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Solving 3x3 Simultaneous Equations

It's been a lot of years since I did any maths like this and I'm struggling to remember from school...

I know three points on a curve and I wish to calculate an equation for the curve
The three points I know are (315, 4.05) (680, 3.50) (1102, 3.03)

My working is shown below. The values I get for x, y and z don't fit when I put them back into the original equations

I'm probably doing something quite fundamentally wrong! Anyone care to point me in the right direction? Or even just tell me what I should be doing?

Much appreciated!

(A)	4.05 = 99225x + 315y + z
(B)	3.5 = 462400x + 680y + z
(C)	3.03 = 1216609x + 1103y + z

----------------------------------------------------------------------------
(A)	315y = 4.05 - 99225x + z
	y = 4.05/315 - 315x + z

(B)	3.5 = 462400x + 680(4.05/315 - 315x + z) + z
(C)	3.03 = 1216609x + 1103(4.05/315 - 315x + z) + z

(B)	3.5 = 462400x + 8.743 - 214200x + 681z
	-5.243 = 248200x + 681z
	248200x = -681z - 5.243
	x = -(681/248200)z - 5.243/248200

(C)	3.03 = 1216609x + 14.181 - 347445x + 1104z
	-11.151 = 869164x + 1104z
	869164x = -1104z - 11.151
	x = -(1104/869164)z - 11.151/869164

(B)=(C)	-x = -x
	(1104/869164)z - 11.151/869164 = (681/248200)z - 5.243/248200
	(1104/869164)z - (681/248200)z = 11.151/869164 - 5.243/248200
	((1104/869164)-(681/248200))z = (11.151/869164)-(5.243/248200)
	z = ((11.151/869164)-(5.243/248200)) / ((1104/869164)-(681/248200))
	z = 0.00563

----------------------------------------------------------------------------
(A)	4.05 = 99225x + 315y + 0.00563
(B)	3.5 = 462400x + 680y + 0.00563
(C)	3.03 = 1216609x + 1103y + 0.00563

(A)	315y = 4.05 - 99225x + 0.00563
	y = 4.05/315 - 315x + 0.00563
(B)	3.5 = 462400x + 680(4.05/315 - 315x + 0.00563) + 0.00563
	3.5 = 462400x + 8.743 - 214200x + 3.834
	-9.077 = 248200x
	x = -(9.077/248200)
	x = -0.0000366

----------------------------------------------------------------------------
(A)	4.05 = 99225x + 315y + z
	4.05 = 99225*(-0.0000366) + 315y + 0.00563
	4.05 = 3.632 + 0.00563 + 315y
	315y = 0.412
	y = 0.00131

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Ken Butters

You didn't say so... but I am going to assume that your "Curve" is a parabola.

The formula for a parabola is Ax(squared) +Bx + C = y

X and y represent the (X,Y) coordinates for the points you are Given.

You need to solve for A, B, C which will be constants.

Plug in X and y for each of the 3 points that you have... and you will have 3 equations... and 3 variables to solve for(A, B and C).

Given these points:
(315, 4.05)
(680, 3.50)
(1102, 3.03)

First point one would yeild the equation : A*315^2+B*315+C = 4.05
or simplified : A*99225 + B*315 + C = 4.05

Through subtraction and/or substituion you should be able to solve for A, B, C.

daz84

ASKER

That's about how far I got, it's the substitution algebra that I fear is letting me down.

I am happy that the three equations I have at the top of the "code" section of my post are correct, it's solving them for x,y,z or a,b,c in the case of your example, that is the bit I am finding difficult.

I thought I could remember how to do it from years ago and was quite happy when I got something that looked reasonable, but on plugging the values back in to the original equations it yeilds garbage. This is the reason I am lost and need some help.

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Ken Butters

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phoffric

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d-glitch

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Ken Butters

Question....

on your third equation... you had this coordinate: (1102, 3.03)

But this equation: 3.03 = 1216609x + 1103y + z

you used 1103 in the equation... but have 1102 as the point.... which is accurate?

daz84

ASKER

That was a typo as I was typing the question buttersk, well spotted! The correct value is 1103

Thank you all for your efforts, especially buttersk! I am going to attempt this with d-glitch's method however, as it looks easier! And easy is what I'm gunning for!

You will all get credit once I have calculated the correct answer!