How to Turn Word Problems into Algebra Equations

Date: 6/6/96 at 7:56:25
From: Anonymous
Subject: Algebra word problems

I am having problems setting up algebra equations from word problems.  
I don't know where to put the letters and numbers  Could you please 
help?
                     -Mike

Date: 6/6/96 at 11:46:18
From: Doctor Anthony
Subject: Re: Algebra word problems

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.


(2) If a man walks to the station at 3 mph he misses his train by 
1 minute. If he runs at 6 mph he has 2 minutes to spare.  Find the 
distance to the station.

Let x = distance to the station.  Now we know that the difference in 
TIME is 3 minutes between the two methods of travel, so we shall need 
to write down an equation involving TIME.

Time = Dist/Speed, so when he walks, time = x/3  (hours)
                       when he runs, time = x/6  (hours)

Subtract these and equate to the time (in hours) between the two 
methods of travel.

               x/3 - x/6 = 3/60

               x(2-1)/6 = 3/60

                    x/6 = 3/60

                      x = 18/60

                      x = 3/10 of a mile.   

So the distance to the station is 3/10 mile.  You can check the result 
now by calculating the actual time to walk, and the time to run.

Time to walk = (3/10)/3 = 1/10 hour = 6 minutes
Time to run  = (3/10)/6 = 1/20 hour = 3 minutes
                         Difference = 3 minutes.

You will note that the important first step is to use letters to 
represent unknown quantities so that you can write down equations.  
Lay out the work clearly, and express any information given in the 
question in terms of the symbols you have introduced.

-Doctor Anthony,  The Math Forum
 Check out our web site!  http://mathforum.org/dr.math/