Some issues solving linear equation

[Dieter]

First of all I have to say that I don't know anything about the HP48 series and its mathematical algorithms. But there are is something that makes me wonder if it could be a clue for finding the solution here.

(06-22-2018 01:06 PM)sasa Wrote:  x + y + 2z = -2
3x - y + 14z = 6
x + 2y = -5

x = -.904761904762
y = -2.04761904762
z = .47619047619

This is only one of solution, as this system have infinite number of solutions, however there is no further notice from the calculator about. The solution actually is in form:
x = -4n + 1
y = 2n - 3
z = n

And that's also what the emulator returned here. In this case for n = 10/21.

May it be possible that you have this value (0,47619...) stored somewhere in a variable so that the emulator evaluates the given general solution for the case where the parameter (you named it n) has this special value? Have you tried clearing all variables?

(06-22-2018 01:06 PM)sasa Wrote:  The following synthetic example is much more obvious, the system have no solution:
x - 5y + 3z = -8
0x + 2y + 1z = 3
0x + 0y + 0z = 7

However, emulator returns:
x = -.150793650794
y = 1.5374603175
z = -6.34920634921E-2

Again the factor 10/21 shows up. Here z = –2/15 · 10/21. Or –4/63 if you prefer.

That's why I suspect that somewhere a variable with such a value may be stored and the calculator / emulator includes this value while generating the solution.

Just an idea.

Dieter