-9x\qquad\\

28-2x+9x&=&-9x+9x\qquad&$(+9x on both sides)$\\

28+7x&=&0\qquad&\\

28+7x-7x&=&0-7x\qquad&$-7x from both sides$\\

28&=&7x\qquad&\\

28&=&7x\qquad&$Divide both sides by 7$\\

\frac{28}{7}&=&\frac{7x}{7}\\

4&=&x\\

x&=&4\\

\end{array}$$

NOW when you added 9x to both sides (and I did the same as you)

it would have been better to add 2x to both sides because then all the xes would have been on the same side from that point.

I am sorry MG. I have made this more dificult than necessary for you.

I should have started out with easier ones and explained it better. :(

At every step you need to think "What do I want to get rid of' That is the real trick here!

Melody May 16, 2015

#41**+18 **

I understand your method and you'd have to think when to add numbers randomnly.

I don't know how i got mines wrong it's not the same as your method but i got 4 in the end and the method got me there...?

MathsGod1 May 16, 2015

#42**+8 **

You need to provide the link to your one MG - they should have all been on the same thread in the first place.

I remember seeing your one - it was correct. We will do more. :)

Melody May 16, 2015

#44**+10 **

Hi MG,

Alright let's work through a little example one step at a time.

I know that you can already do questions like this but I want you to think about **WHY** you are doing each line.

**4x+7=13**

Now, the eventual aim is to get the x all by itself on one side and all the numbers on the other side.

How can this be done?

First you need all the lots of x by itself. In this case you need 4x by itself.

You have to get rid of the tacked on +7

the opposite of adding 7 is minusing 7

**SO you will need to minus 7 from both sides**

4x+7 = 13

-7 -7

4x = 6

Now you need to get rid of the 4

There is an invisable * between the 4 and the x so you have

4*x = 6 which is the same as

x*4 = 6

NOW the opposite of multiplying by 4 is dividing by 4 and divide is the same as a fraction line.

**So divide both sides by 4 (put both sides over 4 )**

$$\\\frac{4x}{4}=\frac{6}{4}$$

Now cancel down both sides

$$\\\frac{\not{4}x}{\not{4}}=\frac{\not{6}^3}{\not{4}^2}\\\\

x=\frac{3}{2}\\\\

x=1\frac{1}{2}$$

----------------------------------------------------

Now here is one a bit similar for you.

Don't let the LaTex get in the way of the mathematics. This question is more about the algebra.

In this question I want to see the procedure. LaTex layout is secondary.

I want to see excellent reasoning - like mine above - for every step.

**5x-8=20**

Melody May 17, 2015

#45**+13 **

Before i start, i want clarify something, when you got to the fraction, you simplied it to a improper fraction then to a mixed number, right?

:)

$$\begin{array}{rrlll}\\\\

5x-8&=&20\qquad&(To\;get\;X\;alone\;+\:8\;so\;-8\;goes\;away)\\\\

5x&=&28\qquad&(5x\:=\;x\times5\;so\;\div\;by\;5)\\\\

\frac{5x}{5}&=&\frac{28}{5}\qquad&(Now\;cancel\;out)\\\\

x&=&\frac{28}{5}\qquad&(Improper\;fraction\;to\;Mixed\;number)\\\\

x&=&5\frac{3}{5}\\\\

\end{array}$$

I don't know if there is anything i have missed out or added, but i think this is right.

CODE:

\begin{array}{rrlll}\\\\5x-8&=&20\qquad&(To\;get\;X\;alone\;+\:8\;so\;-8\;goes\;away)\\\\5x&=&28\qquad&(5x\:=\;x\times5\;so\;\div\;by\;5)\\\\\frac{5x}{5}&=&\frac{28}{5}\qquad&(Now\;cancel\;out)\\\\x&=&\frac{28}{5}\qquad&(Improper\;fraction\;to\;Mixed\;number)\\\\x&=&5\frac{3}{5}\\\\\end{array}

MathsGod1 May 17, 2015

#46**+8 **

Yes - that is excellent MG (the mathematics AND the Latex)

Here is another if you are interested

$$\frac{x}{3}=-4$$

Remember I want the logic behind what you do :)

Melody May 17, 2015

#47**+8 **

A task is always great to do when you're making progress!

(In my head)

$$\begin{array}{rrlll}\\

\frac{x}{3}&=&-4\qquad&(divide\;both\;side\;by\;3)\\

x&=&-12\\\\

\end{array}$$

I'm kind of curious on this question, isn't it already done, like the last question when i was left in that state i just simplified not divide, so the answer is (maybe) -4.

Oops EDIT:

$$\begin{array}{rrlll}\\

\frac{x}{3}&=&-4\qquad&(Divide\;by\;3)\\

x&=&-1.3\\

\end{array}$$

It seems like the answer is -4...

