# Write an Algebraic Expression

IntroductionAlgebraicexpressions are mathematical verbal statements of
real life situations. One must know how to transform a verbal statement and to
write an algebraic expression. An algebraic equation is nothing but an
equivalence of two algebraic expressions. Hence to solve an algebraic equation
it is important to correctly write an algebraic expression on both sides of an equation.

## Write an algebraic expression

Before we go into the details of solving an algebraic equation which states that two algebraic expressions are equal, let us understand how to write
an algebraic expression that appears on both sides of an algebraic equation. To write the same correctly one must understand what exactly is the
given statement. It may be totally verbal, semi verbal or totally algebraic. One must also know how to correctly pick up the required information and
discard the unwanted information.
Let us take an example to write an algebraic expression for the statement ‘ In my school quiz I was asked to write an algebraic expression to
express my age 5 years back’. In this statement, the unwanted information for the purpose to write an algebraic expression is, ‘ In my school quiz
was asked to write’. Hence let us ignore that. The question does not ask you what is your present age and thereby your age 5 years back. May be it
is a general question posted to all the students. Hence, let the present age for anybody be
assumed as ‘x’ and naturally, for anybody, 5 years back the age must have been 5 years less than the present age’. Algebraically it can be written
as ‘x – 5’.
This is one of the concepts of how to write an algebraic expression.

## Find the ways to solve complex algebraic equations

Let us illustrate with an example.
Good quality walnuts are priced at $2.5/lb and the second quality of walnuts are priced at$1.5/lb. A trader found that most people can not afford the rate of $2.5/lb but can pay$2.0/lb, for a reasonable quality. The trader decides to mix 40 pounds of good quality walnuts with some quantity of second quality of walnuts and sell the mixture at \$2.0/lb to remain competitive. How many pounds of second quality of walnuts he must add?
Let us find the ways to solve this  complex algebraic equation.
Basically the fact is that the trader wants to be competitive ‘but’ at the same time he/she would not like to loose the income.
Hence the cost in both type of mixtures must be same.
In other words, the algebraic expressions for the cost of both the mixtures must be same. Let us figure out the same in steps.
We do not how many pounds of second quality of walnuts to be added with 40 pounds of first quality.
Hence, let us say it is ‘x’ pounds.
The immediate conclusion is, the total quantity of the mixture is ‘x + 40’ pounds.
The algebraic expression for the cost of 40 pounds of first quality in dollars = 40*2.5 = 100
The algebraic expression for the cost of ‘x’ pounds of second quality in dollars = x*1.5 = 1.5x
Hence, the total cost of the mixture in dollars is, (100 + 1.5x)
Now let us find the total cost of the mixture from a different way.
The algebraic expression for the cost of (x + 40) pounds of the mixture in dollars = (x + 40)*2 = 2x + 80.
As we said, no trader would like to loose his/her income. Therefore, the derived algebraic expressions form an equation, one is equal to other. That is,
2x + 80 = 100 + 1.5x  or, 0.5x = 20 or x = 40.
Thus, the trader has to mix 40 pounds of second quality of walnuts.