Solving Problems With and Without Algebra

Date: 05/14/2002 at 23:26:42
From: M Hanlon
Subject: Math equation

In a basketball game, Harold and Isaac scored a total of 19 points. 
Isaac and Jacob scored a total of 14 points.  Isaac scored as many 
ponts as Harold and Jacob together.  How many points did each player 
score?

I have not taken algebra yet and do not know how to do this 
equation.  I used various methods but could not come up with the 
answer. Thanks for the help.

Date: 05/17/2002 at 12:43:17
From: Doctor Ian
Subject: Re: Math equation

Hi,

Just for fun, let me show you how you would solve the problem 'by 
algebra'.  Then I'll show you how to solve it without algebra.

If we let H stand for Harold's points, and so on, we get

  H + I = 19

  I + J = 14

      I = H + J

which is a set of three equations with three unknowns.  Note that 
since I is the sum of H and J, we can just get rid of it by 
substituting (H+J) wherever we see I:

  H + (H+J) = 19

  (H+J) + J = 14

      (H+J) = H + J

The third equation no longer tells us anything useful, so we can 
get rid of it:

   2H +  J = 19

    H + 2J = 14

Now we can use the same trick again:  If H + 2J = 14, 
then H = 14 - 2J, so 

   2(14 - 2J) +  J = 19

    (14 - 2J) + 2J = 14

Again, the final equation tells us nothing useful, so we can drop 
it.  Now we have one equation, with one variable, which is good, 
because that's the kind of thing we can solve:
  
   2(14 - 2J) +  J = 19

       28 - 4J + J = 19

           28 - 3J = 19

           28 - 19 = 3J

                 9 = 3J

                 3 = J

So Jacob scored 3 points, and you can use that information to 
figure out what the other players scored. 

But what if you _haven't_ had algebra yet?  Can you still solve a 
problem like this?  

Let's start with the fact that Harold and Isaac scored 19 points 
together.  Here are the possibilities:

  Harold   Isaac
  -----    -----
     19        0
     18        1
     17        2
     16        3
     15        4
     14        5
     13        6
     12        7
     11        8
     10        9
      9       10
      8       11
      7       12
      6       13
      5       14
      4       15
      3       16
      2       17
      1       18
      0       19

Now, Isaac and Jacob scored a total of 14 points.  So we can add 
another column to our table:

  Harold   Isaac   Jacob
  -----    -----   -----
     19        0      14
     18        1      13
     17        2      12
     16        3      11
     15        4      10
     14        5       9
     13        6       8
     12        7       7
     11        8       6
     10        9       5
      9       10       4
      8       11       3
      7       12       2
      6       13       1
      5       14       0
      4       15      -1
      3       16      -2
      2       17      -3
      1       18      -4
      0       19      -5

Now assuming that Jacob can't score negative points, we can 
eliminate the final 6 rows of the table!

Now for our final piece of information:  Isaac scored as many 
points as Harold and Jacob put together.  Let's add one more 
column to the table:

  Harold   Isaac   Jacob   Harold+Jacob
  -----    -----   -----   ------------
     19        0      14             33 
     18        1      13             31 
     17        2      12             29
     16        3      11              ?
     15        4      10              ?
     14        5       9              ?
     13        6       8              ?
     12        7       7              ?
     11        8       6              ?
     10        9       5              ?
      9       10       4              ?
      8       11       3              ?
      7       12       2              9
      6       13       1              7

I'll leave the other values for you to fill in.  But notice that 
in only _one_ of these rows will the values in the 2nd and 4th 
columns be the same.  That row satisfies all the conditions in 
the problem. 

The thing about a problem like this is that it doesn't really 
illustrate the power of algebra, because the table we had to 
build to work the problem without algebra wasn't all that big.  
But suppose that we had been working with numbers like 196 and 
147 instead of numbers like 19 and 14?  

The great thing about a solution using algebra is that it's about 
the same size regardless of how large or small the numbers in the 
problem are.  

What that means in practice is that for small problems, it often 
makes more sense to give trial-and-error a chance to work before 
going to the trouble of setting up equations.  A big part of 
becoming a good problem solver is learning to recognize when you 
need steam roller to crush something, and when you can just use a 
hammer... or your hand.   

- Doctor Ian, The Math Forum
  http://mathforum.org/dr.math/ 

Date: 05/17/2002 at 14:46:32
From: M Hanlon
Subject: Math equation

Thanks so much for your help and time indoing this for me.  I can 
now understand both ways of doing it.

M Hanlon