EDIT: I'm sure the answer is not -4.

I'll be back I need to have breakfast

(I wake up to maths)

(PS Its a good thing :D)

MathsGod1 May 17, 2015

#48**+8 **

Nope not right MG - lets think about this

$${\frac{{\mathtt{x}}}{{\mathtt{3}}}} = -{\mathtt{4}}$$

is the same as x divided by 3

you need to get rid of the divided by 3 to get x by itself.

what is the opposite of dividing by 3?

If I am confusing you, think about this.

$$6\div3 \;\;\fbox{\begin{minipage}{0.6cm}\hfill\vspace{0.6cm}\end{minipage}}\;\;3=6$$ What symbol goes into the box to make this statement true?

Melody May 17, 2015

#49**+5 **

I have a question for **Nauseated or Heureka**

I just made a LaTex box for MathsGod1 but there was a lot of coding involved that I just copied from our LaTex thread. I seemed to recall that Nauseated made a box like this quite recently and that the coding for his was much simpler. Do either of you know the coding that I am talking about?

Melody May 17, 2015

#50**+13 **

Sorry for taking so long i had to clean.

Well, now that you put it that way it sounds a lot easier!

$$\begin{array}{rrlll}\\

\frac{x}{3}&=&-4\qquad&(Times\;by\;3,\;because\;the\;inverse\;of\;divide\;is\;times)\\

x&=&-12\\

\end{array}$$

Hmm..Also i'm not sure about the coding.

MathsGod1 May 17, 2015

#51**+8 **

Yes that is good.

Your coding is also good.

Untill I am totally convinced that you are not fluking it I would like you to put ALL the working in.

So lets have a look at you last answer

$$\begin{array}{rrlll}\\\frac{x}{3}&=&-4\qquad&(Times\;by\;3,\;because\;the\;inverse\;of\;divide\;is\;times)\\x&=&-12\\\end{array}$$

I would like to see you put an extra line of working in...like this

$$\begin{array}{rlll}\\\frac{x}{3}&=&-4\qquad&(Times\;by\;3,\;because\;the\;inverse\;of\;divide\;is\;times)\\\\

\frac{x}{3}\times 3&=&-4\times 3\\\\

x&=&-12\\

\end{array}$$

-----------------------------------------

If you want to get fancy with your LaTex you could even show the cancelling

$$\frac{x}{\not{3}}\times \not{3}&=&-4\times 3\\\\$$

\frac{x}{\not{3}}\times \not{3}&=&-4\times 3\\\\

Here is another one. There is no nedd for you to rqce back with an answer, answer when (and if) you want :)

$$-6x+4=-2$$

remember, you get rid of tack ons first, then you get x on its own second. :)

Melody May 17, 2015

#52**+13 **

$$$$$$\setlength{\fboxsep}{20pt}

\fbox{Hi Melody. Here are a few ways to do "quick" boxes} \\\

\setlength{\fboxsep}{20pt}

\fbox{} \

\setlength{\fboxsep}{15pt}

\fbox{} \

\setlength{\fboxsep}{10pt}

\fbox{} \

\setlength{\fboxsep}{5pt}

\fbox{} \\\

\setlength{\fboxsep}{20pt}

\fbox{ } \leftarrow ${Comds are the same, except for} \

\setlength{\fboxsep}{15pt}

\fbox{ } \leftarrow ${ one space for additional [x]pt width}

\setlength{\fboxsep}{10pt}

\fbox{ } \

\setlength{\fboxsep}{5pt}

\fbox{ } \\\

\setlength{\fboxrule}{3pt}

\setlength{\fboxsep}{6pt}

\fbox{ } \leftarrow ${ fboxrule Comds affect border boldness}\\

\setlength{\fboxrule}{5pt}

\setlength{\fboxsep}{15pt}

\fbox{? } \leftarrow ${ Text (if used) is Center justified } \\

\setlength{\fboxrule}{3pt}

\setlength{\fboxsep}{15pt}

\fbox{ End of Examples from }

\setlength{\fboxrule}{5pt}

\setlength{\fboxsep}{20pt}

\fbox{ Your friendly neighborhood Troll }$$

Nauseated May 17, 2015

#53**+13 **

Ok I will try it.

First of all, that is so **COOL** Nauseated but it must take in alot of Coding.

EDITED

$$\begin{array}{rrlll}\\

-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\

-6x-4&=&-2-4\\\\

-6x&=&-6\qqquad&Divide\;6\;on\;both\;sides\\\\

\frac{-6x}{6}&=&\frac{6}{6}\qquad&simplify\\\\

\frac{-6x}{\not{6}}&=&\frac{6}{\not{6}}\\\\

-x&=&\frac{1}{1}

\end{array}$$

I think...Sorry i had to rush this because i was half way and my sis needs the laptop i'll think about it after.

\begin{array}{rrlll}\\

-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\

-6x-4&=&-2-4\\\\

-6x&=&-6\qqquad&Divide\;6\;on\;both\;sides\\\\

\frac{-6x}{6}&=&\frac{6}{6}\qquad&simplify\\\\

\frac{-6x}{\not{6}}&=&\frac{6}{\not{6}}\\\\

-x&=&\frac{1}{1}

\end{array}

MathsGod1 May 17, 2015

#55**+5 **

It does not pay to rush MG

You said you were going to -4 but then you added 4 instead!

