Basic Approach to Solving Algebraic Word Problems

Algebra is used to solve problems in science, industry, business,and the home. Algebraic equations can be used to describe laws of motion, pressures of gases, electric circuits, and nuclear facility operations, just to name a few. They can be applied to all sortof problems such as:

1) The ages of people.

2) The cost of articles.

3) football scores.

4) And other everyday matters.

The basic approach to solving word problems is the same. First, condense the available information into algebraic equations, and, second, solve the equations. Of these two basic steps, the first is frequently the most difficult to master because there are no clearly defined rules such as those that exist for solving equations. However, in writing algebraic expressions for the various quantities given in word problems, it is helpful to look for certain words that indicate mathematical operations. The words "sum" and "total" signify addition; the word "difference" signifies subtraction; the words "product," "times," and "multiples of" signify multiplication; the words "quotient," "divided by," "per," and "ratio" signify division; and the words "same as" and "equal to" signify equality. When quantities are connected by these words and others like them, these quantities can be written as algebraic expressions.

Sometimes you may want to write equations initially using words. For example, Bob is 30 years older than Joe. Express Bob’s age in terms of Joe’s. Bob’s age = Joe’s age plus 30 years If we let Bob’s age be represented by the symbol *B *and Joe’s age by the symbol* J*, this becomes:

Example:

A man is three times as old as his son was four years ago.

Let *M *= man’s age and *S *= son’s age. Then *M *= 3 (*S* - 4).

4. A car travels in one hour 40 miles less than twice as far as it

travels in the next hour.

Let *x*1 be the distance it travels the first hour and *x*2 the distance it travels

the second then, *x*1 = (2) (*x*2) - 40.