Thread: 3x3 arithmetic equations

Been playing with 3x3 equations with coefficients in an arithmetic series..
E.g

x+2y+3z=4
5x+6y+7z=8
9x+10y+11z= 12

Or even this

x+2y+3z=4
9x+11y+13z=15
4x+1y+-2z= -5

( where each separate line is an arithmetic series).

It appears that these types of equations ( been playing with wolfram) appear have no solutions, always??

Can anyone see an easy way why?

Originally Posted by rodders

( where each separate line is an arithmetic series).

That makes no sense...

Clarify your problem.

Originally Posted by DenisB

That makes no sense...

Clarify your problem.

Sorry, I mean the coeff are in an arithmetic series.
So in the first set of equations the coeff are 1,2,3,4 ( first equation), second equation (5,6,7,8), third equation (9,10,11,12).

In the second set of equations the coeff are 1,2,3,4 ( first eqn), second equation are (9,11,13,15), and third equation (4,1,-2,-5)

So these sequences are arithmetic and I am interested in the equations formed by using these sequences?

Does that help?

Not really.
You're trying to solve a system of 3 equations.

Can you solve this one:
x + y + z = 9
x + y - z = 1
2x - y + 3z = 13

The first system

x+2y+3z=4
5x+6y+7z=8
9x+10y+11z= 12

has a solution $x=0, y=-1, z=2$

actually it has infinitely many solutions

Originally Posted by rodders

x+2y+3z=4
5x+6y+7z=8
9x+10y+11z= 12

Nothing wrong with coefficients being arithmetic sequences,
but you can't simply assign at random any results, like your 4,8,12 above.

With following realistic values, you'll get a unique solution:
1x+ 2y + 3z = 14
5x +6y + 7z = 38
9x + 10y + 11z = 62

OK??

Originally Posted by Idea

The first system
x+2y+3z=4
5x+6y+7z=8
9x+10y+11z= 12
has a solution $x=0, y=-1, z=2$
actually it has infinitely many solutions

Sure does! Another is a=1, b=-3, c=3

Thanks for all this. I think I have worked out what I was doing wrong!