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

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. :)

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!

Hi MG

Your understanding of equations and your LaTex presentations are coming along brilliantly.

All the adults on the forum a are very pleased that you are learning here with us!

Your first post here was incorrect (you did not add 9x correctly) and you have redone it so you should delete your first post so it will not be confusing.

Better comments for the correct post would have been  

(-5 from both sides)

(+2x to both sides)

(divide both sides by -7)

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}

I had to google FigNewton to find out what you were talking about Chris.   

It is a pity I was not on the forum today.  You have obviously been in a very fun mood :)   

Add 4 to both sides    

I agree MG. This was a good question, thanks Civonamzuk.

And Alan has given a great answer.  Thanks Alan

Your English is just fine Dbg :)

We can all understand you perfectly and that is the whole point of any language!

  ROF  LOL     .....     ROF  LOL

That is hilarious Chris.  I've only just seen it, I have been out almost all day.  

You have outdone yourself - how are you going to top this one next time  ?