Make your own free website on
Chapter 3

1st Degree Polynomial Equations

3.1. The Equation ax+b=0

Let's see how we solve the above equation with the help of the properties of the operations, for the various values of a,b.

We have ax+b = 0 <=> ax+b - b = - b <=> ax = - b

Now we distinct the following possibilities:
A.    If  the equation has exactly one solution. That is: 

B.   If a=0  the equation becomes 0x = - b and:
            B1. If  the equation does not have a solution in R.

            B2. If   b=0 the equation takes the form 0x=0, which is true for every 

Our conclusions can be summarized in the table bellow:

The equation ax+b=0
It has unique solution 
If a=0 and  
It is has no solutions in R
If a=0 and b=0
It is true for every 
The solution of an equation, that is the procedure that we follow in order to solve an equation is an algorithm. Its steps can be seen in the following example:

Solve the equation: 


<=>  <=>
Step 1: We eliminate the denominators multiplying the terms of the equation with the lowest common multiple of the denominators.
<=> 6(x-4) - 15(x-3) = 10(4x-1) <=>
Step 2: Applying the distributive law a(b+c)=ab+ac we eliminate the parenthesis  
<=> 6x - 24 - 15x +45=40x - 10 <=>
Step 3: We take all the unknown terms to the one side and the known terms to the other side 
<=> 6x - 24 -15x + 45 = 40x - 10 <=>
Step 4: We do the relevant operations
<=> -49x = -31 <=>
Step 5: We divide both sides by the factor of the unknown variable(here -49)
<=> x = -31/-49 <=> 

We have to make clear the following: