Chapter 3

1st Degree Polynomial Equations
ax+b=0

3.5 Problems that can be solved with the help of polynomial equations of the 1st degree.

In order to solve a problem with the help of equations we work as following

• We read carefully the problem in order to understand what we are given and what we are asked to find.
• We choose which one of the unknowns we are going to name x and we try to express the rest of the unknowns as functions of x.
• We transform the expressions of the problem, which are in natural language, into mathematical expressions.
• We create the equation that we are going to solve in respect with x.
• We examine if the solution we found satisfies the problem. If this is not the case the solution is rejected.
• We must be sure that the members of the equations represent entities of the same kind.
Example 1

The price of a product increased 20%  the  first year and 25% the following year. The final price is £930.
What was the starting price of the product?

Solution

Lets name x the starting price of the product. The first year the price increased 20% and it became:
x + (20/100)x = x + (1/5)x = 6x/5
The following year it increased 25% and it became:
6x/5 + (25/100)(6x/5) = 6x/5 + 3x/10 =15x/10=3x/2
Finally we have the equation
3x/2 = 9303x=2*930x=(2*930)/3x=620

So, the starting price of the product was £620.

Exercise 1

The age of a father is three times bigger than the age of his son. After 12 years the age of the father is going to be double the age of the son. What are the ages of the father and son today?