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How to Solve Word Problems

The most difficult homework assignment for most math students is working story/word problems. Solving word problems requires excellent reading comprehension and translating skills.

Students often have difficulty substituting English terms for algebraic symbols and equations. But once an equation is written, it is usually easily solved. To help you solve word problems follow these 10 steps:

Step 1 - Read the problem three times. Read the problem quickly the first time as a scanning procedure. As you are reading the problem the second time, answer these three questions:

  1. What is the problem asking me? (Usually at the end of the problem)
  2. What is the problem telling me that is useful? (Cross out unneeded information).
  3. What is the problem implying? (Usually something you have been told to remember). Read the problem a third time to check that you fully understand its meaning.

Step 2 - Draw a simple picture of the problem to make it more real to you (e.g., a circle with an arrow can represent travel in any form - by train, by boat, by plane, by car, or by foot).

Step 3 - Make a table of information and leave a blank space for information you are not told.

Step 4 - Use as few unknowns in your table as possible. If you represent all the unknown information in terms of a single letter do so! When using more than one unknown, use a letter that reminds you of that unknown. Then write down what your unknown represents. This eliminates the problem of assigning right answer to the wrong unknown. Remember, you have to create as many separate equations as you have unknowns.

Step 5 - Translate the English terms into an algebraic equation using the list of terms in (Translating English Terms into Algebraic Symbols), and (Translating English Words into Algebraic Expressions). Remember the English terms are sometimes stated in a different order than the algebraic terms.

Step 6 - Immediately retranslate the equation, as you now have it, back into English. The translation will not sound like a normal English phrase, but the meaning should be the same as the original problem. If the meaning is not the same, the equation is incorrect and needs to be rewritten. Rewrite the equation until it means the same as the English phrase.

Step 7 - Review the equation to see fit is similar to equations from your homework and if it makes sense. Some formulas dealing with specific word problems may need to be rewritten. Distance problems, for example, may need to be written solving for each of the other variables in the formula. Distance = Rate x Time; therefore, Time = Distance/Rate, and Rate = Distance/Time. Usually, a distance problem will identify the specific variable to be solved.

Step 8 - Solve the equation using the rules of algebra

Remember: Whatever is done to one side of the equation must be done to the other side of the
equation. The unknown must end up on one side of the equation, by itself. If you have more than one unknown, then use the substitution or elimination method to solve the equations.

Step 9 - Look at your answer to see if it makes common sense.

Example: If tax was added to an item, it should cost more or if a discount was applied to an item it should cost less. Is there more than one answer? Does your answer match the original question? Does your answer, have; the correct units?

Step 10 - Put your answer back into the original equation to see if it is correct If one side of the equation equals the other side of the equation, then you have the correct answer. If you do not have the correct answer, go back to Step 5.

Translating English Terms Into Algebraic Symbols

Sum +
Add +
In addition +
More than +
Increased +
In excess +
Greater +
Decreased by -
Less than -
Subtract -
Difference -
Diminished Reduce -
Remainder -
Times as much x
Percent of x
Product x
Interest on x
Per /
Divide /
Quotient /
Quantity ( )
Is =
Was =
Equal =
Will be =
Results =
Greater than >
Greater than or equal to
Less than <
Less than or equal to

Translating English Words Into Algebraic Expressions

Ten more than x x + 10
A number added to 5 5 + x
A number increased by 13 x + 13
5 less than 10 10 - 5
A number decreased by 7 x - 7
Difference between x and 3 x - 3
Difference between 3 and x 3 - x
Twice a number 2x
Ten percent of x 0.10x
Ten times x 10x
Quotient of x and 3 x/3
Quotient of 3 and x 3/x
Five is three more than a number 5 = x + 3
The product of 2 times a number is 10 2x = 10
One half a number is 10 x/2 = 10
Five times the sum of x and 2 5(x + 2)
Seven is greater than x 7 > x
Five times the difference of a number and 4 5(x - 4)
Ten subtracted from 10 times a number is
that number plus 5
10x - 10 = x + 5
The sum of 5x and 10 is equal to the product of x and 15 5x + 10 = 15x
The sum of two consecutive integers (x) + (x + 1)
The sum of two consecutive even integers (x) + (x + 2)
The sum of two consecutive odd integers (x) + (x + 2)


Paul D. Nolting, Ph.D., Winning at Math, 1997. 1989 by Academic Success Press, Inc.


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