Main ways used at solving of equations
Transferring terms of equation from one side to another.
Multiplication and division by non-zero expression (number).
Raising to a power. Extraneous roots of equation.
Extracting of a root. Loss of roots of equation.
Solving of equation is a process, consisting mainly in a replacement of the given equation by another, equivalent equation. This replacement is called an identical transformation . Main identical transformations are the following.
| 1. | Replacement of one expression by  another, identically equal to it.
     
     For example, the equation ( 3
     
      x+
     
     2 )
     
      
       2
      
     
     = 15
     
      x +
     
     10  may be replaced by the     next equivalent equation:  9
     
      x
     
     
      
       2
      
     
     
      +
     
     12
     
      x +
     
     4 = 15
     
      x +
     
     10
     
      . | 
| 2. | Transferring terms of equation from  one side to another with back signs.
     
     So, in the previous equation we can transfer  all terms from the right-hand side 	to the left with the sign "minus":
     
     
     9
     
      x
     
     
      
       2
      
     
     
      +
     
     12
     
      x +
     
     4
     
      –
     
     15
     
      x  –
     
     10 = 0, after this we receive: 9
     
      x
     
     
      
       2
      
     
     
      –
     
     3
     
      x –
     
     6 = 0
     
      . | 
| 3. | Multiplication or division of both sides of equation by the same expression	( number ), not equal to zero.
     
     This is very important, because
     
      a new equation 	can be not equivalent to previous,  if the expression, by which we multiply or	divide, can be equal to zero. E x a m p l e : The equation x – 1 = 0 has the single root x = 1 . Multiplying it by x – 3 , we receive the equation ( x – 1 )( x – 3 ) = 0, which has two roots: x = 1 and x = 3 . The last value isn’t a root for the given equation x – 1 = 0 . This value is so called an extraneous root. And vice versa, division can result to a loss of roots . In our case, if ( x – 1 )( x – 3 ) = 0 is the origin equation, then the root x = 3 will be lost at division of this equation by x – 3 . 
      In the last equation (p.2) we can divide all terms by 3 (not zero!) and finally receive:
       | 
  This equation is equivalent to an original  one: