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!