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:
 What is the problem asking me? (Usually at the end of the problem)
 What is the problem telling me that is useful? (Cross out unneeded
information).
 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 
³ 
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) 
Reference:
Paul D. Nolting, Ph.D., Winning at Math, 1997. 1989 by Academic Success
Press, Inc.
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