Definition: An equation is an algebraic statement that two expressions are equal. An equation consists of:

20 x^{ 3} + 25 x^{ 2} + 5 x = 0has three solutions: x = 0, x = −¼ and x = −1. This is called a solution set. We can verify that all three values of x satisfy the equation by substituting each of them into the equation and getting the true statement that 0 = 0.
x^{ 2} = −9.There is no real number which when squared will yield −9 but there are two complex numbers, namely 3 i and −3 i.
x = x + 1.There is no number that is unchanged when you add 1 to it.
Perform the same mathematical operation on both sides of the equation. 
Invert (undo) the operations that were applied to x in the reverse order in which they were applied. 
x − 5 = 13.Solution: In the expression on the lefthandside of the equation, x has had 5 subtracted from it so we must add 5 back on to get x alone. (Adding 5 is the inverse operation of subtracting 5.) Adding 5 to both sides gives:
x − 5 + 5 = 13 + 5.Simplifying both sides then gives:
x = 18.This last equation is just a manipulated form of the original equation, but because it has x alone on the lefthandside and a number on the righthandside, it is the solution. We must still check the solution by substituting it into the original equation. This gives 18 − 5 = 13 or 13 = 13, so the lefthandside equals the righthandside and the solution checks out.
Solution: In the expression on the lefthandside of the equation, x has had this sequence of operations applied to it in this order:
Why these operations and why this order? Here is an analogy. Picture how you get dressed in the morning (to be polite we will only talk about socks and shoes):
Also note that since your shoes were put on last they have to come off first. 
,is a formula that is used to solve oblique triangles. It is a relation between the lengths of sides a, b and c and the angle C which is opposite side c. Given any three of these quantities, the fourth can be found. Solve the cosine law for angle C.
Algebra Coach Exercises 
Suppose that a and b are any expressions. Then the following statement is true: If a · b = 0 then either a = 0, or b = 0, or both.In other words a product can equal zero only if one of its factors is zero. This fact can be used to replace a “complicated” equation a · b = 0 by two simpler equations a = 0 and b = 0. 
(x − 1)(x + 2) = 0.Solution: There are two factors on the lefthandside of the equation:
5 x (x + 1)(4 x + 1) = 0.Solution: There are four factors on the lefthandside of the equation:
x (x + 1)(4 x + 1) = 0.But note that you can’t divide by any of the other factors, say by x, because that would amount to dividing by zero, which is an undefined operation. (Also it would cause you to lose that corresponding solution.)
20 x^{ 3} + 25 x^{ 2} + 5 x = 0.We didn’t solve it but we did verify that it has the same three solutions as the present example, namely x = 0, x = −1 and x = −¼, by substituting them back in and checking. In fact the only difference between that previous example and our present example is that lefthandside of the equation in this present example is in factored form and in the previous example it is not. In fact, manipulating equations into the form a · b = 0 is one of the most important applications of factoring.
Algebra Coach Exercises 