Solving Linear Equations in Two Variables with Geometric Sequences

[Nev3rforev3r]

Note: I didn't use the template because I feel it did not fit the question well enough.

This is concerning a system of linear equations in two variables where its constants in " ax+by=c " form show a geometric sequence, i.e. " nx + any = a²n ".
Another way of putting this is " y=(-1/a)x + a ". Note that in this second form, the constants are negative reciprocals of each other. I found that the general solution to any two equations following this form is (-ab,a+b) where a is the y-intercept of one equation and b is the y-intercept of the other.

If I graph a bunch of equations which follow one of the above general forms (like " y=(-1/5)x+5 " ) then they make a graph which looks like this: http://sphotos.ak.fbcdn.net/hphotos-ak-ash2/hs601.ash2/155353_1654251350058_1050271369_1853547_7385644_n.jpg"
As you can see, it forms a sort of sideways parabola. After some trial and error it appears to follow y^2=-4x.

Basically, I need to be able to mathematically show why that graphical pattern is there, preferably using that general solution (-ab,a+b) which I found.

Some notable information may include:
While solving for (-ab,a+b), I get to the point where y=(a+b)(a-b)/(a-b), indicating that there is a hole of some sort where a=b.
I think that where a=b there are infinite solutions to a set of any two equations, due to the fact that if a=b and the equations follow y=(-1/a)+a and y=(-1/b)+b, the lines are coincident.
All of these seems to be suggesting to me some sort of limit which is demonstrated by y^2=-4x or that all these lines are tangent to y^2=-4x, but I have been unable to prove this.
It has also occurred to me that it may be a separation of where a solution can be and where a solution cannot be.

Any hints? I don't really want just the answer, but if someone could tilt me in the right direction where I can figure it out mostly myself, it would be great.

Thank you all very much. I've worked hours and hours on this.

Big edit: I just tested a point (-1,0) in " nx + any = a²n ", yielding an answer of a= sqrt(-1)
WOAAAH

Another edit: If you take y²=-4x and input (-ba,a+b) it yields a²+b²=2ab. This is really freaky because of how close it is to the Pythagorean theorem and the law of cosines.

 

[desi_zwingli]

Hi. I'm also trying to solve this problem, but I'm stuck. How did you find the solution to the general 2x2 system??

I have both patterns: ax+(ar)y=(ar²) and y= ax-1/a
I also have the graphical display of many equations, which forms a sideways parabola. I've solved two 2x2 systems, which follow the patterns and I can't find any relation.

What do I have to do next?

Hope you can help me

Regards :)

 

[epenguin]

You seem to have discovered for yourself an example of envelopes. Here http://en.wikipedia.org/wiki/Envelope_(mathematics) there is a nice simulation but I don't know how you find the explanation. Maybe there is a more accessible one at Dr. Math or somewhere like that, or someone can point you to one.

 

[Nev3rforev3r]

Thank you for the link!

I solved it a while ago and basically found what is described as an "envelope," then proved it in the general case, etc. ,etc. I got the highest grade of 20/20 :D (very difficult, apparently only 6 people from my school have gotten it in like.. 5 years?). Its nice to know the formal and extended version of what the assignment was about.

 

[epenguin]

I am not sure what you want - in the first post you ask how to go on with showing the envelope is a parabola, but in your last post you seem to have solved it? Congratulations anyway.

But anyway whether for help or confirmation: knowing where two lines meet is not much help. On your fig. you can see lots of meetings not on the envelope and not of much interest. But there is your blank area where there are no lines at all. So think if you put yourself at a given x fixed and you vary a. Your fig. is saying that for your positive y and a fixed (negative) x you cannot get the y below a certain value no matter how you vary a, in other words, for fixed x, y as you vary x has a minimum. (It also has a maximum value in your negative quadrant). Usually you use calculus in finding this value, but in this case you could do it by quadratic equation theory. You could find the a that gives you that - and from that find the corresponding y in terms of x alone. Or, I think, you might be able to make a smart shortcut and find the y without ever finding the a. Anyway a should not appear in the final result which is just a relation between x and y.

If we hear further from you we might tell you about some other things.

 

[nestea]

Nev3rforev3r, how did you end up proving the graphical pattern in the general case? Did you explain what an envelope was and solve for the curve using variables instead of numbers? A push in the right direction would be greatly appreciated! :)

 

[Char. Limit]

Another edit: If you take y²=-4x and input (-ba,a+b) it yields a²+b²=2ab. This is really freaky because of how close it is to the Pythagorean theorem and the law of cosines.

Note also that if a²+b²=2ab, then a²-2ab+b²=(a-b)²=0, or a=b.

 

[Nev3rforev3r]

I did not explain that it was an "envelope," (didnt know what one was) but i did explain what the lines had to do with the parabola thingy as best as I could. Note where the intersections of two lines can be, and note what would make the shape happen. Yes, I did prove it in the general case with variables instead of numbers. I used both calculus and stuff from like... 9th grade or earlier haha. I can't really say much more without giving it away. It is supposed to be difficult, after all.

Also, what Char said is significant.

 

[nestea]

i see that the lines are all tangent to the horizontal parabola, but is proving that these lines are tangent to a horizontal parabola with the formula y²=-4x enough to count as proof for the graphical pattern?

what char. said proves that when a=b (and the lines coincide), there is only one point shared between the parabola and the lines - thus the line is tangent to the parabola... right?

i'm getting kind of lost...

 

[Nev3rforev3r]

nestea, keep working at it. Go as far as you think you need to go with this in order to do what is asked by the assignment. Don't take what I said in the first post as the correct answer or the correct path, as it took me A LOT more work to finally figure it out after I made the post and received no replies. These posts do, however, have some correct and relevant elements to them. Plus, there's almost definitely more than one correct path and solution. It does a person good to find their own way.

 

[epenguin]

Yes, I did prove it in the general case with variables instead of numbers. I used both calculus and stuff from like... 9th grade or earlier haha. I can't really say much more without giving it away. It is supposed to be difficult, after all.

Watch out, if you don't tell us then there is a chance that someone else may work it out and announce it and rob you of the credit for being the first person to discover this principle!