Plus you used the latex for cancelling when there was no cancelling to do!

You need to edit this one - or redo it and delete the original. :)

Melody May 17, 2015

#56**+13 **

It has been edited.

Also Melody, do you know when the next system update is so the points update.

Or can only Admin do that.

MathsGod1 May 18, 2015

#57**+18 **

Best Answer

Longest CODE Yet.

\\1)\;Look\;at\;the\;\textcolor[rgb]{1,0,0}{denominators}\;of\;the\;fractions:\qquad\quad\textcolor[rgb]{0,0,1}{3,\;4\;and\;6}\\\\

2)\;They\;are\;all\;factors\;of\;12\;so\;change\;each\;fraction\;(make\;\textcolor[rgb]{1,0,0}{equivalent}\;fractions)\;so\;the\;\textcolor[rgb]{1,0,0}{denominator\;is\;12}\qquad\quad\frac{5}{6}\;=\;\frac{10}{12}\;,\;\frac{3}{4}\;=\;\frac{9}{12}\;,\;\frac{2}{3}\;=\;\frac{8}{12}\\\\

3)\;Now\;order\;the\;fractions\;by\;comparing\;their\;\textcolor[rgb]{1,0,0}{numerators}.\qquad\quad\frac{8}{12}\;,\;\frac{9}{12}\;,\;\frac{10}{12}\\\\

4)\;Put\;the\;fractions\;back\;into\;their\;\textcolor[rgb]{1,0,0}{original form}.\qquad\quad\textcolor[rgb]{1,0,0}{\frac{2}{3}\;,\;\frac{3}{4}\;,\;\frac{5}{6}}

$$\\1)\;Look\;at\;the\;\textcolor[rgb]{1,0,0}{denominators}\;of\;the\;fractions:\qquad\quad\textcolor[rgb]{0,0,1}{3,\;4\;and\;6}\\\\

2)\;They\;are\;all\;factors\;of\;12\;so\;change\;each\;fraction\;(make\;\textcolor[rgb]{1,0,0}{equivalent}\;fractions)\;so\;the\;\textcolor[rgb]{1,0,0}{denominator\;is\;12}\qquad\quad\frac{5}{6}\;=\;\frac{10}{12}\;,\;\frac{3}{4}\;=\;\frac{9}{12}\;,\;\frac{2}{3}\;=\;\frac{8}{12}\\\\

3)\;Now\;order\;the\;fractions\;by\;comparing\;their\;\textcolor[rgb]{1,0,0}{numerators}.\qquad\quad\frac{8}{12}\;,\;\frac{9}{12}\;,\;\frac{10}{12}\\\\

4)\;Put\;the\;fractions\;back\;into\;their\;\textcolor[rgb]{1,0,0}{original form}.\qquad\quad\textcolor[rgb]{1,0,0}{\frac{2}{3}\;,\;\frac{3}{4}\;,\;\frac{5}{6}}$$

MathsGod1 May 18, 2015

#58**+10 **

That is really impressive coding MG, the maths for it is good too. :))

-----------------------

Back to your equation though it is still not quite right.

$$\begin{array}{rrlll}\\

-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\

-6x-4&=&-2-4\\\\

-6x&=&-6\qqquad&Divide\;6\;on\;both\;sides\\\\

\frac{-6x}{6}&=&\frac{6}{6}\qquad&simplify\\\\

\frac{-6x}{\not{6}}&=&\frac{6}{\not{6}}\\\\

-x&=&\frac{1}{1}

\end{array}$$

In your second row you 'forgot' to put the +4 in on the LHS (left hand side)

The 3rd row is good but you could have divided by -6 to get x on its own straght away.

Row 4 Your cancelling is not correct.

$$\\\frac{-\not{6}x}{\not{6}}=\frac{\not{6}}{\not{6}}\\\\

-x=\frac{1}{1}\\\\

-x=1\qquad $Multiply both sides by -1$\\

-x*-1=1*-1\\

x=-1$$

Mine is not lined up beautifully like yours but I have done some corrections :)

You must end up with x by itself on the LHS when you are finished :)

Melody May 19, 2015

#60**0 **

Oh. Ok. I get it, well slightly more now!

The thing I don't get is you said there's something wrong with my seconds row in +4. But i thought you have to minus 4 to get rid of of it on the one side.

