Patterns in Linear Systems

Date: 11/12/2001 at 22:04:24
From: Chris
Subject: Patterns in linear systems

In math class, we were asked to see if we could find a pattern in a a 
certain type of linear system. We were given the equation 2x+5y=8, 
and it was pointed out that the coefficients increase by 3 each time. 
So, we were asked to create linear systems that are similar to that, 
using equations such as 3x+4y=5, or 4x+6y=8 (any equation where the 
difference between consecutive numbers is constant).

I set up systems like this and noticed the solution was always 
(-1, 2). I was surprised by this, so I wanted to see if I could find 
some way to determine that all systems like this would have that 
solution. I let c equal the number in front of the and d represent the 
distance between consecutive numbers. I formed the following equation 
that, I think, expresses how all of the equations could be:

(c)x + (c + d)y = c + 2d

I then let x = -1 and y = 2, and subsituted it into the left and right 
side to see if this made the equations true.

Left side = (c)-1 + (c + d)2
          = -c + 2c + 2d
          = c + 2d

Right side = c + 2d

Since both sides are equal, than the solution to all systems in this 
form must be (-1, 2) because this satisfies all equations in that 
form.

My question is, is that adequate proof that linear systems using 
equations like the one I described? Would there be an easier way to 
explain why the solution of systems with equations like that is 
always (-1, 2)? I am in grade 10 and the way I went about explaining 
why does not seem like anything that we have done in class, so I am 
thinking that there might be a different way to explain it. Did I 
make any mistakes in my explanation? 

Thank you.

Date: 11/13/2001 at 16:55:49
From: Doctor Peterson
Subject: Re: Patterns in linear systems

Hi, Chris.

I don't know what you were expected to do, but you did just what I 
would do. Good work! Using parameters c and d, and plugging in (-1,2) 
to check that what you had discovered is true, is the most 
straightforward way I can see to prove it.

Now I'm inclined to pursue this a little further. To start with, I'd 
like to see whether I could have discovered this fact algebraically 
rather than by experimentation. So I take your equation

    cx + (c + d)y = c + 2d

and try rearranging it to see what I find. For example, how about if 
we rewrite it as if c and d were the variables and x and y were mere 
parameters, by collecting all terms containing each of them together 
and factoring them out? Then we get

    c(x + y - 1) + d(y - 2) = 0

That's interesting, because it means that if (x+y-1) and (y-2) are 
both zero, then this will be true for ALL c and d. That is, the point 
(-1,2) is on all lines in this family. There's your fact, dropping 
right out of the equation (after some experienced massaging). There's 
nothing "better" about this proof; it's just a little more "elegant" 
in that it doesn't LOOK like I had to magically know about (-1,2) in 
order to do it.

Now what else can we do with this? We're looking at a parametrized 
family of lines, and we've seen that for some reason they all pass 
through this one point. What is special about that point? And how does 
each parameter affect the lines?

The fact that my rearranged equation has no actual constant term tells 
me something about how c and d affect the line; it turns out that the 
line is entirely determined by the ratio c/d. You could therefore 
reduce the problem to a single parameter by arbitrarily setting d = 1, 
which is equivalent to dividing the equation through by d. You might 
like to play with that.

Now, what makes (-1,2) special, and is there any way we could make a 
similar family that passes through a different point? One thing I see 
is that my form of the equation makes it look like you've added 
together the lines x+y = 1 and y = 2. That leads to some interesting 
thoughts.

I'll leave it to you to see if you can find any more interesting facts 
about your "family."

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