http://www.youtube.com/watch?v=cxxdTshrM3k

http://www.youtube.com/profile?user=AlgebraWordProblems

http://www.youtube.com/watch?v=Oc58x95Pmcw

 Algebraic Word Problems with Variables

The algebraic solution to a word problem includes the following steps:

• Read the problem carefully

• Organize the problem on paper

Substitute values into a given formula. If no formula is

given, identify the relationship between the various

quantities and express that relationship as an equation.

• Solve for the unknown variable

• Check the solution

http://studentservices.fgcu.edu/learning/clast/algebra/tsld037.htm

What Key Words to look for when answering a word problem:

Problem Solving Plan in 4 Steps:

Clues: 

• Read the problem carefully.

• What facts are you given?

• What do you need to find out?

  Game Plan:

• Define your game plan.

• Define your strategies to solve this problem.

• Try out your strategies. (Using formulas, simplifying, use sketches, guess and check, look for a pattern, etc.)

      Solve:

• Use your strategies to solve the problem.

• Reflect:

• Did you answer the question? Are you sure?

• Did you answer using the language in the question? Same units?

Clue Words:

When deciding on methods or procedures to use to solve problems, the first thing you will do is look for clues which is one of the most important skills in solving problems in mathematics.

If you begin to solve problems by looking for clue words, you will find that these 'words' often indicate an operation. For instance:

Clue Words for Addition:

• sum

• total

• in all  

Clue Words for Subtraction:

• difference

• how much more

• exceed

Clue Words for Multiplication:

• product

• total  

• times

Clue Words for Division

• share

• distribute

• quotient

• average

http://math.about.com/library/weekly/aa041503a.htm

What you do to one side of the equation, make sure to do to the other.

Algebra Word Problems

How to Solve Algebraic word Problems? 

Word problems (or story problems) usually strike fear into the hearts of young and old math students alike!  In all reality, they aren't that bad.

To solve these problems, you look for statements in the problems that describe quantities that are

equal.  Then, you use algebra to write an equation that can be solved.  It is customary to use

variables that make it easier to remember what you're looking for, therefore, you don't use x or y

in most cases.      
http://library.thinkquest.org/20991/alg/word.html

Solving: 

A word problem in algebra is the equivalent of a story problem in math. When you solved story problems in your math class you had to decide what information you had and what you needed to find out. Then you decided what operation

to use. Addition was used to find a totals and subtraction was used to find changes in values.

The approach to solve problems with algebra is usually quite different. Word problems are solved by separating information from the problems into two equal groups, one for each side of an equation. Examine this problem.

Sara has 15 apples and 12 oranges. How many pieces of fruit does she have?

We know that the sum of 15 and 12 is equal to the total amount of fruit. As explained in the Basics of the Equation lesson, an unknown number or value is

represented by a letter. The total number of pieces of fruit is unknown, so we will represent that amount with x. When the value that a particular variable will represent is

determined, it is defined by writing a statement like,

Let x = Total Amount of Fruit

Once again, the sum of 15 apples and 12 oranges is equal to the total amount of fruit. This can be used to translate the problem into an equation, like the following:

15 + 12 = x

The next step is to solve this equation.

Now solve the equation which was created in the last step.

Let x = Total Pieces of Fruit

Initial Equation	15 + 12	=	x
After combining like terms	27	=	x

The answer is then rewritten as a sentence.

There are 27 Total Pieces of Fruit.

By using simple arithmetic, this problem probably could have been solved faster without setting up an algebra equation. But, knowing how to use an equation for this problem builds awareness of concepts which are useful,

and sometimes critical to solving much harder problems. One such problem will be presented in the next example.

Examine this word problem.

Two consecutive numbers have a sum of 91. What are the numbers?

Take notice, this problem has two numbers which are unknown, unlike the previous one which only had one unknown value. In order for this problem to be solved using basic algebra methods, we must set up an equation that

has only one variable (such as x). Proceed to the next page to find out how this is done. 

Now solve the equation which was created in the last step.

Let x = Total Pieces of Fruit

Initial Equation	15 + 12	=	x
After combining like terms	27	=	x

The answer is then rewritten as a sentence.

There are 27 Total Pieces of Fruit.

http://www.algebrahelp.com/lessons/wordproblems/basics/pg2.htm

Showing Your Work:     With word problems, you have to express the information in symbolic form, and that means you use a letter to represent some quantity that you want to find or need to use in writing down an equation.  For example you could use 't' to represent time and x to represent distance.  I will do a couple of examples to illustrate the method, but of course, every problem will be slightly different, and there is no ONE way of doing word problems.   (1) A has three times as many sweets as B.  If he gives B six sweets, he will then have twice as many as B then has.  How many sweets did they each have to start with?   Let x = number of sweets that B has initially; then 3x is the number that A has.  If now A gives 6 sweets to B then A has 3x-6 sweets and B has x+6 sweets.  Now we are told that after this transfer, A has twice as many sweets as B, so we can write down an equation to represent this fact, i.e.        3x-6 = 2(x+6)      3x-6 = 2x + 12      3x-2x = 12 + 6          x = 18 So initially B had 18 sweets and A had 3*18 = 54 sweets. Check: After transfer B has 18+6 = 24, A has 54-6 = 48, and 48 is twice 24. http://mathforum.org/library/drmath/view/57278.html