MathsGod1 May 19, 2015

#61**+10 **

Mmm,

lets see

$$\begin{array}{rrlll}\\-6x+4&=&-2\qquad&(-4\;to\;get\;X\;alone)\\\\

-6x\textcolor[rgb]{1,0,0}{+4}-4&=&-2-4 &\textcolor[rgb]{1,0,0}{\mbox{You still need the +4 here because you have not cancelled it out yet} }\\\\

-6x&=&-6\\\\

etc

\\\\

\end{array}$$

Melody May 19, 2015

#62**+15 **

Just did a Maths test today i think i got 90-100%

Those questions came up, but i aced them with your help!

Also, i checked the inverse and they were right.

Also, with the second step, it's not the answer it's just me showing how it would look when taking away 4.

(To take it, like you said)

:D

MathsGod1 May 19, 2015

#63**+5 **

That is excellent MG I am really please.

I am wondering...

Do you get much maths homework from school? Do you have a maths text book that you can use at home?

Melody May 19, 2015

#64**+10 **

Nope...We get it atleast once a week which isn't much in my opinion for a core subject.

It's on this website called MyMaths.

It is used for schools and teachers can assign homework their, there is also games- maths games.

I have a math book, the CGP one fr year 7 & 8 but it's all the math basics.

I'm wanting to get a harder one just to look more into maths.

MathsGod1 May 19, 2015

#65**+5 **

Ok MG.

It is really good to work in advance but it is even more important to make sure you understand and can do all the work you have already covered. This basic work is the building blocks that you will need for always.

The more thoroughly you understand them the easier future work will be and the further you will be able to go with mathematics in your future :)

Use forum questions - like you have been - to not only discover new and exciting mathematics but also to practice all the basic maths that you already have.

I still have not homed in properly on your level.

Here are a few questions. I want to work out what you do and do not understand.

You can copy my questions and then put the answers with the questions.

Don't get help, I want to see what you can do by yourself :)

$$\\1) 3x-2x^2 - 8xy+3yx-7\\

2) 5x*6xy\\

3) 3(x+4)\\

4) -2(x+4)\\

5) 17-4x-x+5\\

6) 3-4(5x-6)\\

7) \frac{3x}{2}+\frac{4x}{5} \qquad $(Make an entire fraction)$\\

8) 0.06*0.3 \qquad $ (No calculator)$\\

9) 0.06\div 0.2 \qquad $ (No calculator)$\\

10) 3x^2+2x+5\\

11) \frac{5x+2}{6y}\times 9xy^2$$

\\1) 3x-2x^2 - 8xy+3yx-7\\ 2) 5x*6xy\\ 3) 3(x+4)\\ 4) -2(x+4)\\ 5) 17-4x-x+5\\ 6) 3-4(5x-6)\\ 7) \frac{3x}{2}+\frac{4x}{5} \qquad $(Make an entire fraction)$\\ 8) 0.06*0.3 \qquad $ (No calculator)$\\ 9) 0.06\div 0.2 \qquad $ (No calculator)$\\ 10) 3x^2+2x+5\\11) \frac{5x+2}{6y}\times 9xy^2

Melody May 20, 2015

#66**+5 **

Can you do this one MG

How do you solve 2(x-3) + 5x = 4(x+2) - 5 ?

Zac's answer (Zac gives great answers))

http://web2.0calc.com/questions/how-do-you-solve-2-x-3-5x-4-x-2-5#r1

(no cheating now )

Melody May 20, 2015

#67**+10 **

Wait are their any catch to these question or are they just normal question?

I will answer these when i get back..Someone's coming back home from a 7-week holiday

MathsGod1 May 20, 2015

#68**+10 **

I typed in a page worth of Latex for Q1 but it just all dissapeard.

In the end i got no here. So I'm kinda stuck.

MathsGod1 May 20, 2015

#69**+5 **

Sometimes I lose a whole page of LasTex too.

I know it is extremely frustrating.

Those are just normal questions. If they are easy that is good, I just want to get a better idea of what you can do on your own, that's all.

Melody May 21, 2015

#70**+10 **

It is very frustrating...

$$\\3x-2x^2-8xy+3yx-7\\\\

1(-2x^2-8yx+3yx-7)\\

I\;have\;no\;idea\;whether\;the\;task\;was\;even\;to\;simplify\\\\

My\;workings?\\\\

Honestly,\;I\;just\;was\;staringz;at\;it\;and\;figured\;it\;out\;(not\;that\;it\;is\;right\;or\;not)\\\\

At\;first\;I\;put\;x\;outside\;until\;i\;realised\;you\;can't\;times\;x\;by\;something\;to\;get\;7,\;as\;they're\;not\;like\;terms\\\\

So\;then\;I\;put\;1\;since\;that\;is\;the\;HCF\\\\

It\;looks\;really\;cheap\;since\;it's\;just\;the\;same\;just\;in\;brackets...\\\\\

But\;thats\;for\;you\;to\;judge.$$

MathsGod1 May 21, 2015

#71**+5 **

I didn't want you to factorise MG.

