[Blindmath] Solving systems by Graphing

[Tim in 't Veld via Blindmath]

Elise,
I've seen a few of your messages and you do seem rather stuck, here's my 
reaction to your question. Mainly drawing it up to show you what sort of 
notation (LaTeX) I use and how I think this problem could be approached. 
I've failed badly in advanced mathematics myself. I first studied 
physics and then computer science, despite years of trying I never got 
the mathematical skills required to succeed (but I did learn a lot 
trying and ended up in information science - see www.facespeaker.org for 
the sort of thing I'm doing now).

I'm curious if you understand the below and what would be your reaction. 
The sort of frustration you voice is just so common with blind students 
trying to grasp mathematics. I deliberately keep the below equations 
very simple, but if it gets more complicated I'd lose track immediately 
and writing  this down already took me some time. Before you know it you 
get a - sign wrong or something, even with this simple example.

Answer to your question.
If you mean a problem like:
y_1  = 1 + x^2
  y_2 = -1 - x^2
Find an x such that y_1 = y_2

No such x exists, as there is no intersection point of those 2 equations.
The typical sighted student hides the fact that he has no idea what is 
going on by simply sketching 2 graphs (or, rather, letting his graphing 
calculator do that for him) and just saying "see, the lines don't 
intersect!". But hey, I can't draw graphs!

Having some computer science background I first turn to logic.
Logically, we can see that for any x y_1 is positive and y_2 is 
negative. That is because x^2 >= 0. Put in LaTeX:
\forall x y_1 >= 0 \land y_2 <= 0 (logic notation: for any x y_1 is 
greater than or equal to zero and y_2 is less than or equal to zero) 
Obviously, a positive number cannot be equal to a negative number. You 
understand this proof?

But this is still a bit of a trick which I could use because this 
problem is simple and I have logic background. What is necessary here is 
a substitution approach.

Note that I formulated the question as:
Find an x such that y_1 = y_2
What it really says there is
Find an x such that x^2 + 1 = -x^2 -1
We can reformulate this equation to:
2x^2 + 2 = 0
That's a quadratic equation of the form
ax^2 +bx + c = 0 (a = 2, b = 0, c = 2)
The solution of this equation is given by the formula
x = \frac{b - \sqrt{b^2 -4ac}}{2a}
(standard algebra knowledge - won't go into the derivation here)
\sqrt{b^2 - 4ac} means square root of b^2 - 4 ac.
b^2 - 4ac = 0^2 - 4 * 2 * 2 = -16.
\sqrt{-16} is an irrational number,
therefore this equation has no solution (y_1 and y_2 don't intersect).

OK I spent an hour and some googling on this message, mainly due to lack 
of mathematical routine (it was a good and necessary exercise, this is 
ridiculously basic stuff)

But the sighted guy had a much easier time with his graphing calculator! 
As a blind student, the simplest thing I could do is appeal to logic. 
But that was because I have a lot of background in the field and this 
problem is simple enough that I have an intuitive solution. To solve 
this correctly requires me to understand the business with quadratic 
equations. Any high school student should (do they really??), but my 
point is if things get a little more complicated the blind student will 
need tremendous background and skill to solve mathematical problems. And 
they lack the "intermediate step" of training using graphs. Well, maybe 
succeeding in mathematics as a blind student requires one to be born a 
great mathematician?

Sorry if this is a bit unstructured / oversimplified; but hope it gets 
the point across.

Tim
On 24-5-2014 11:15, Elise Berkley via Blindmath wrote:
>
 >
 > Quick question, everyone:
 >
 > When I solve a system by graphing, is it inconsistent (no solution)
 > because they do not intercept at some point?  Thanks - studying for
 > my final.  elise
 >
 >
 >
 > Elise Berkley
 >
 > "The joy of the Lord is my strength."
 >
 >
 >
 >
 >
 >
 > _______________________________________________ Blindmath mailing
 > list Blindmath at nfbnet.org
 > http://nfbnet.org/mailman/listinfo/blindmath_nfbnet.org To
 > unsubscribe, change your list options or get your account info for
 > Blindmath:
 > http://nfbnet.org/mailman/options/blindmath_nfbnet.org/tim%40dvlop.nl
 >
 >
 >
BlindMath Gems can be found at 
<http://www.blindscience.org/blindmath-gems-home>