Dr. Algebra

I have talked to so many students that absolutely hate word or story problems.  The reason they are so disliked is because many students don’t really know how to approach and solve them.

The key to solving word problems is to identify the relevant information and write it in a form that you know how to solve. Once you are at that point, then its just a matter of doing the algebra and finding the answer by using skills you already have.

General Solution Process:

Write down and label everything you know.

    1. Read the problem and see what you know and don’t know.  Think about what relationships the various parts have to each other.
    2. Determine what the unknown value is and how it can be used to find the solution.  (It might not be the same thing).
    3. Write known values in terms of the unknown value.
    4. Solve the resulting equation.

Example:
A river flows at a constant speed of 8 mph  (Miles Per Hour).  Bob has a boat that can  go 12 mph at top speed on still water.  If Bob is in his boat, at max speed, and going with the river, how far will he go in three hours.

  1. We can immediately see this is a rate/speed/distance problem because the problem asks for a distance traveled in some amount of time and at some amount of speed.  We also know that distance = rate x time.  The time is given as three hours so we need the rate.  The rate, or speed, is then what we are looking for.  Its the unknown.  
  2. We know the boat can go 12 mph in still water and the river flows at a rate of 8 mph.  If they are going the same direction, then the total rate is the boat rate plus the river rate.  Total Rate = Boat Rate + River Rate,  or T=B+R.  (Note we have assigned variables in a somewhat arbitrary manner).  So the total rate of the boat on the river is B + R which is 8 mph + 12 mph = 20 mph.
  3. Now its easy to write distance = rate x time and then substituting what we know gives us:  [katex]Distance=\frac{20miles}{hour}3 hours[/katex] and then the hours cancel leaving 60 miles.