I wanted you to simplify. That means to collect the like terms. I will explain better later I'm on my phone now and it is too difficult.

Do you know if this expression has any like terms?

Melody May 21, 2015

#72**+15 **

Yes I am familiar with like terms.

$$\\3x-2x^2-8xy+3yx-7\\\\

3x\;is\;alone.\\\\

2x^2\;is\;alone\\\\

8xy\;and\;3yx\;are\;a\;like\;term\\\\

-7\;is\;alone\;$$

I have a question, when you asked me the question you had asked me to solve it, so I wasn't sure was either to simplify or factorise. Was there something in the question that told me to simplify?

(Just to make sure for future algerbraic questions)

MathsGod1 May 23, 2015

#73**+15 **

Hi Melody, I want to tell you something about LaTeX, (you probably know this, but none has told me about this i figured out on my own)

It Isn't really that big, but it's quicker.

When doing \\ for extra lines when you do it on the first line you don't have to do it for the rest, only at the end.

E.g.

\\Hi\\\\

What's\;up\\\\

I\;Like\;this\;forum\\\\

$$\\Hi\\\\

What's\;up\\\\

I\;Like\;this\;forum\\\\$$

And the $\\{text}\\$ put texts in on the side\\

$$$\\LaTeX\\$

LaTex\;a\;way\;of\;displaying\;text\\

It\;is\;very\;neat\;and\;easy\;to\;understand\\\\

$\\Formula\\$

A\;formula\;is\;a\;rule\;to\;something\\$$

It doesn't work for if you do it again.

MathsGod1 May 23, 2015

#74**+5 **### Maybe **Heureka** can tell us why :/

#### LaTex is always doing unexpected things :))

Hi MG,

\\ at the beginning of the first line tops the indentation happening.

The first line will be indented otherwise because it is like having a new paragraph.

A second line space in the coding also starts a new paragraph and if you do not want the insert you would need the \\ at the beginning there to.

Otherwise \\ just means start a new line but not a new paragraph.

**I'll just try to check what I am saying.**

$$\mbox{1. no backslashes so this row should be indented}\\

\mbox{2. This one should not be indented}

\mbox{3. I just skipped 2 rows so this one should be indented}

\\\mbox{4. I just skipped 2 rows BUT the I put two backslashes SO this comment should NOT be indented}$$

**That is interesting - the last one did not work the way I expected it to.**

**------------------------------------------------------------------------------**

**\mbox{1. no backslashes so this row should be indented}\\ **

**\mbox{2. This one should not be indented}**

** \mbox{3. I just skipped 2 rows so this one should be indented}**

** \\\mbox{4. I just skipped 2 rows BUT the I put two backslashes SO this comment should NOT be indented}**

Melody May 24, 2015

#75**+10 **

Now for the algebra. Yes I wanted you to simplify.

Yes I should have stated this implicitly.

You can probably do this easily but I would still like you to answer them please. :)

Just put the answer after the question as I have shown you here.

Thay way questions and answers stay together and it is easy for both of us to check.:)

$$\\1) 3x-2x^2 - 8xy+3yx-7=$ put answer here$\\\\

2) 5x*6xy\\\\

3) 3(x+4)\\\\

4) -2(x+4)\\\\

5) 17-4x-x+5\\\\

6) 3-4(5x-6)\\\\

7) \frac{3x}{2}+\frac{4x}{5} \qquad $(Make an entire fraction)$\\\\

8) 0.06*0.3 \qquad $ (No calculator)$\\\\

9) 0.06\div 0.2 \qquad $ (No calculator)$\\\\

10) 3x^2+2x+5\\\\

11) \frac{5x+2}{6y}\times 9xy^2\\\\$$

This thread is starting to get difficult to work with, you should consider a new one.

***** Maybe AFTER these questions are answered.**

When you start a new one make sure the new one references the old one in the first post AND

makes sure the old one references the new one in your last post. **That way it will STILL all stay together!**

Happy answering :)

----------------------------------------------------------------------------

There was also this one that I asked you to try on an earlier post in this thread :)

Can you solve this equation MG

How do you solve 2(x-3) + 5x = 4(x+2) - 5 ?

Zac's answer (Zac gives great answers))

http://web2.0calc.com/questions/how-do-you-solve-2-x-3-5x-4-x-2-5#r1

(no cheating now )

Melody May 24, 2015

#76**+15 **

Thanks for the tip.

