# Solving system of equations, Gaussian Elimination

I am hoping to gain some insight on the characteristic differences between the following two distinct sets [(1) and (2)] of systems of equations. In particular, Gaussian elimination doesn't produce the correct result for the second.

(1)
Gaussian elimination does work for a system of equations like this (Wikipedia example):
2x + y - z = 8
-3x - y + 2z = -11
-2x + y + 2z = -3

However, I have so far been unable to use Gaussian elimination to solve the below example. Context: given 3 coordinates, I want to solve the resulting system of equations to produce a parabola that intersects each point. Three example (x,y) coordinates: (-1.5, 1) , (-0.4, 1.5) , (0.8, 2.7)

(2)
Resulting system of equations:
1 = a(-1.5)^2 + b(-1.5) + c
1.5 = a(-0.4)^2 + b(-0.4) + c
2.7 = a(0.8)^2 + b(0.8) + c

My gut tells me it has to do with the distinction that occurs as a result of plugging in these values into a quadratic formula template, wherein the unknown values are "not the typical" x, y, and z variables, but instead what I would call the coefficients. My mathematical understanding and vocabulary isn't very deep here; any instruction or explanation would be appreciated.

Thanks so much,
Mike

Comments & criticism welcome - I'm new here.

• I am not sure whether your definition of a parabola is unique. en.wikipedia.org/wiki/Five_points_determine_a_conic as you see, you need 5 points to determine a specific conic, but maybe i'm missing the point. Futhermore please type out your word by word, so we can see what you have done, whether it was correct and where you got stuck. Welcome to stackexchange and don't take the downvotes personally! :) Commented Feb 6, 2021 at 5:16
• If you want to use the $(x-x_0)(x-x_1)$ template, you'll need to find the specific two points where the parabola intersects the $x$-axis. And plug in those $x_0, x_1$ values. Commented Feb 6, 2021 at 5:20
• What is your question ?
– user65203
Commented Feb 6, 2021 at 14:03
• "Gaussian elimination doesn't produce the correct result for the second": can you expand ?
– user65203
Commented Feb 6, 2021 at 14:03
• @YvesDaoust excuses, should have been 'obvious' Commented Feb 7, 2021 at 5:12

## 1 Answer

I am not sure whether your definition of a parabola is unique.

As you see here , you need 5 points to determine a specific conic, but maybe i'm missing the point.

To determine a specific parabola you need 3 points.

Futhermore please type out your word by word, so we can see what you have done, whether it was correct and where you got stuck. Welcome to stackexchange and don't take the downvotes personally! :)

If you want to use the $$(x−x_0)(x−x_1)$$ template, you'll need to find the specific two points where the parabola intersects the $$x$$-axis. And plug in those $$x_0$$,$$x_1$$ values.

• Thanks for the help here. I definitely unknowingly followed very non-standard procedure and question clarity. Thank you for making up the difference. I will consider your comments and suggestions as best I can for future posts. Cheers.
– Mike
Commented Feb 27, 2021 at 17:47