Algebraic Equations
Introduction
The fundamental idea on which algebraic equation are based is that an equation is exactly like a set of balance scales. If you do exactly the same thing to both sides of the equation then the equation will always balance. If you add the same amount to both sides or subtract the same amount to both sides the equation will balance. If you multiply the Left Hand Side (LHS) by the same amount as the Right Hand Side (RHS), then the equation will still balance, and so on. This is all analogous with a set of balance scales. With an algebraic equation we can go further and do exotic things and, for example, take the square root of both side and the equation will still balance.
Once you have this fundamental idea in your mind, we can go on to talk about how to use these allowed opersations systematically to solve an equation.
Another thing in passing  since we commonly use the letter 'x' in algebra, the use of the 'X' sign for multiplication is strongly discouraged.
Basic Equations  No 1
Considerx + 5 = 8
Although you might be able to guess what the answer is, I would to explain the algebraic method. The algebraic method would be (as always) to do exactly the same thing to both LHS and RHS.
Here I will subtract 5 from both sides
x = 3
which is the answer.
As a second example, consider
x 4 = 6
The algebraic method would be to do exactly the same thing to both LHS and RHS.
Here I will add 4 to both sides
x = 10
which is the answer.
Try these examples before proceeding
a) x + 3 = 8

Basic Equations  No 2
Consider
2x = 8
Again, although you might be able to guess what the answer is, I would to explain the algebraic method. The algebraic method would be (as always) to do exactly the same thing to both LHS and RHS.
Here I will divide both sides by 2
x = 4
which is the answer.
As a second example, consider
\[ \frac{x}{6} = 6 \]
The algebraic method would be to do exactly the same thing to both LHS and RHS.
Here I will multiply both sides by 6
x = 36
which is the answer.
Try these examples before proceeding
a) \[ 4x = 8 \] b)\[ \frac{x}{3} = 5 \] c) \[ 3x = 12 \] d)\[ \frac{x}{4} = 2 \] e) \[ 12x = 48 \]

Basic Equations  No 3
If I can now make use of the methods used in the last two sections simultaneously in order to show how to tackle a more 'complicated' expression in a systematic way.
Consider
3x  7 = x  5
Firstly, I am going to use the methods of addition or subtraction to get every term containing an x on to the same side, and everything which is just a number on to the other side.
Secondly, I am going to use the operations of multiplication or division to obtain x
The first process involves two steps
Subtract x from both sides
2x  7 =  5
Add 7 to both sides
2x = 2
The second process involves just one step. We are told what 2x equals but I actually want to know what x equals, so
Divide both sides by 2
x = 1
which is the answer.
As a second example, consider
6x + 9 = 25  2x
The first process involves two steps
Add 2x to both sides
8x + 9 = 25
Subtract 9 from both sides
8x = 16
The second process involves just one step. We are told what 8x equals but I actually want to know what x equals, so
Divide both sides by 8
x = 2
which is the answer.
This method is extremely important  it will occur again and again. 1. Get all the terms involving 'x' on one side of the equation and all the 'numbers' (terms involving no 'x's on the other side. This will be achieved by addition and/or subtraction. 2. Derive the x value by dividing or multiplying the expression from the previous step (expect it be division in most cases). (I have mentioned 'x' but we could use any letter we want to and the rules would be the same) 
It is important to stress the following (to prevent a common mistake): A term cannot disappear from one side of the equation and appear unchanged on the other. Exs. A 9 cannot disappear from the LHS and appear on the RHS in the next line. But a 9 can disappear from the LHS and a +9 can appear on the RHS on the next line. 
Try these examples before proceeding
a) 3x + 7 = 9  2x 