I will try answer it, but I haven't done Simplifying in some time now- I'll give it all I've got (lol this is probably so easy for you, it is for me, but I can't remember much, since for the past couple of month 2-3 we've been doing catching up on fraction, percentages and decimals).

$$\\3x-2x^2-8xy+3yx-7=\\\\

Firstly,\;to\;point\;out\;3yx\;and\;8xy\;are\;the\;same\;\\

So\;I'll\;do\;xy\;as\;followed\;in\;the\;alphabet\\\\

Like\;Terms\;are\;highlighted\\\\

3x\textcolor[rgb]{1,0,0}{-2x^2}-\textcolor[rgb]{0,1,0}{-8xy}\textcolor[rgb]{0,1,0}{+3xy}\textcolor[rgb]{0,0,1}{-7}\\\\

The\;only\;like\;terms\;are\;8xy\;and\;3xy\\\\

-8xy\;+\;(+3xy)\;=-5xy\\\\

Now\;it\;is\;3x-2x^2-5xy-7\\\\$$

\\3x-2x^2-8xy+3yx-7=\\\\

Firstly,\;to\;point\;out\;3yx\;and\;8xy\;are\;the\;same\;\\

So\;I'll\;do\;xy\;as\;followed\;in\;the\;alphabet\\\\

Like\;Terms\;are\;highlighted\\\\

3x\textcolor[rgb]{1,0,0}{-2x^2}-\textcolor[rgb]{0,1,0}{-8xy}\textcolor[rgb]{0,1,0}{+3xy}\textcolor[rgb]{0,0,1}{-7}\\\\

The\;only\;like\;terms\;are\;8xy\;and\;3xy\\\\

-8xy\;+\;(+3xy)\;=-5xy\\\\

Now\;it\;is\;3x-2x^2-5xy-7\\\\

I think that's the final simplified form, but it was only one step (only like term).

EDIT: (Accidental extra minus sign in coloured equation)

MathsGod1 May 24, 2015

#78**0 **

Our LaTex is only displaying some of the time.

Maybe it will display better if you start a new thread :/

Melody May 24, 2015

#79**0 **

Hang on a tick - you seem to have an extra minus sign in the working:/

oh sorry you explained that. It is the black minus sign that I'd like to see gone.

The minus sign belongs to whatever term is behind it, in this case 8xy, so it is best if they are all green :)

Melody May 24, 2015

#80**+5 **

Yes this post is getting very slow.

I will continue question two in a new post.

MathsGod1 May 24, 2015

#81**0 **

They are only suppost to be simple questions. Can't you just do them all at once?

Melody May 24, 2015

#82**+10 **

ok, i thought i had to answer them in latex and workings showed i'll try them quicker under.

Q2) 5x*6yx=30xy

Q3) 3(x+4)=12x - (3*x)+(3*4)=12x

Q4) -2(x+4)=-10x - (-2*x)+(-2*4)=-10x

Q5) 17 - 4x -x +5=22+(-5x) - (like terms -x & -4x and 17 & +5) 17+5=22 / -4x-x=-5

Q6)3-4(5x-6)=23x+24 - (Not sure but...) (-4*5x)+(-4*-6)=-20x+24 + (3*5x)+(3*-6)=15x+(-18x) 15x+(-18x)-20x+24=-23x+24

Q7) 3x/2 + 4x/5 = 12x+10

Q8) 0.06*0.3=0.018 (6*3=18/1000=0.018 )

Q9) 0.06/0.2 (6/2=3/1000=0.003)

Q10) 3x^2 +2x +5= 3x^2+2x+5 (Like terms none) the answer is given

Q11) 5x+2/6y * 9xy^2= ....Have no idea.

WAIT. 9xy^2 as a fraction...ah

MathsGod1 May 24, 2015

#83**+15 **

$$\begin{array}{rrlll}\\\\

2(x-3)+5x&=&4(x+2)-5\qquad&(2x-3)=2x-6\;and\;4(x+2)=4x+8\\\\

2x-6+5x&=&4x+8-5\qquad&(+5\;on\;both\;sides)\\\\

2x-1+5x&=&4x+8\qquad&(+1\;on\;each\;side)\\\\

2x+5x&=&4x+9\qquad&(-2x\;on\;each\;side)\\\\

5x&=&2x+9\qquad&(-2x\;on\;each\;side)\\\\

3x&=&9\qquad&(Divide\;both\;side\;by\;3)\\\\

\frac{3x}{3}&=&\frac{9}{3}\qquad&(Cancel\;out)\\\\

3&=&x\\\\

x&=&3

\end{array}$$

This is my answer and i **swear**** **I haven't looked at Zacs post yet, i care about my education and if i don't make mistakes then i'll make progress, if i get it wrong and don't cheat it's better than cheating and getting it right.

I think i got it wrong at the first step, but then I got a result.

Took me about 15-20min to do.

Code:

\begin{array}{rrlll}\\\\

2(x-3)+5x&=&4(x+2)-5\qquad&(2x-3)=2x-6\;and\;4(x+2)=4x+8\\\\

2x-6+5x&=&4x+8-5\qquad&(+5\;on\;both\;sides)\\\\

2x-1+5x&=&4x+8\qquad&(+1\;on\;each\;side)\\\\

2x+5x&=&4x+9\qquad&(-2x\;on\;each\;side)\\\\

5x&=&2x+9\qquad&(-2x\;on\;each\;side)\\\\

3x&=&9\qquad&(Divide\;both\;side\;by\;3)\\\\

\frac{3x}{3}&=&\frac{9}{3}\qquad&(Cancel\;out)\\\\

3&=&x\\\\

x&=&3

\end{array}

MathsGod1 May 24, 2015

#84**+5 **

ok lets look at this

Q2) 5x*6yx=30xy **Nope**

Q3) 3(x+4)=12x - (3*x)+(3*4)=12x **nope **

Q4) -2(x+4)=-10x - (-2*x)+(-2*4)=-10x **nope**

Q5) 17 - 4x -x +5=22+(-5x) **(good)****=22-5x**

(like terms -x & -4x and 17 & +5) 17+5=22 / -4x-x=-5 ** -4x-x=-5x **

Q6)3-4(5x-6)=23x+24 - (Not sure but...) (-4*5x)+(-4*-6)=-20x+24 **good + ** (3*5x)+(3*-6)=15x+(-18x) 15x+(-18x)-20x+24=-23x+24** now you have lost it **

Q7) 3x/2 + 4x/5 = 12x+10 **No**

Q8) 0.06*0.3=0.018 (6*3=18/1000=0.018 ) ** good**

Q9) 0.06/0.2 (6/2=3/1000=0.003) **please try to spot your own error **

Q10) 3x^2 +2x +5= 3x^2+2x+5 (Like terms none) the answer is given **great**

Q11) 5x+2/6y * 9xy^2= ....Have no idea. ** fair enough **

WAIT. 9xy^2 as a fraction...ah

Can you fix any of your own errors MG ?

Melody May 24, 2015

#86**+10 **

I think your equation solution is correct but you havwe done it the very long way :)

Before you think about what to get rid of you have to think about whether either side can be simplified.

So it is

Can i simplify the sides independently? If so do it.

What do I want to get rid of? How? Do it - to both sides.

Can i simplify the sides independently? If so do it.

What do I want to get rid of? How? Do it - to both sides.

Can i simplify the sides independently? If so do it.

etc etc

This is why I wanted to see if you can simplify terms properly :)

$$\begin{array}{rrlll}\\\\

2(x-3)+5x&=&4(x+2)-5\qquad&(2x-3)=2x-6\;and\;4(x+2)=4x+8\\\\

2x-6+5x&=&4x+8-5\textcolor[rgb]{0,0,1}{\;\;excellent\; start!\;}\qquad&(+5\;on\;both\;sides)\\\\

&&&\textcolor[rgb]{0,0,1}{\mbox{but now, before you start adding and subtracting things you need to simpliey each side seperately}}\\\\

2x+5x-6&=&4x+8-5\\\\

7x-6&=&4x+3&Now\; add\; 6 \;to \;both \;sides\\\\

7x-6+6&=&4x+3+6&Now\; simplify\\\\

7x&=&4x+9&Now\; subtract\;4x\;from\;each\;side\;\\\\

7x-4x&=&4x-4x+9&Now\; simplify\;\\\\

3x&=&9&Now\; divide\; both \;sides\; by\; 3\;\\\\

\frac{3x}{3}&=&\frac{9}{3}&Now\; simplify\;\\\\

x&=&3&\;\;\\

\end{array}$$

Melody May 24, 2015

#87**+5 **

Hi Zegroes, this is a thread that I am using just to teach MathsGod1 on. You gotta behave yourself and stay outa here.

Melody May 24, 2015

#89**+5 **

Because you don't want this type of special attention. This would be your idea of a severe punishment. I don't mind that, I like MathsGod the way he is and I like you the way you are. Now do not respond in here or I'll clip your ear!!

Melody May 24, 2015

#90**+10 **

I'll take a CLOSE look at these now because i slept, before i had a high fever.Ok, I'll start off at question 2)

Ok, I'll start off at question 2)

5x*6xy? hmm...

5*6=30

OH!!!(took like 3min ha)

Just realised lol.

There are two x's being times!

so, **30x^2y** ...i think

Q3)3(x+4)=12x

so, 3*x=3x and 3*4=12

3x and 12 aren't like terms so the answer is ah, 3x+12, makes a lot of sence when it's looked close at.

Q4) 2(x+4)=

again, 2*x=2x 2*4=8

Not like terms,** 2x+8**

Q6) 3-4(5x-6)= Not sure...But I'm taking a hard stare.

I understand it now.Sorry, i had check online, not once in my life have I ever been invited to a question like this I didn't know how to solve it.

Simplifying 3 + -4(5x + -6) = 0 Reorder the terms: 3 + -4(-6 + 5x) = 0 3 + (-6 * -4 + 5x * -4) = 0 3 + (24 + -20x) = 0 Combine like terms: 3 + 24 = 2727 + -20x= 0

I thought you'd have to times everything by 3 rather than add.

(Darn, I can't get out of this text form!)

3x/2 + 4x/5 =

The reason I did this is because in school, the way i learnt to times fraction is

Numerator*Numerator

Denominator*Denominator

12x/10

Maybe simplify it6x/5?

0.06/0.2

Well I've never divided a decimal by decimal before...Or actually considered it.

so, I'm not sure what my mistake is all i know is that my brother told me a fact, that i nver thought about dividing in decimals will make it smaller, multiplying- bigger...

Honestly, I have no idea....With Google helps.

**To divide decimal numbers:**- If the divisor is not a whole number, move decimal point to right to make it a whole number and move decimal point in dividend the same number of places.
- Divide as usual. ...
- Put decimal point directly above decimal point in the dividend.
- Check your answer.

1. so 0.2 = 2 0.06 = 0.6

2.2/0.6=3.333333

3.so like 33.3

Not sure...Hard to do it on computer but I'm positive It's wrong...

MathsGod1 May 24, 2015

#91**+10 **

Ok you are getting there MG

Q2,3,4,8 all good

Q5 almost good

Rest need work.

Q6 You answer is good except that +- must be changed back to a - and there is not =0 in the question.

Q7 This is how it should be done

$$\\\frac{3x}{2}+\frac{4x}{5}\\\\

$You must get a common denominator that will be 10$\\\\

=\frac{3x*5}{2*5}+\frac{4x*2}{5*2}\\\\

=\frac{15x}{10}+\frac{8x}{10}\\\\

=\frac{15x}{10}+\frac{8x}{10}\\\\

=\frac{23x}{10}\qquad or\\\\

=2.3x\\\\$$

Q9) 0.06/0.3

you cannot divide by a decimal so you have to change at least the bottom (or both) into a whole number

$$\\\frac{0.06}{0.3}=\frac{0.06*10}{0.3*10}=\frac{0.6}{3}=0.2\qquad \\\\

or\\\\

\frac{0.06}{0.3}=\frac{0.06*100}{0.3*100}=\frac{6}{30}=\frac{1}{5}\quad or\;=0.2$$

Q11)

This was my question

$$\frac{5x+2}{6y}\times9xy^2$$

but when you rewrote it you wrote

5x+2/6y*9xy^2 this is entirely different expression! I will show you the expression you have written.

$$5x+\frac{2}{6}*y*9xy^2=5x+\frac{2y}{6}*9xy^2\\\\$$

So before you try to simplify my Q11 I would like you to write it properly without the aid of LaTex.

In other words, where do you need to put brackets! [ always remember PEDMAS ]

Melody May 25, 2015

#92**+5 **

Q7) DOH!

I thought it was times, otherwise i would have done what you done

Q9) Thanks for the explanation :D , i thought the answer had be divided back .

Q11) Sorry about that, i forgot about it and how text can be easily confused wih something else.

$$\\5x+\frac{2}{6}*y*9xy^2=5x+\frac{2y}{6}*9xy^2\\\\$$

.MathsGod1 May 25, 2015

#93**+5 **

You did not answer my question for number 11 MG

The question was write this correctly with brackets and NO latex.

$$\frac{5x+2}{6y}\times9xy^2$$

.Melody May 25, 2015

#95**+5 **

No not quite MG because the first y would be on the top.

The division would be done BEFORE the multiplication

Melody May 25, 2015

#96**+10 **

But i have done the division before the times.

In the fraction then * 9xy^2

MathsGod1 May 25, 2015

#97**+5 **

your answer is

(5x+2)/6y *9xy^2

this is

$$\frac{(5x+2)}{6}*y *9xy^2$$

See, the first y will be in the wrong place - it is supposed to be on the bottom!

Melody May 25, 2015

#98**+5 **

Unless you want to ask questions on these, would you like to start fresh?

(Take a look at my previous posts first though)

Can you simplify this one for me ? Do it with cancelling.

Can you remember how to do that or do you want me to show you?

$$\frac{21x}{y}\times \frac{8y^2x}{14t}$$

Perhaps you should copy this over to a new thread :)

Melody May 25, 2015

#99**+10 **

Ok I'll get a new thread started once I'm off the tablet and i get hold of the laptop it's too hard on this.

Let me just answer the fraction question.

So thank you for correcting me :

(5x+2)/(6y)*9xy^2

MathsGod1 May 25, 2015

#101**+5 **

The post has been made.

http://web2.0calc.com/questions/latex-form-part-3

You can now take this off your watchlist.

MathsGod1 May 26, 2015